This deck is a guided tour that ties the whole course together. It covers the three pillars - vectors and the geometry of space, multivariable differentiation, and multivariable integration - along with the family of big theorems, and gives a decision guide for choosing a coordinate system and choosing a theorem. It targets the exam-killing traps: a missing Jacobian or volume factor, the wrong coordinate system, and the wrong big theorem.
Subject: Calculus III · 110 slides · symbolic lesson
Open the interactive version of this deck · Homework for this lesson
Objectives
By the end of this review you will be able to:
1. See the whole course as three pillars plus one family of big theorems.
2. Pick the right coordinate system for a region on sight.
3. Pick the right big theorem for a line, surface, or flux problem.
4. Set up mixed problems that pull from several units, then avoid the classic traps: missing factors and the wrong tool.
Warm-up
Discussion prompt
Before we open Week 16 - Comprehensive Review: without looking back, what was the main idea of Week 15 - Divergence & Stokes's Theorems, and what could you do by the end of it that you could not do before?
Hint: One sentence for the idea, one for the skill. If the second one is blank, that is the part to revisit.
Answer:
That deck covers parametric surfaces and surface area, scalar surface integrals, and oriented surfaces and flux, then reaches the two capstone theorems: the Divergence Theorem of Gauss, and Stokes's Theorem. It targets the classic errors: applying the Divergence Theorem to an open surface, orienting the normal the wrong way, and confusing when to reach for Stokes and when for Divergence.
Concept
Everything in Calculus III sits on three pillars.
Pillar 1 - Geometry of space. Vectors, lines, planes, surfaces, and curves that move through space.
Pillar 2 - Differentiation. Partial derivatives, the gradient, extrema, and Lagrange multipliers.
Pillar 3 - Integration. Double and triple integrals, in rectangular, polar, cylindrical, and spherical coordinates.
On top of the pillars sits one family of big theorems that connects derivatives and integrals across regions.
Counterexample
Discussion prompt
Everything in Calculus III sits on three pillars.
That is stated as though it always holds. Do one of two things: produce a case where it fails, or say precisely what rules such a case out. "It just does" is not on the menu.
Hint: Hunt at the extremes first — zero, one, negative, empty, equal. If every extreme survives, the reason they survive is the proof.
Answer:
On top of the pillars sits one family of big theorems that connects derivatives and integrals across regions.
Intuition
Single-variable calculus had one derivative and one integral. Here the same two ideas simply grow more inputs and more directions.
A vector points a direction. A partial derivative measures change in one direction. The gradient bundles all those directions into one arrow.
An integral still adds up tiny pieces - now the pieces are little areas, volumes, or bits of a curve or surface.
The big theorems are the grand finale: they say that adding up a derivative over a region equals reading the original function on the boundary. That is the Fundamental Theorem of Calculus, grown up.
Analogy
Discussion prompt
Explain It is really one continuous story by analogy to something with no Calculus III in it at all — a queue, a recipe, a map, a bank balance, whatever fits. Then say where your analogy breaks.
Hint: An analogy that never breaks is not an analogy, it is the same idea wearing a hat. Find the seam — that is the part that is actually new.
Answer:
Single-variable calculus had one derivative and one integral. Here the same two ideas simply grow more inputs and more directions.
Concept
A vector carries both size and direction. In space we write it in components.
\[ \mathbf{v} = \langle v_1, v_2, v_3 \rangle = v_1\mathbf{i} + v_2\mathbf{j} + v_3\mathbf{k} \]
Its length is the distance formula in disguise.
\[ \lVert \mathbf{v} \rVert = \sqrt{v_1^2 + v_2^2 + v_3^2} \]
A unit vector keeps the direction but has length one. Divide by the magnitude to get it.
\[ \mathbf{u} = \frac{\mathbf{v}}{\lVert \mathbf{v} \rVert} \]
Explain it
Discussion prompt
Explain Pillar 1: vectors and the geometry of space to a student a year behind you. No notation, no jargon they have not met — and it still has to be true.
Hint: If your explanation needs a symbol they have never seen, you are describing the notation rather than the idea.
Answer:
A vector carries both size and direction. In space we write it in components.
Concept
The distance between two points in space is the length of the vector joining them.
\[ d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2} \]
A sphere is every point a fixed distance from a center. Its equation is that distance formula squared.
\[ (x-a)^2 + (y-b)^2 + (z-c)^2 = R^2 \]
Completing the square on a messy quadratic turns it back into this clean center-radius form.
Socratic
Discussion prompt
The distance between two points in space is the length of the vector joining them.
Suppose that were not true. What is the first thing in Week 16 - Comprehensive Review that would stop working?
Hint: Follow it one step downstream. The answer is whatever was quietly relying on it.
Answer:
A sphere is every point a fixed distance from a center. Its equation is that distance formula squared.
Concept
The dot product multiplies two vectors and returns a single number.
\[ \mathbf{a} \cdot \mathbf{b} = a_1 b_1 + a_2 b_2 + a_3 b_3 = \lVert \mathbf{a}\rVert\,\lVert \mathbf{b}\rVert \cos\theta \]
So it measures the angle between them. When it is zero, the vectors are perpendicular.
\[ \cos\theta = \frac{\mathbf{a}\cdot\mathbf{b}}{\lVert\mathbf{a}\rVert\,\lVert\mathbf{b}\rVert} \]
It also builds projections and work. The key fact: a dot product is a number, never a vector.
Concept
The cross product of two space vectors returns a new vector, perpendicular to both.
\[ \mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix} \]
Its length is the area of the parallelogram the two vectors span.
\[ \lVert \mathbf{a} \times \mathbf{b} \rVert = \lVert \mathbf{a}\rVert\,\lVert \mathbf{b}\rVert \sin\theta \]
Direction follows the right-hand rule, and order matters: swapping the inputs flips the sign.
\[ \mathbf{a} \times \mathbf{b} = -\,\mathbf{b} \times \mathbf{a} \]
Trap
A plane and a line both need a vector, so students grab whichever one is handy. That is the wrong move.
Using a direction vector as if it were the plane's normal:
\[ \text{plane through } P \text{ with } \mathbf{d}=\langle 1,1,0\rangle:\quad x + y = \text{const}\;? \]
That equation describes a plane whose normal is the direction of the line, so the plane ends up perpendicular to the line you meant to describe.
A line is built from a direction vector; a plane is built from a normal vector. They play opposite roles.
\[ \text{line: } \mathbf{r}(t)=\mathbf{r}_0 + t\,\mathbf{d} \]
\[ \text{plane: } \mathbf{n}\cdot(\mathbf{r}-\mathbf{r}_0)=0 \]
The normal is perpendicular to the plane; the direction lies along the line. Never swap their jobs.
Ranking
Put in order
Put the moves of Worked example: cross product and area into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. The cross product of two space vectors is the perpendicular vector we want.
Worked example
Find a vector perpendicular to both, and the area of the parallelogram they span.
\[ \mathbf{a} = \langle 1,2,3\rangle, \qquad \mathbf{b} = \langle 4,5,6\rangle \]
Set up the determinant for the cross product
Why: The cross product of two space vectors is the perpendicular vector we want.
\[ \mathbf{a}\times\mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & 3 \\ 4 & 5 & 6 \end{vmatrix} \]
Expand along the top row
Why: Each minor drops its own row and column; remember the middle term carries a minus sign.
\[ = \mathbf{i}(2\cdot 6 - 3\cdot 5) - \mathbf{j}(1\cdot 6 - 3\cdot 4) + \mathbf{k}(1\cdot 5 - 2\cdot 4) \]
Simplify each component
Why: Arithmetic only.
\[ \mathbf{a}\times\mathbf{b} = \langle -3,\; 6,\; -3\rangle \]
Take the magnitude for the area
Why: The length of the cross product equals the parallelogram area.
\[ \lVert \mathbf{a}\times\mathbf{b}\rVert = \sqrt{9+36+9} = \sqrt{54} = 3\sqrt{6} \]
Verify perpendicularity with a dot product
Why: The result must be orthogonal to a, so their dot product must be zero.
\[ \mathbf{a}\cdot(\mathbf{a}\times\mathbf{b}) = 1(-3)+2(6)+3(-3) = -3+12-9 = 0 \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "cross product and area", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The result must be orthogonal to a, so their dot product must be zero.
Concept
A line needs a point and a direction. A plane needs a point and a normal.
\[ \text{symmetric line: } \frac{x-x_0}{a}=\frac{y-y_0}{b}=\frac{z-z_0}{c} \]
\[ \text{plane: } a(x-x_0)+b(y-y_0)+c(z-z_0)=0 \]
To build a plane through three points, cross two edge vectors to get the normal, then plug one point in.
Concept
Quadric surfaces are the graphs of second-degree equations. Read the signs to name them.
\[ \text{ellipsoid: } \frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1 \quad (\text{all plus}) \]
\[ \text{one-sheet hyperboloid: two plus, one minus}=1 \]
\[ \text{two-sheet hyperboloid: one plus, two minus}=1 \]
A single squared variable set equal to the other two gives a paraboloid; equal to zero gives a cone. Trace the surface by slicing with coordinate planes.
Concept
A vector-valued function traces a path through space as time runs.
\[ \mathbf{r}(t) = \langle x(t),\, y(t),\, z(t)\rangle \]
Differentiate and integrate component by component. The derivative is velocity; its magnitude is speed.
\[ \mathbf{r}'(t) = \langle x'(t),\, y'(t),\, z'(t)\rangle, \qquad \text{speed} = \lVert \mathbf{r}'(t)\rVert \]
Integrating a vector function adds a vector constant, not a scalar one - one constant per component.
Concept
Arc length adds up speed over the time interval.
\[ L = \int_a^b \lVert \mathbf{r}'(t)\rVert \, dt \]
Curvature measures how fast the unit tangent turns. Two handy formulas:
\[ \kappa = \frac{\lVert \mathbf{T}'(t)\rVert}{\lVert \mathbf{r}'(t)\rVert} = \frac{\lVert \mathbf{r}'(t)\times \mathbf{r}''(t)\rVert}{\lVert \mathbf{r}'(t)\rVert^3} \]
The unit tangent points along motion; the principal normal points toward the center of the turn. Always normalize the tangent before differentiating for the normal.
Socratic
Discussion prompt
Arc length adds up speed over the time interval.
Suppose that were not true. What is the first thing in Week 16 - Comprehensive Review that would stop working?
Hint: Follow it one step downstream. The answer is whatever was quietly relying on it.
Answer:
Curvature measures how fast the unit tangent turns. Two handy formulas:
Concept
With more inputs, one derivative becomes many. Each partial derivative asks: how does the output change if only one input moves?
\[ f_x = \frac{\partial f}{\partial x}, \qquad f_y = \frac{\partial f}{\partial y} \]
Bundle all the partials into one vector and you get the gradient, the master object of this pillar.
Concept
To take a partial derivative, treat every other variable as a constant and differentiate normally.
\[ f(x,y) = x^2 y + y^3 \;\Rightarrow\; f_x = 2xy, \quad f_y = x^2 + 3y^2 \]
Mixed second partials are equal for the nice functions we meet - that is Clairaut's theorem.
\[ f_{xy} = f_{yx} \]
Concept
When the inputs themselves depend on other variables, change flows along every path - and you must add up all of them.
\[ \frac{dz}{dt} = \frac{\partial z}{\partial x}\frac{dx}{dt} + \frac{\partial z}{\partial y}\frac{dy}{dt} \]
Draw a tree diagram: one branch per path from the output down to the variable, one product per branch, then sum. Dropping a branch is the classic mistake.
Concept
The gradient packs all the first partials into a single vector.
\[ \nabla f = \left\langle f_x,\; f_y,\; f_z \right\rangle \]
The directional derivative in a direction is just the gradient dotted with a unit vector.
\[ D_{\mathbf{u}} f = \nabla f \cdot \mathbf{u}, \qquad \lVert \mathbf{u}\rVert = 1 \]
Setting the gradient to zero locates critical points. The gradient shows up everywhere in this pillar.
Intuition
Stand on a hillside described by a function of two variables. The gradient at your feet points in the direction of steepest climb, and its length is how steep that climb is.
Turn ninety degrees from the gradient and the ground is level - so the gradient is perpendicular to the level curves and level surfaces.
That single picture explains directional derivatives (dot with your walking direction), tangent planes (the gradient is the normal), and why extrema happen where the slope is flat in every direction.
Step zero
Discussion prompt
Worked example: directional derivative — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Compute the gradient
Answer:
Worked example
Find the rate of change of this function at the point, in the direction of the given vector.
\[ f(x,y) = x^2 y + y^3, \quad P=(1,2), \quad \mathbf{v}=\langle 3,4\rangle \]
Compute the gradient
Why: The directional derivative is built from the gradient dotted with a unit direction.
\[ \nabla f = \langle 2xy,\; x^2 + 3y^2\rangle \]
Evaluate the gradient at P
Why: Plug in the point before dotting - the gradient is a function of position.
\[ \nabla f(1,2) = \langle 2(1)(2),\; 1 + 3(4)\rangle = \langle 4,\; 13\rangle \]
Normalize the direction vector
Why: The formula demands a unit vector; skipping this is the number-one error here.
\[ \lVert \mathbf{v}\rVert = \sqrt{9+16}=5, \qquad \mathbf{u} = \left\langle \tfrac{3}{5},\, \tfrac{4}{5}\right\rangle \]
Dot the gradient with the unit vector
Why: The directional derivative is the projection of the gradient onto the direction.
\[ D_{\mathbf{u}} f = \langle 4,13\rangle \cdot \left\langle \tfrac{3}{5},\tfrac{4}{5}\right\rangle = \tfrac{12}{5}+\tfrac{52}{5} = \tfrac{64}{5} \]
Verify the direction was truly a unit vector
Why: If the length of u is one, the rate is trustworthy.
\[ \left\lVert \left\langle \tfrac{3}{5},\tfrac{4}{5}\right\rangle\right\rVert = \sqrt{\tfrac{9}{25}+\tfrac{16}{25}} = 1 \checkmark, \quad D_{\mathbf{u}} f = \tfrac{64}{5} \]
Picture it
Animation
Shows: Each line of the worked example "directional derivative", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: If the length of u is one, the rate is trustworthy.
Trap
The direction vector is right there, so students dot the gradient with it as given.
\[ \text{with } \mathbf{v}=\langle 3,4\rangle \text{ (length 5): } \langle 4,13\rangle\cdot\langle 3,4\rangle = 64 \]
That answer is five times too big, because the vector was five times too long. A rate of change cannot depend on how long you drew the arrow.
Normalize first, then dot. Only a unit vector gives an honest rate per unit distance.
\[ \mathbf{u} = \tfrac{1}{5}\langle 3,4\rangle, \qquad \langle 4,13\rangle\cdot\mathbf{u} = \tfrac{64}{5} \]
The correct value is the wrong one divided by the length of v. Always divide the direction by its own magnitude before using it.
Concept
Write a surface as a level surface of some function, then the gradient is its normal vector.
\[ F(x,y,z)=0 \;\Rightarrow\; \mathbf{n} = \nabla F \]
The tangent plane uses that normal at the point of tangency.
\[ F_x(P)(x-x_0)+F_y(P)(y-y_0)+F_z(P)(z-z_0)=0 \]
The normal line runs along the same gradient direction through the point. Both come straight from the gradient.
Socratic
Discussion prompt
Write a surface as a level surface of some function, then the gradient is its normal vector.
Suppose that were not true. What is the first thing in Week 16 - Comprehensive Review that would stop working?
Hint: Follow it one step downstream. The answer is whatever was quietly relying on it.
Answer:
The tangent plane uses that normal at the point of tangency.
Hypothesis
Predict first
Worked example: tangent plane to a surface is about to be worked. State your hypothesis first: which rule or definition decides this one, and what is the first move it forces? Then watch whether the example agrees with you.
Correct: Write the surface as F equals zero
Why: A level surface lets the gradient serve as the normal.
A hypothesis you wrote down is falsifiable; a vague sense of how it will go is not. If the example opens somewhere else, that gap is the thing worth chasing.
Worked example
Find the tangent plane to this surface at the given point.
\[ x^2 + y^2 + z^2 = 9, \qquad P=(2,1,2) \]
Write the surface as F equals zero
Why: A level surface lets the gradient serve as the normal.
\[ F(x,y,z) = x^2+y^2+z^2-9 \]
Compute the gradient
Why: The gradient of F is normal to the level surface everywhere.
\[ \nabla F = \langle 2x,\; 2y,\; 2z\rangle \]
Evaluate the normal at P
Why: We need the specific normal at the point of tangency.
\[ \nabla F(2,1,2) = \langle 4,\; 2,\; 4\rangle \]
Write the plane through P with that normal
Why: A plane is a point plus a normal vector.
\[ 4(x-2)+2(y-1)+4(z-2)=0 \;\Rightarrow\; 4x+2y+4z=18 \]
Verify the point satisfies the plane
Why: P must lie in its own tangent plane.
\[ 4(2)+2(1)+4(2)=8+2+8=18 \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "tangent plane to a surface", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: A plane is a point plus a normal vector.
Concept
Relative extrema of a surface hide where the tangent plane is flat - where the gradient is the zero vector.
\[ \nabla f = \mathbf{0}: \quad f_x = 0 \text{ and } f_y = 0 \]
To sort each critical point, use the second-partials test with the discriminant.
\[ D = f_{xx}f_{yy} - (f_{xy})^2 \]
Read the result:
\[ \begin{aligned} D>0,\, f_{xx}>0 &\Rightarrow \text{local min}\\ D>0,\, f_{xx}<0 &\Rightarrow \text{local max}\\ D<0 &\Rightarrow \text{saddle}\\ D=0 &\Rightarrow \text{inconclusive} \end{aligned} \]
Step zero
Discussion prompt
Worked example: classify the critical points — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Set both first partials to zero
Answer:
Worked example
Locate and classify every critical point.
\[ f(x,y) = x^3 - 3xy + y^3 \]
Set both first partials to zero
Why: Critical points are where the gradient vanishes.
\[ f_x = 3x^2 - 3y = 0, \qquad f_y = 3y^2 - 3x = 0 \]
Solve the system
Why: From the first, y equals x squared; substitute into the second.
\[ y=x^2 \Rightarrow x^4 - x = 0 \Rightarrow x(x^3-1)=0 \Rightarrow x=0,\,1 \]
List the critical points
Why: Pair each x with its y value.
\[ (0,0) \quad \text{and} \quad (1,1) \]
Build the discriminant
Why: Second partials give the D-test.
\[ f_{xx}=6x,\; f_{yy}=6y,\; f_{xy}=-3 \;\Rightarrow\; D = 36xy - 9 \]
Test each point
Why: Plug each critical point into D and check the sign of f_xx.
\[ D(0,0)=-9<0 \Rightarrow \text{saddle}; \quad D(1,1)=27>0,\, f_{xx}=6>0 \Rightarrow \text{local min} \]
Verify the minimum value
Why: Evaluate f at the classified minimum to report the actual value.
\[ f(1,1) = 1 - 3 + 1 = -1 \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "classify the critical points", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Evaluate f at the classified minimum to report the actual value.
Concept
On a closed, bounded region the biggest and smallest values are guaranteed to exist. They live in one of two places.
Inside: critical points found from the gradient being zero.
On the edge: check the boundary by parameterizing it or using Lagrange multipliers.
Evaluate f at every candidate and compare. Forgetting the boundary is a classic way to miss the true maximum.
Concept
When a constraint ties the variables together, the extremum happens where the two gradients line up.
\[ \nabla f = \lambda \nabla g \]
Geometrically the level curve of f just kisses the constraint curve, so their normals are parallel.
You must also keep the constraint equation itself - it is one of the equations you solve, not an afterthought.
\[ g(x,y) = k \]
Ranking
Put in order
Put the moves of Worked example: a Lagrange problem into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. Set the gradient of f equal to lambda times the gradient of g.
Worked example
Maximize the product on the given circle.
\[ f(x,y)=xy \quad \text{subject to} \quad x^2+y^2=8 \]
Write the Lagrange equations
Why: Set the gradient of f equal to lambda times the gradient of g.
\[ \langle y, x\rangle = \lambda \langle 2x, 2y\rangle \;\Rightarrow\; y = 2\lambda x,\; x = 2\lambda y \]
Eliminate lambda
Why: Substitute one equation into the other to relate x and y.
\[ x = 2\lambda(2\lambda x) = 4\lambda^2 x \;\Rightarrow\; \lambda = \pm\tfrac{1}{2} \]
Use the constraint for each case
Why: Never drop the constraint - it pins down the actual points.
\[ \lambda=\tfrac12: y=x,\; 2x^2=8 \Rightarrow (2,2),(-2,-2) \]
Evaluate f at the candidates
Why: Compare the objective across all constrained critical points.
\[ f(2,2)=4,\; f(-2,-2)=4; \quad \lambda=-\tfrac12 \Rightarrow f=-4 \]
Verify the winning points satisfy the constraint
Why: The maximum must actually lie on the circle.
\[ 2^2+2^2=8 \checkmark, \quad \text{maximum } f = 4 \]
Picture it
Animation
Shows: Each line of the worked example "a Lagrange problem", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The maximum must actually lie on the circle.
Concept
Integration now adds up tiny pieces of area or volume across a region or solid.
\[ \iint_R f\, dA, \qquad \iiint_E f\, dV \]
Two decisions drive every integral: what order to integrate, and which coordinate system fits the region. Choosing well turns a nightmare into a one-liner.
Concept
A double integral is worked as an iterated integral: do the inside one first, holding the outer variable fixed.
\[ \iint_R f\, dA = \int_a^b \int_{g_1(x)}^{g_2(x)} f(x,y)\, dy\, dx \]
The inside limits may depend on the outer variable, but the outside limits must be plain constants.
Fubini's theorem lets you swap the order - but only after you redraw the region and re-read the limits from it.
Concept
Sketch the region first, always. Then decide whether to sweep with vertical or horizontal strips.
Type I: vertical strips. The outer variable is x over constants; y runs between two curves of x.
Type II: horizontal strips. The outer variable is y over constants; x runs between two curves of y.
Area itself is a double integral of the constant one.
\[ \text{Area}(R) = \iint_R 1\, dA \]
Socratic
Discussion prompt
Sketch the region first, always. Then decide whether to sweep with vertical or horizontal strips.
Suppose that were not true. What is the first thing in Week 16 - Comprehensive Review that would stop working?
Hint: Follow it one step downstream. The answer is whatever was quietly relying on it.
Answer:
Area itself is a double integral of the constant one.
Concept
For circles, disks, and anything with radial symmetry, switch to polar.
\[ x = r\cos\theta, \quad y = r\sin\theta, \quad x^2+y^2 = r^2 \]
The area element is not just dr times d-theta. It carries an extra factor of r.
\[ dA = r\, dr\, d\theta \]
That extra r is the single most forgotten thing in the whole course. Write it before anything else.
Intuition
Picture a little tile in polar coordinates, cut by two nearby radii and two nearby circles.
Its radial side has length one step of r. But its curved side is an arc, and the arc length grows with how far out you are - it equals r times the angle step.
So the tile is not a unit square; its area is radius times both steps. That radius is the missing r, and it comes straight from the geometry, not from a rule to memorize.
\[ dA \approx (r\, d\theta)(dr) = r\, dr\, d\theta \]
Step zero
Discussion prompt
Worked example: a polar double integral — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Convert the integrand and the region to polar
Answer:
Worked example
Integrate over the disk of radius two centered at the origin.
\[ \iint_D (x^2+y^2)\, dA, \qquad D: x^2+y^2 \le 4 \]
Convert the integrand and the region to polar
Why: The disk is radial, so polar limits are clean and the integrand simplifies.
\[ x^2+y^2 = r^2, \qquad 0 \le r \le 2, \quad 0 \le \theta \le 2\pi \]
Write the integral with the extra r
Why: The area element in polar is r dr d-theta - the r is not optional.
\[ \int_0^{2\pi}\int_0^2 r^2 \cdot r\, dr\, d\theta = \int_0^{2\pi}\int_0^2 r^3\, dr\, d\theta \]
Do the inner integral
Why: Integrate r cubed in r from 0 to 2.
\[ \int_0^2 r^3\, dr = \left[\tfrac{r^4}{4}\right]_0^2 = \tfrac{16}{4} = 4 \]
Do the outer integral
Why: A constant integrated over a full turn.
\[ \int_0^{2\pi} 4\, d\theta = 8\pi \]
Check with a symmetry sanity test
Why: The integrand is positive over a full disk, so the answer must be positive and scale with area - it does.
\[ 8\pi > 0 \checkmark, \qquad \iint_D (x^2+y^2)\, dA = 8\pi \]
Picture it
Animation
Shows: Each line of the worked example "a polar double integral", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The integrand is positive over a full disk, so the answer must be positive and scale with area - it does.
Trap
After converting to polar, students integrate as if the area element were just dr times d-theta.
\[ \int_0^{2\pi}\int_0^2 r^2\, dr\, d\theta = \int_0^{2\pi}\tfrac{8}{3}\, d\theta = \tfrac{16\pi}{3} \]
That is the wrong answer. The missing r changed r squared into r squared, not r cubed, and the value is off.
Include the Jacobian factor r that comes with polar coordinates.
\[ dA = r\, dr\, d\theta \]
\[ \int_0^{2\pi}\int_0^2 r^2\cdot r\, dr\, d\theta = 8\pi \]
The right answer is 8 pi. Any coordinate change carries a Jacobian; in polar it is simply r.
Break the constraint
Discussion prompt
The rule this trap just fixed:
Include the Jacobian factor r that comes with polar coordinates.
Now break it on purpose. Build a case that violates it and follow the consequences until something visibly fails. Where does the failure first show up — and would you have noticed it if you had not been looking?
Hint: The dangerous rules are the ones whose violation still produces an answer. If yours fails loudly, try to find one that fails quietly.
Concept
A triple integral sweeps a solid. Integrating the constant one gives its volume.
\[ V = \iiint_E 1\, dV \]
Weight each piece by a density and you get mass; weight by position and you get moments and the center of mass.
\[ m = \iiint_E \rho(x,y,z)\, dV \]
Set the innermost limits first - they may depend on the outer variables. Then project the solid onto a plane to get the outer double-integral limits.
Concept
Cylindrical is polar in the plane with an untouched height axis. It fits cylinders, cones, and paraboloids with axial symmetry.
\[ x=r\cos\theta,\; y=r\sin\theta,\; z=z \]
The volume element inherits the same extra r as polar.
\[ dV = r\, dz\, dr\, d\theta \]
If your solid looks the same when you spin it around the z-axis, cylindrical is usually the fast lane.
Concept
Spherical uses distance from the origin, the angle down from the positive z-axis, and the azimuth around it.
\[ x=\rho\sin\phi\cos\theta,\; y=\rho\sin\phi\sin\theta,\; z=\rho\cos\phi \]
Its volume element carries a bigger factor - remember this one cold.
\[ dV = \rho^2 \sin\phi\, d\rho\, d\phi\, d\theta \]
The angle down from the axis runs from zero to pi; the azimuth runs from zero to two pi. Spheres, balls, and cone-and-sphere regions love spherical.
\[ 0 \le \phi \le \pi, \qquad 0 \le \theta \le 2\pi \]
Estimation
Predict first
Find the volume of the ball of radius a using spherical coordinates.
Commit before you compute: what does Worked example: volume of a ball come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify against the known formula
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. This must match the classic volume of a sphere.
Worked example
Find the volume of the ball of radius a using spherical coordinates.
\[ E: x^2+y^2+z^2 \le a^2 \]
Set the spherical limits
Why: A ball is the cleanest possible spherical region: every angle, radius up to a.
\[ 0\le\rho\le a,\quad 0\le\phi\le\pi,\quad 0\le\theta\le 2\pi \]
Write the integral with the correct volume element
Why: The spherical factor rho squared sin phi is mandatory.
\[ V = \int_0^{2\pi}\int_0^{\pi}\int_0^{a} \rho^2 \sin\phi\, d\rho\, d\phi\, d\theta \]
Separate the three independent integrals
Why: The limits are constants and the integrand factors, so each variable integrates on its own.
\[ V = \left(\int_0^{2\pi} d\theta\right)\left(\int_0^{\pi}\sin\phi\, d\phi\right)\left(\int_0^{a}\rho^2\, d\rho\right) \]
Evaluate each factor
Why: Standard antiderivatives.
\[ V = (2\pi)(2)\left(\tfrac{a^3}{3}\right) = \tfrac{4}{3}\pi a^3 \]
Verify against the known formula
Why: This must match the classic volume of a sphere.
\[ V = \tfrac{4}{3}\pi a^3 \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "volume of a ball", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: This must match the classic volume of a sphere.
Trap
In spherical, students often drop the sin phi, or use just rho squared, or mix up which angle runs to pi.
\[ \int_0^{2\pi}\int_0^{\pi}\int_0^{a}\rho^2\, d\rho\, d\phi\, d\theta = 2\pi\cdot\pi\cdot\tfrac{a^3}{3} = \tfrac{2\pi^2 a^3}{3} \]
That has a stray pi squared and no hope of being right - the missing sin phi broke it.
Cylindrical carries r; spherical carries rho squared sin phi. Write the factor before you write the limits.
\[ dV_{\text{cyl}} = r\, dz\, dr\, d\theta, \qquad dV_{\text{sph}} = \rho^2\sin\phi\, d\rho\, d\phi\, d\theta \]
\[ \int_0^{2\pi}\int_0^{\pi}\int_0^{a}\rho^2\sin\phi\, d\rho\, d\phi\, d\theta = \tfrac{4}{3}\pi a^3 \]
Keep the angle down from the axis in zero to pi and the azimuth in zero to two pi, and the factor gives the right volume.
Concept
Let the shape of the region choose the coordinates for you.
| Region looks like | Best system | Volume factor |
|---|---|---|
| Box or simple curves | Rectangular | dA or dV |
| Disk, circle, annulus | Polar | r |
| Cylinder, cone, paraboloid | Cylindrical | r |
| Ball, sphere, cone-and-sphere | Spherical | rho squared sin phi |
The wrong system is not incorrect math - it is a self-inflicted mess of ugly limits. Match the symmetry and the factor follows automatically.
Discrimination
Sort into buckets
Sort these by Volume factor, from memory, without looking back at Which coordinate system? A decision guide. Telling them apart on the spot is the skill; the table is only where the answer happens to be written down.
Concept
The last stretch of the course is one family of theorems that connect a derivative inside a region to values on its boundary.
There are four members: the Fundamental Theorem of Line Integrals, Green's theorem, Stokes's theorem, and the Divergence theorem.
They look different but say the same thing. Learn the shared idea and the pieces stop competing for space in your head.
Socratic
Discussion prompt
The last stretch of the course is one family of theorems that connect a derivative inside a region to values on its boundary.
Suppose that were not true. What is the first thing in Week 16 - Comprehensive Review that would stop working?
Hint: Follow it one step downstream. The answer is whatever was quietly relying on it.
Answer:
They look different but say the same thing. Learn the shared idea and the pieces stop competing for space in your head.
Concept
A vector field attaches an arrow to every point - think wind, water flow, or a force.
\[ \mathbf{F}(x,y,z) = \langle P,\, Q,\, R\rangle \]
Divergence measures how much the field spreads out from a point - a scalar.
\[ \operatorname{div}\mathbf{F} = \nabla\cdot\mathbf{F} = P_x + Q_y + R_z \]
Curl measures how much it swirls - a vector.
\[ \operatorname{curl}\mathbf{F} = \nabla\times\mathbf{F} \]
Concept
A scalar line integral adds up a function along a curve, weighting by arc length.
\[ \int_C f\, ds = \int_a^b f(\mathbf{r}(t))\,\lVert \mathbf{r}'(t)\rVert\, dt \]
A vector line integral adds up the field along the curve - this is work.
\[ \int_C \mathbf{F}\cdot d\mathbf{r} = \int_a^b \mathbf{F}(\mathbf{r}(t))\cdot \mathbf{r}'(t)\, dt \]
The speed factor is the length of the derivative. Reversing the curve flips the sign of the vector integral but leaves the scalar one alone.
Concept
If a field is the gradient of some potential, the line integral only cares about the endpoints.
\[ \int_C \nabla f\cdot d\mathbf{r} = f(B) - f(A) \]
This is the single-variable Fundamental Theorem of Calculus, promoted to curves in space.
Two immediate payoffs: the path does not matter, and any closed loop gives zero.
\[ \oint_C \nabla f\cdot d\mathbf{r} = 0 \]
Step zero
Discussion prompt
Worked example: evaluate via the FTLI — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Test whether the field is conservative
Answer:
Worked example
Evaluate the work done by this field along any path from the first point to the second.
\[ \mathbf{F} = \langle 2xy,\; x^2\rangle, \quad A=(0,0),\; B=(1,3) \]
Test whether the field is conservative
Why: The FTLI only applies to a gradient field; check the mixed-partial condition.
\[ P_y = 2x, \qquad Q_x = 2x \;\Rightarrow\; P_y = Q_x \checkmark \]
Find a potential function
Why: Integrate P in x, then fix the leftover function using Q.
\[ f = \int 2xy\, dx = x^2 y + g(y), \quad f_y = x^2 + g'(y) = x^2 \Rightarrow g'(y)=0 \]
Write the potential
Why: With g constant we can take it zero.
\[ f(x,y) = x^2 y \]
Evaluate at the endpoints
Why: The FTLI says the answer is the potential at B minus at A.
\[ f(1,3) - f(0,0) = (1)(3) - 0 = 3 \]
Verify the potential reproduces the field
Why: Its gradient must equal F.
\[ \nabla(x^2 y) = \langle 2xy,\; x^2\rangle = \mathbf{F} \checkmark, \quad \text{work} = 3 \]
Picture it
Animation
Shows: Each line of the worked example "evaluate via the FTLI", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The FTLI says the answer is the potential at B minus at A.
Concept
In the plane, a field is conservative on a simply connected domain exactly when the cross partials match.
\[ \frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} \]
In space the same idea reads as zero curl.
\[ \operatorname{curl}\mathbf{F} = \mathbf{0} \]
When you rebuild the potential by integrating, never forget the leftover function of the other variables - it is where a whole term can hide.
Concept
Green's theorem trades a loop integral around a closed plane curve for a double integral over the region it encloses.
\[ \oint_C P\, dx + Q\, dy = \iint_R \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dA \]
The curve must be closed and oriented counterclockwise. A clockwise loop just flips the sign.
A neat corollary computes area from the boundary alone.
\[ \text{Area} = \tfrac{1}{2}\oint_C (x\, dy - y\, dx) \]
Concept
Stokes's theorem lifts Green's theorem out of the plane and onto a surface in space.
\[ \oint_C \mathbf{F}\cdot d\mathbf{r} = \iint_S (\operatorname{curl}\mathbf{F})\cdot \mathbf{n}\, dS \]
The loop integral around the boundary curve equals the flux of the curl through any surface capping that curve.
The boundary orientation and the surface normal must agree by the right-hand rule. This is the curl-and-boundary theorem.
Socratic
Discussion prompt
Stokes's theorem lifts Green's theorem out of the plane and onto a surface in space.
Suppose that were not true. What is the first thing in Week 16 - Comprehensive Review that would stop working?
Hint: Follow it one step downstream. The answer is whatever was quietly relying on it.
Answer:
The loop integral around the boundary curve equals the flux of the curl through any surface capping that curve.
Concept
The Divergence theorem trades the flux out of a closed surface for a triple integral of divergence over the solid inside.
\[ \iint_S \mathbf{F}\cdot \mathbf{n}\, dS = \iiint_E \operatorname{div}\mathbf{F}\, dV \]
The surface must be closed - it has to fully enclose the solid - and the normal points outward.
This is the flux-and-closed-surface theorem. Total outflow equals total spreading inside.
Explain it
Discussion prompt
Explain The Divergence theorem to a student a year behind you. No notation, no jargon they have not met — and it still has to be true.
Hint: If your explanation needs a symbol they have never seen, you are describing the notation rather than the idea.
Answer:
The Divergence theorem trades the flux out of a closed surface for a triple integral of divergence over the solid inside.
Ranking
Put in order
Put the moves of Worked example: flux by the Divergence theorem into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. The Divergence theorem only applies to a surface that fully encloses a solid - the sphere does.
Worked example
Find the outward flux of this field through the unit sphere.
\[ \mathbf{F} = \langle x,\, y,\, z\rangle, \qquad S: x^2+y^2+z^2 = 1 \]
Confirm the surface is closed
Why: The Divergence theorem only applies to a surface that fully encloses a solid - the sphere does.
\[ E: x^2+y^2+z^2 \le 1 \]
Compute the divergence
Why: Convert the hard surface flux into an easy volume integral.
\[ \operatorname{div}\mathbf{F} = 1 + 1 + 1 = 3 \]
Integrate the divergence over the solid
Why: A constant integrand pulls out; the remaining integral is just the volume of the unit ball.
\[ \iiint_E 3\, dV = 3\cdot \tfrac{4}{3}\pi (1)^3 = 4\pi \]
Verify the volume factor used
Why: The unit-ball volume is four-thirds pi, so the flux is three times that.
\[ 3\cdot\tfrac{4}{3}\pi = 4\pi \checkmark, \qquad \text{flux} = 4\pi \]
Picture it
Animation
Shows: Each line of the worked example "flux by the Divergence theorem", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The unit-ball volume is four-thirds pi, so the flux is three times that.
Intuition
Every member of the family says the same sentence: adding up a derivative over a region equals reading the original object on the region's boundary.
The plain Fundamental Theorem of Calculus already did this: the integral of a derivative over an interval equals the function at the two endpoints - the boundary of the interval.
\[ \int_a^b f'(x)\, dx = f(b) - f(a) \]
FTLI does it for a curve (boundary is two endpoints). Green and Stokes do it for a surface (boundary is a loop). Divergence does it for a solid (boundary is a closed surface).
Once you see the pattern, you never have to guess which theorem exists - you only choose which region you are on.
Analogy
Discussion prompt
Explain All four theorems are one idea by analogy to something with no Calculus III in it at all — a queue, a recipe, a map, a bank balance, whatever fits. Then say where your analogy breaks.
Hint: An analogy that never breaks is not an analogy, it is the same idea wearing a hat. Find the seam — that is the part that is actually new.
Answer:
Every member of the family says the same sentence: adding up a derivative over a region equals reading the original object on the region's boundary.
Concept
Line them up by the dimension of the region and its boundary.
| Theorem | Integrate over | Evaluate on boundary |
|---|---|---|
| FTC | interval | two endpoints |
| FTLI | curve | two endpoints |
| Green / Stokes | region or surface | closed boundary curve |
| Divergence | solid | closed boundary surface |
Same idea, one dimension higher each time. The derivative on the inside matches the raw object on the boundary.
Discrimination
Sort into buckets
Sort these by Evaluate on boundary, from memory, without looking back at The unifying table. Telling them apart on the spot is the skill; the table is only where the answer happens to be written down.
Concept
Read the problem for two clues: what kind of integral is asked, and what geometry is given.
| You are given | Reach for |
|---|---|
| Gradient field, endpoints | FTLI |
| Closed loop in the plane | Green's theorem |
| Closed loop bounding a surface, curl | Stokes's theorem |
| Flux through a closed surface | Divergence theorem |
Curl plus a boundary curve points to Stokes; divergence plus a closed surface points to Divergence. Match the words to the tool.
Comparison
Comparison matrix
From Which theorem? A decision guide: refill the Reach for column from what you know. The rest of the table is as it appeared.
| You are given | Reach for |
|---|---|
| Gradient field, endpoints | FTLI |
| Closed loop in the plane | Green's theorem |
| Closed loop bounding a surface, curl | Stokes's theorem |
| Flux through a closed surface | Divergence theorem |
Trap
A flux problem appears, so the student fires off the Divergence theorem on a surface that is only a cap, not a closed shell.
\[ S: \text{hemisphere only} \;\Rightarrow\; \iint_S \mathbf{F}\cdot\mathbf{n}\, dS \ne \iiint_E \operatorname{div}\mathbf{F}\, dV \]
The equation is false because the hemisphere does not enclose a solid. The theorem simply does not apply, and the answer is wrong.
Check the hypotheses first. Divergence needs a closed surface; Stokes needs a boundary curve for a surface; Green needs a closed plane loop.
For an open cap, either close it with a disk and subtract that piece, or apply Stokes using its boundary curve instead.
\[ \iint_{S_{\text{closed}}} = \iint_{S_{\text{cap}}} + \iint_{S_{\text{disk}}} = \iiint_E \operatorname{div}\mathbf{F}\, dV \]
The right move is to verify the geometry matches the theorem before computing anything.
Two truths and a lie
Sort into buckets
Some of these hold up and some are the exact mistakes this lesson is built to prevent. Sort them.
Concept
Almost every lost point on the final comes from one of three habits. Catch them before they catch you.
One - missing factors. The extra r in polar and cylindrical, and the rho squared sin phi in spherical. Also the speed factor in a scalar line integral.
Two - wrong coordinate system. Forcing a ball into rectangular limits, or a box into spherical. The math is fine but the limits become unworkable.
Three - wrong theorem. Using Divergence on an open surface, or Stokes without a boundary curve. Check the hypotheses first, always.
Counterexample
Discussion prompt
Almost every lost point on the final comes from one of three habits. Catch them before they catch you.
That is stated as though it always holds. Do one of two things: produce a case where it fails, or say precisely what rules such a case out. "It just does" is not on the menu.
Hint: Hunt at the extremes first — zero, one, negative, empty, equal. If every extreme survives, the reason they survive is the proof.
Explain it to yourself
Discussion prompt
In Pattern: choosing a coordinate system this move is made:
3. Write the volume factor immediately
Why is that legal? Name the rule or definition it rests on before you read on.
Hint: If you can only say "because that is what you do", the rule is the thing to go and find.
Answer:
Record r, or r again, or rho squared sin phi before you touch the limits - so it is never forgotten.
Pattern
1. Sketch the region and name its symmetry
Why: The shape, not the integrand, decides the coordinates. Look for disks, cylinders, cones, and balls.
2. Match symmetry to the system
Why: Radial in a plane goes polar; axial in space goes cylindrical; spherical symmetry goes spherical.
3. Write the volume factor immediately
Why: Record r, or r again, or rho squared sin phi before you touch the limits - so it is never forgotten.
4. Read the limits off the sketch
Why: Inner limits may depend on outer variables; outer limits must be constants.
Explain it to yourself
Discussion prompt
In Pattern: choosing the right theorem this move is made:
2. Look at the geometry of the boundary
Why is that legal? Name the rule or definition it rests on before you read on.
Hint: If you can only say "because that is what you do", the rule is the thing to go and find.
Answer:
Endpoints point to FTLI; a closed plane loop to Green; a boundary curve of a surface to Stokes; a closed surface to Divergence.
Pattern
1. Name the integral you are handed
Why: Line integral, surface flux, or a loop? The type narrows the family fast.
2. Look at the geometry of the boundary
Why: Endpoints point to FTLI; a closed plane loop to Green; a boundary curve of a surface to Stokes; a closed surface to Divergence.
3. Check the hypotheses before computing
Why: Closed surface for Divergence, simply connected domain for a conservative field, matching orientation for Stokes.
4. Convert to the easier side
Why: The whole point is to trade a hard boundary integral for an easy region integral, or the reverse.
Explain it to yourself
Discussion prompt
In Pattern: setting up any problem this move is made:
3. Write factors and limits before integrating
Why is that legal? Name the rule or definition it rests on before you read on.
Hint: If you can only say "because that is what you do", the rule is the thing to go and find.
Answer:
Jacobian factor first, then limits read from the sketch. Setup errors dwarf arithmetic errors.
Pattern
1. Identify the object and the region
Why: Vector or scalar? Curve, surface, or solid? This chooses the whole toolbox.
2. Pick the coordinate system or theorem
Why: Use the two decision guides - symmetry for coordinates, boundary geometry for theorems.
3. Write factors and limits before integrating
Why: Jacobian factor first, then limits read from the sketch. Setup errors dwarf arithmetic errors.
4. Evaluate, then verify
Why: Sanity-check sign, symmetry, units, and known formulas. A quick check catches most slips.
Real world
Discussion prompt
Outside this lesson: where does Week 16 - Comprehensive Review actually turn up? Name one concrete situation — a job, a piece of software someone ships, a decision somebody has to make — and say which part of Pattern: setting up any problem is doing the work in it.
Hint: Vague is the failure mode here. "Engineering" is not a situation; "deciding whether this build is fast enough to ship" is.
Answer:
That deck is a guided tour that ties the whole course together. It covers the three pillars - vectors and the geometry of space, multivariable differentiation, and multivariable integration - along with the family of big theorems, and gives a decision guide for choosing a coordinate system and choosing a theorem. It targets the exam-killing traps: a missing Jacobian or volume factor, the wrong coordinate system, and the wrong big theorem.
Check
You must find the outward flux of a vector field through a closed cylinder - its top, bottom, and curved side together.
Check your understanding
Which tool is the natural fit?
Answer: A
Why: Flux through a closed surface is exactly the Divergence theorem's setup: it converts the surface flux into a triple integral of divergence over the solid the cylinder encloses.
Elimination
Eliminate the wrong options
Which coordinate system makes this integral cleanest?
3 of these 4 are wrong. Strike them one at a time, and say what rules each one out before you strike the next. The survivor is the answer.
Survives elimination: A
Why: A ball has full spherical symmetry, so spherical coordinates give constant limits (rho from 0 to 3, phi from 0 to pi, theta from 0 to two pi) and the factor rho squared sin phi.
Check
You need to integrate a function over the solid ball defined by the region below.
\[ x^2 + y^2 + z^2 \le 9 \]
Check your understanding
Which coordinate system makes this integral cleanest?
Answer: A
Why: A ball has full spherical symmetry, so spherical coordinates give constant limits (rho from 0 to 3, phi from 0 to pi, theta from 0 to two pi) and the factor rho squared sin phi.
Prediction
Predict first
The volume element dV in spherical coordinates equals which of these?
Answer it in your own words, now, with nothing to choose from. The options are on the next slide — and picking the right one off a list is an easier skill than producing it.
Correct: rho squared times sin phi, d rho d phi d theta
Why: The spherical Jacobian is rho squared times sin phi, so dV equals rho squared sin phi d rho d phi d theta. Dropping any piece of that factor gives a wrong volume.
Check
You are converting a triple integral into spherical coordinates.
Check your understanding
The volume element dV in spherical coordinates equals which of these?
Answer: A
Why: The spherical Jacobian is rho squared times sin phi, so dV equals rho squared sin phi d rho d phi d theta. Dropping any piece of that factor gives a wrong volume.
Prediction
Predict first
What is the directional derivative at P in the direction of v?
Answer it in your own words, now, with nothing to choose from. The options are on the next slide — and picking the right one off a list is an easier skill than producing it.
Correct: 22/5
Why: The gradient at P is the vector 2,4. Normalizing v gives 3/5, 4/5, and the dot product is 6/5 plus 16/5, which is 22/5.
Check
Find the rate of change at the point in the direction of the given vector.
\[ f(x,y)=x^2+y^2, \quad P=(1,2), \quad \mathbf{v}=\langle 3,4\rangle \]
Check your understanding
What is the directional derivative at P in the direction of v?
Answer: A
Why: The gradient at P is the vector 2,4. Normalizing v gives 3/5, 4/5, and the dot product is 6/5 plus 16/5, which is 22/5.
Check
At a critical point of a function of two variables you compute the discriminant and the second partial in x.
\[ D > 0 \quad \text{and} \quad f_{xx} < 0 \]
Check your understanding
What does the second-partials test conclude?
Answer: A
Why: When D is positive the point is a genuine extremum, and a negative f_xx means the surface curves downward there, so it is a local maximum.
Elimination
Eliminate the wrong options
What is the value of the line integral of the gradient of f from A to B?
3 of these 4 are wrong. Strike them one at a time, and say what rules each one out before you strike the next. The survivor is the answer.
Survives elimination: A
Why: By the FTLI the integral is f(B) minus f(A). Here f(2,1) equals 4 times 1, which is 4, and f(0,0) is 0, so the value is 4.
Check
The field is the gradient of the potential below. Evaluate the line integral along any path from the first point to the second.
\[ f(x,y) = x^2 y, \quad A=(0,0), \quad B=(2,1) \]
Check your understanding
What is the value of the line integral of the gradient of f from A to B?
Answer: A
Why: By the FTLI the integral is f(B) minus f(A). Here f(2,1) equals 4 times 1, which is 4, and f(0,0) is 0, so the value is 4.
Connect it up
Draw it
One page, no notation unless you need it: draw how these connect — Pattern: choosing a coordinate system · Pattern: choosing the right theorem · Pattern: setting up any problem · The course is three pillars · It is really one continuous story. Put an arrow wherever one of them is what makes another possible, and label the arrow with why.
Recap
The course stands on three pillars: the geometry of space, multivariable differentiation, and multivariable integration - crowned by one family of big theorems.
The gradient is the master object of differentiation; it drives directional derivatives, tangent planes, extrema, and Lagrange multipliers.
For integration, let the region's symmetry pick the coordinate system, and write the volume factor before the limits.
| System | Factor |
|---|---|
| Polar / cylindrical | r |
| Spherical | rho squared sin phi |
The four big theorems say one thing: integrate a derivative over a region and it equals the original object read on the boundary.
| Region | Theorem |
|---|---|
| Curve | FTLI |
| Plane region / surface | Green / Stokes |
| Solid | Divergence |
Guard against the three exam-killers: missing factors, the wrong coordinate system, and the wrong theorem. Set up carefully, then verify.
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