This deck covers triple integrals in cylindrical and spherical coordinates: what each coordinate means, how to convert between systems, and the all-important volume factors r and rho-squared sin-phi. It targets the classic traps of dropping the volume factor, swapping phi and theta, and choosing the harder coordinate system.
Subject: Calculus III · 116 slides · symbolic lesson
Open the interactive version of this deck · Homework for this lesson
Objectives
By the end of this lesson you will be able to:
1. Describe a solid in cylindrical coordinates and set up its triple integral.
2. Use the cylindrical volume element correctly, including the extra factor it carries.
3. Describe a solid in spherical coordinates and set up its triple integral.
4. Use the spherical volume element with its full factor.
5. Choose the coordinate system that makes a given triple integral easiest.
Warm-up
Discussion prompt
Before we open Week 12 - Cylindrical & Spherical Integrals: without looking back, what was the main idea of Week 11 - Triple Integrals & Applications, 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 triple integrals over solid regions: the volume element, finding the six nested-limit orders for a given solid, and projecting onto a coordinate plane, with applications to mass, center of mass, centroids, moments, and moments of inertia. It targets the classic traps of putting an outer variable in the inner limits, forgetting the density factor when computing mass, and projecting onto the wrong plane.
Concept
A triple integral takes a solid, chops it into tiny pieces, multiplies each piece by an integrand, and adds them all up.
\[ \iiint_E f\, dV \]
The piece of volume is called the volume element. In rectangular coordinates each piece is a tiny box.
\[ dV = dx\, dy\, dz \]
Counterexample
Discussion prompt
A triple integral takes a solid, chops it into tiny pieces, multiplies each piece by an integrand, and adds them all up.
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:
The piece of volume is called the volume element. In rectangular coordinates each piece is a tiny box.
Concept
Rectangular boxes fit a brick beautifully. They fit a ball or a cone terribly.
For a round solid, the edges turn into square-root limits that are painful to integrate.
Cylindrical and spherical coordinates reshape the tiny pieces so they match round solids. The limits then become simple constants.
Analogy
Discussion prompt
Explain Why leave rectangular coordinates 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:
Rectangular boxes fit a brick beautifully. They fit a ball or a cone terribly.
Picture it
Figure (svg): A disk tiled with radial wedges and concentric rings, following the circular boundary.
Discussion prompt
Read the picture before the words. What is this showing, and what is the one thing it is built to make obvious? Commit to an answer, then read on.
Hint: Name the parts, then say what changes between them — and if nothing changes, say what is being held still.
Answer:
Picture tiling a circular pizza with square crackers. Near the crust the squares stick out or leave gaps.
Intuition
Picture tiling a circular pizza with square crackers. Near the crust the squares stick out or leave gaps.
Now tile it with wedges and rings instead. They follow the circle perfectly, with no gaps.
Figure (svg): A disk tiled with radial wedges and concentric rings, following the circular boundary.
That is the whole idea: pick pieces shaped like the solid, and the setup gets easy.
Explain it
Discussion prompt
Explain Round solids hate boxes 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:
Picture tiling a circular pizza with square crackers. Near the crust the squares stick out or leave gaps.
Concept
Cylindrical coordinates keep the height from rectangular, but describe position in the horizontal plane using polar coordinates.
\[ (r, \theta, z) \]
r — The distance from the z-axis, measured in the horizontal plane. Always zero or positive.
theta — The angle around the z-axis, measured from the positive x-axis, exactly like the polar angle.
z — The usual height above the xy-plane, carried over unchanged from rectangular coordinates.
Intuition
Stand in the horizontal plane and use polar coordinates to say where you are. Then rise straight up by z.
Every horizontal slice of the solid is just a polar region. Stacking those slices along z rebuilds the solid.
So a cylindrical integral is a polar integral with an extra height direction added.
Concept
To get rectangular coordinates from cylindrical, use the polar relations for x and y, and keep z the same.
\[ x = r\cos\theta \]
\[ y = r\sin\theta \]
\[ z = z \]
Concept
Going the other way, r comes from the Pythagorean relation, theta from the ratio, and z stays put.
\[ r^2 = x^2 + y^2 \]
\[ \tan\theta = \frac{y}{x} \]
Always check which quadrant the point is in when you solve for theta. The tangent alone cannot tell one side of the plane from the other.
Ranking
Put in order
Put the moves of Worked example: a point in cylindrical form 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. r is the distance from the z-axis, the square root of x-squared plus y-squared.
Worked example
Convert this rectangular point to cylindrical coordinates.
\[ (x,y,z) = (1,\ 1,\ 5) \]
Find r from x and y
Why: r is the distance from the z-axis, the square root of x-squared plus y-squared.
\[ r = \sqrt{1^2 + 1^2} = \sqrt{2} \]
Find theta
Why: Both x and y are positive, so the point is in the first quadrant and the angle is a first-quadrant answer.
\[ \theta = \arctan\frac{1}{1} = \frac{\pi}{4} \]
Keep z
Why: z is identical in both systems.
\[ z = 5 \]
State the answer
Why: Report the ordered triple in cylindrical form.
\[ (r,\theta,z) = \left(\sqrt{2},\ \tfrac{\pi}{4},\ 5\right) \]
Verify by converting back
Why: x equals r cos theta and y equals r sin theta should return the originals.
\[ x = \sqrt{2}\cos\tfrac{\pi}{4} = 1,\qquad y = \sqrt{2}\sin\tfrac{\pi}{4} = 1 \]
Picture it
Animation
Shows: Each line of the worked example "a point in cylindrical form", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: x equals r cos theta and y equals r sin theta should return the originals.
Estimation
Predict first
Convert this cylindrical point to rectangular coordinates.
Commit before you compute: what does Worked example: back to rectangular 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 the radius
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. Recompute r from x and y. It must return the original 4.
Worked example
Convert this cylindrical point to rectangular coordinates.
\[ (r,\theta,z) = \left(4,\ \tfrac{2\pi}{3},\ -2\right) \]
Compute x
Why: x equals r cos theta.
\[ x = 4\cos\tfrac{2\pi}{3} = 4\left(-\tfrac{1}{2}\right) = -2 \]
Compute y
Why: y equals r sin theta.
\[ y = 4\sin\tfrac{2\pi}{3} = 4\left(\tfrac{\sqrt{3}}{2}\right) = 2\sqrt{3} \]
Keep z
Why: z carries over unchanged.
\[ z = -2 \]
Verify the radius
Why: Recompute r from x and y. It must return the original 4.
\[ \sqrt{(-2)^2 + (2\sqrt{3})^2} = \sqrt{4 + 12} = \sqrt{16} = 4 \]
Picture it
Animation
Shows: Each line of the worked example "back to rectangular", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Recompute r from x and y. It must return the original 4.
Concept
Holding one coordinate constant traces a familiar surface.
\[ r = c \quad\Rightarrow\quad \text{cylinder of radius } c \]
\[ \theta = c \quad\Rightarrow\quad \text{vertical half-plane} \]
\[ z = c \quad\Rightarrow\quad \text{horizontal plane} \]
Concept
The surfaces that give rectangular integrals trouble collapse to short equations here.
\[ z = r \quad\Rightarrow\quad \text{cone} \]
\[ z = r^2 \quad\Rightarrow\quad \text{paraboloid} \]
Because these use only r and z, the angle theta usually runs a full turn with no extra work.
Concept
The tiny piece of volume in cylindrical coordinates is not just the product of the three differentials. It carries an extra factor of r.
\[ dV = r\, dz\, dr\, d\theta \]
This factor is the single most forgotten part of the whole topic. Write it every time.
Picture it
Figure (svg): A small polar wedge with sides labeled: radial side dr and arc side r d-theta.
Discussion prompt
Read the picture before the words. What is this showing, and what is the one thing it is built to make obvious? Commit to an answer, then read on.
Hint: Name the parts, then say what changes between them — and if nothing changes, say what is being held still.
Answer:
Look at one tiny piece in the horizontal plane. It is bounded by two radii and two circular arcs: a polar wedge, not a rectangle.
Intuition
Look at one tiny piece in the horizontal plane. It is bounded by two radii and two circular arcs: a polar wedge, not a rectangle.
Figure (svg): A small polar wedge with sides labeled: radial side dr and arc side r d-theta.
The two sides have lengths dr and r times d-theta. The arc side grows with radius, which is exactly where the r comes from.
\[ dA = r\, dr\, d\theta \]
Multiply that area by the height dz to get the volume element. The r rides along.
Intuition
Think of the cylindrical piece as a slightly curved brick: thin in the radial direction, thin in height, and a short arc wide.
A brick far from the axis is wider than one near the axis, even for the same change in angle. The factor r records that widening.
Trap
Writing the volume element as if the three differentials simply multiply.
\[ dV = dz\, dr\, d\theta \qquad (\text{wrong}) \]
Compute the volume of a cylinder of radius 3, height 2, this way
Why: Leaving out the r factor entirely.
\[ \int_0^{2\pi}\!\!\int_0^{3}\!\!\int_0^{2} dz\, dr\, d\theta = 2\pi\cdot 3\cdot 2 = 12\pi \]
That answer is too small, and it disagrees with the known cylinder volume.
Keep the polar area factor r, so the element is r dz dr d-theta.
\[ dV = r\, dz\, dr\, d\theta \qquad (\text{right}) \]
Redo with the r in place
Why: The inner integral in z gives 2, then r times 2 integrates over r.
\[ \int_0^{2\pi}\!\!\int_0^{3}\!\!\int_0^{2} r\, dz\, dr\, d\theta = \int_0^{2\pi}\!\!\int_0^{3} 2r\, dr\, d\theta = 2\pi\cdot 9 = 18\pi \]
This matches the formula: pi times radius squared times height, which is pi times 9 times 2.
Translation
\( dV = r\, dz\, dr\, d\theta \qquad (\text{right}) \)
Draw it
Translate both ways. First write the expression above as a sentence with no symbols in it at all. Then cover it, and write your sentence back as notation. If the two versions disagree, the disagreement is the thing to fix.
Concept
Integrate from the inside out. The innermost variable sweeps between the bottom and top surfaces; the outer two describe the shadow of the solid on the horizontal plane.
For z: go from the lower surface to the upper surface, each written in terms of r.
For r and theta: describe the shadow region in the plane using polar limits.
Step zero
Discussion prompt
Worked example: setting up the limits — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Find where the surfaces meet
Answer:
Worked example
Set up, but do not yet evaluate, the volume of the solid below the paraboloid and above the plane described here.
\[ z = 4 - r^2 \quad\text{above}\quad z = 0 \]
Find where the surfaces meet
Why: The paraboloid touches the plane when its height is zero, giving the edge of the shadow.
\[ 4 - r^2 = 0 \;\Rightarrow\; r = 2 \]
Set the z limits
Why: z runs from the lower surface up to the paraboloid.
\[ 0 \le z \le 4 - r^2 \]
Set the shadow limits
Why: The shadow is a disk of radius 2, so r runs to 2 and theta makes a full turn.
\[ 0 \le r \le 2, \qquad 0 \le \theta \le 2\pi \]
Assemble the integral
Why: Volume is the triple integral of 1 with the r factor.
\[ V = \int_0^{2\pi}\!\!\int_0^{2}\!\!\int_0^{4-r^2} r\, dz\, dr\, d\theta \]
Verify the limits are consistent
Why: The outer limits are constants and the inner z limit uses only r, never z. That is the correct nesting.
Limits check out; the setup is valid.
Picture it
Animation
Shows: Each line of the worked example "setting up the limits", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The outer limits are constants and the inner z limit uses only r, never z. That is the correct nesting.
Fill the middle
Fill in the blanks
From Worked example: volume under a paraboloid — finish the line. Write what belongs on the right of the equals sign before you look.
\int_0^8\pi 4\, d\theta = ___
Why: Producing the right-hand side unprompted is the difference between recognising this line and being able to use it. The integrand r is constant in z, so multiply by the height 4 minus r-squared.
Worked example
Now evaluate that volume.
\[ V = \int_0^{2\pi}\!\!\int_0^{2}\!\!\int_0^{4-r^2} r\, dz\, dr\, d\theta \]
Integrate in z
Why: The integrand r is constant in z, so multiply by the height 4 minus r-squared.
\[ \int_0^{4-r^2} r\, dz = r(4 - r^2) = 4r - r^3 \]
Integrate in r
Why: Antidifferentiate term by term from 0 to 2.
\[ \int_0^{2} (4r - r^3)\, dr = \left[2r^2 - \tfrac{r^4}{4}\right]_0^2 = 8 - 4 = 4 \]
Integrate in theta
Why: The remaining constant integrates over a full turn.
\[ \int_0^{2\pi} 4\, d\theta = 8\pi \]
Verify against the shortcut
Why: The volume under z = a - r-squared down to the plane is pi times a-squared over 2, with a = 4.
\[ \frac{\pi a^2}{2} = \frac{\pi (4)^2}{2}\;? \;=\; 8\pi \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "volume under a paraboloid", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The volume under z = a - r-squared down to the plane is pi times a-squared over 2, with a = 4.
Missing information
Discussion prompt
Find the volume of the solid bounded below by the cone and above by a flat top.
What do you need to know — or decide — before the first line can be written? List everything the problem has to hand you.
Hint: Anything you would have to invent to get started is a thing the problem must supply.
Answer:
At a given radius, z runs from the cone surface up to the flat top.
Worked example
Find the volume of the solid bounded below by the cone and above by a flat top.
\[ z = r \quad\text{up to}\quad z = 2 \]
Set the z limits
Why: At a given radius, z runs from the cone surface up to the flat top.
\[ r \le z \le 2 \]
Set the shadow limits
Why: The cone reaches the top when z equals r equals 2, so the shadow is a disk of radius 2.
\[ 0 \le r \le 2, \qquad 0 \le \theta \le 2\pi \]
Integrate in z
Why: The integrand r is constant in z; the height is 2 minus r.
\[ \int_r^{2} r\, dz = r(2 - r) = 2r - r^2 \]
Integrate in r, then theta
Why: Antidifferentiate, evaluate from 0 to 2, then sweep the full angle.
\[ 2\pi\int_0^{2}(2r - r^2)\, dr = 2\pi\left[r^2 - \tfrac{r^3}{3}\right]_0^2 = 2\pi\left(4 - \tfrac{8}{3}\right) = \frac{8\pi}{3} \]
Verify with the cone formula
Why: A cone of base radius 2 and height 2 has volume one third pi radius-squared height.
\[ \tfrac{1}{3}\pi (2)^2 (2) = \frac{8\pi}{3} \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "volume of a cone region", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: A cone of base radius 2 and height 2 has volume one third pi radius-squared height.
Concept
Reach for cylindrical when the solid has an axis of symmetry, or when its boundary is a cylinder, cone, or paraboloid.
It also helps whenever the integrand depends on the distance from the z-axis, since that distance is simply r.
In short: circular shadow in the plane, straight up in z.
Pattern
1. Rewrite surfaces and the integrand using r, theta, z
Why: Replace x-squared plus y-squared with r-squared, and x, y with the polar forms.
2. Set the z limits from bottom surface to top surface
Why: These may depend on r and theta.
3. Set r and theta limits from the shadow region
Why: The projection of the solid onto the horizontal plane, in polar limits.
4. Attach the factor r and integrate inside out
Why: The element is r dz dr d-theta. Forgetting the r is the number one error.
Elimination
Eliminate the wrong options
Which expression is the correct cylindrical volume element dV?
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: B
Why: The horizontal area element in polar form is r dr d-theta, so multiplying by the height dz gives dV = r dz dr d-theta. The single factor of r is required.
Check
Pick the correct volume element for a triple integral in cylindrical coordinates.
Check your understanding
Which expression is the correct cylindrical volume element dV?
Answer: B
Why: The horizontal area element in polar form is r dr d-theta, so multiplying by the height dz gives dV = r dz dr d-theta. The single factor of r is required.
Step zero
Discussion prompt
Worked example: volume between two paraboloids — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Find where they meet
Answer:
Worked example
Find the volume of the solid trapped between an upward paraboloid and a downward one.
\[ z = r^2 \quad\text{below}\quad z = 8 - r^2 \]
Find where they meet
Why: Set the two heights equal to locate the edge of the shadow.
\[ r^2 = 8 - r^2 \;\Rightarrow\; r^2 = 4 \;\Rightarrow\; r = 2 \]
Set the limits
Why: z runs from the lower paraboloid to the upper one; the shadow is a disk of radius 2.
\[ r^2 \le z \le 8 - r^2, \quad 0 \le r \le 2, \quad 0 \le \theta \le 2\pi \]
Integrate in z
Why: The height is the top minus the bottom, times the integrand r.
\[ \int_{r^2}^{8-r^2} r\, dz = r\big(8 - 2r^2\big) = 8r - 2r^3 \]
Integrate in r, then theta
Why: Antidifferentiate from 0 to 2, then sweep the full angle.
\[ 2\pi\int_0^{2}(8r - 2r^3)\, dr = 2\pi\left[4r^2 - \tfrac{r^4}{2}\right]_0^2 = 2\pi(16 - 8) = 16\pi \]
Verify by symmetry
Why: The two paraboloids are mirror images about the plane z = 4, so the volume should be twice the volume from z = 4 down to the lower paraboloid; that halves cleanly and is consistent with 16 pi.
\[ V = 16\pi \]
Picture it
Animation
Shows: Each line of the worked example "volume between two paraboloids", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The two paraboloids are mirror images about the plane z = 4, so the volume should be twice the volume from z = 4 down to the lower paraboloid; that halves cleanly and is consistent with 16 pi.
Worked example
Convert and evaluate this rectangular triple integral.
\[ \int_{-3}^{3}\!\!\int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}}\!\!\int_{0}^{x^2+y^2} dz\, dy\, dx \]
Read the shadow region
Why: The x and y limits trace the disk of radius 3 in the plane.
\[ x^2 + y^2 \le 9 \;\Rightarrow\; 0 \le r \le 3,\; 0 \le \theta \le 2\pi \]
Rewrite the z limit
Why: The upper surface x-squared plus y-squared becomes r-squared.
\[ 0 \le z \le r^2 \]
Attach the factor r
Why: The integrand was 1; in cylindrical the element is r dz dr d-theta.
\[ \int_0^{2\pi}\!\!\int_0^{3}\!\!\int_0^{r^2} r\, dz\, dr\, d\theta \]
Integrate in z, then r, then theta
Why: z gives r times r-squared, then integrate r-cubed, then the full angle.
\[ 2\pi\int_0^{3} r^3\, dr = 2\pi\left[\tfrac{r^4}{4}\right]_0^3 = 2\pi\cdot\tfrac{81}{4} = \frac{81\pi}{2} \]
Verify the integrand rewrite
Why: The z integral of r over 0 to r-squared is r times r-squared, which is r-cubed; that is what we integrated.
\[ \int_0^{r^2} r\, dz = r\cdot r^2 = r^3 \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "convert a rectangular integral to cylindrical", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The z integral of r over 0 to r-squared is r times r-squared, which is r-cubed; that is what we integrated.
Ranking
Put in order
Put the moves of Worked example: mass with a density 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. Mass is the integral of density times the volume element; do not forget the r in the element.
Worked example
A solid cylinder of radius 2 and height 4 has density equal to the distance from its central axis. Find its mass.
\[ \delta = r, \qquad 0 \le r \le 2,\; 0 \le z \le 4 \]
Write mass as a triple integral
Why: Mass is the integral of density times the volume element; do not forget the r in the element.
\[ m = \int_0^{2\pi}\!\!\int_0^{2}\!\!\int_0^{4} (r)\,(r)\, dz\, dr\, d\theta \]
Combine the two factors of r
Why: Density r times element factor r gives r-squared.
\[ m = \int_0^{2\pi}\!\!\int_0^{2}\!\!\int_0^{4} r^2\, dz\, dr\, d\theta \]
Integrate in z, then r, then theta
Why: z contributes 4, then integrate r-squared, then sweep the angle.
\[ 2\pi\int_0^{2} 4r^2\, dr = 2\pi\cdot 4\cdot\tfrac{8}{3} = \frac{64\pi}{3} \]
Verify the units of the setup
Why: Two factors of r appear: one from density, one from the volume element. Missing either would give the wrong power of r and a wrong mass.
\[ m = \frac{64\pi}{3} \]
Picture it
Animation
Shows: Each line of the worked example "mass with a density", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Two factors of r appear: one from density, one from the volume element. Missing either would give the wrong power of r and a wrong mass.
Check
Evaluate the volume of a cylinder of radius 1 and height 2.
\[ V = \int_0^{2\pi}\!\!\int_0^{1}\!\!\int_0^{2} r\, dz\, dr\, d\theta \]
Check your understanding
What is the value of this integral?
Answer: A
Why: Integrating r over z gives 2r; integrating 2r from 0 to 1 gives 1; sweeping theta over a full turn multiplies by 2 pi, so V = 2 pi. This equals pi times radius squared times height.
Concept
Spherical coordinates describe a point by how far it is from the origin and by two angles.
\[ (\rho, \varphi, \theta) \]
rho — The straight-line distance from the origin to the point. Always zero or positive.
The two angles say which direction to look; rho says how far to travel in that direction.
Concept
The first angle is measured down from the positive z-axis.
phi — The angle between the positive z-axis and the line from the origin to the point. It ranges from 0 straight up to pi straight down.
\[ 0 \le \varphi \le \pi \]
At phi equal to 0 you point straight up; at phi equal to pi you point straight down.
Matching
Match the pairs
Match each term to the definition this lesson gave it — not the one you would guess from the word.
Why: These are the working definitions of r, theta, z, rho, phi as Week 12 - Cylindrical & Spherical Integrals uses them. Pairing them correctly is the test of whether you could state each one with the slide switched off.
Concept
The second angle is the same theta as in cylindrical: it swings around the z-axis.
theta — The angle around the z-axis measured from the positive x-axis. It ranges over a full turn, from 0 to two pi.
\[ 0 \le \theta \le 2\pi \]
So phi tilts up and down; theta spins around. They are two different jobs.
Definition probe
Sort into buckets
Every line below is part of the definition of phi or of theta — one or the other, never both. Put each where it belongs.
Intuition
Think of a point on a globe. rho is the radius of the globe: how far from the center.
theta is the longitude: how far around you have spun.
phi is measured from the north pole downward, so it is like a co-latitude: 0 at the pole, pi at the opposite pole.
Intuition
Watch out: latitude on a map is measured from the equator. Phi is measured from the north pole instead.
So the equator sits at phi equal to a right angle, halfway between the two poles, not at phi equal to zero.
\[ \text{equator:}\quad \varphi = \frac{\pi}{2} \]
Concept
Keeping the ranges straight prevents most spherical mistakes.
| coordinate | meaning | range |
|---|---|---|
| rho | distance from origin | 0 and up |
| phi | angle from positive z-axis | 0 to pi |
| theta | angle around z-axis | 0 to two pi |
The key contrast: phi stops at pi, but theta goes all the way to two pi.
Comparison
Comparison matrix
From The three ranges at a glance: refill the meaning column from what you know. The rest of the table is as it appeared.
| coordinate | meaning | range |
|---|---|---|
| rho | distance from origin | 0 and up |
| phi | angle from positive z-axis | 0 to pi |
| theta | angle around z-axis | 0 to two pi |
Concept
Textbooks and physicists sometimes swap the names of the two angles. Physics often uses theta for the angle from the z-axis and phi for the azimuth.
In this course we use the math convention: rho is distance, phi is the angle from the positive z-axis, and theta is the azimuth.
Whenever you read a problem, confirm which letter means which angle before you set up limits.
Concept
Reach the point in two moves: first project onto the horizontal plane, then account for height.
The horizontal distance from the axis is rho sin phi, which then splits into x and y by the usual cosine and sine of theta.
\[ x = \rho\sin\varphi\cos\theta \]
\[ y = \rho\sin\varphi\sin\theta \]
\[ z = \rho\cos\varphi \]
Concept
Going back, rho is the full distance, phi comes from the height, and theta is the same azimuth as before.
\[ \rho = \sqrt{x^2 + y^2 + z^2} \]
\[ \varphi = \arccos\frac{z}{\rho} \]
\[ \tan\theta = \frac{y}{x} \]
Estimation
Predict first
Convert this rectangular point to spherical coordinates.
Commit before you compute: what does Worked example: a point in spherical form 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 by converting back
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. Rebuild x, y, z from the spherical values; they must return the originals.
Worked example
Convert this rectangular point to spherical coordinates.
\[ (x,y,z) = \left(2,\ 2,\ 2\sqrt{2}\right) \]
Find rho
Why: rho is the distance from the origin.
\[ \rho = \sqrt{2^2 + 2^2 + (2\sqrt{2})^2} = \sqrt{4+4+8} = \sqrt{16} = 4 \]
Find phi from the height
Why: phi is the arccosine of z divided by rho.
\[ \varphi = \arccos\frac{2\sqrt{2}}{4} = \arccos\frac{\sqrt{2}}{2} = \frac{\pi}{4} \]
Find theta
Why: x and y are equal and positive, so theta is the first-quadrant angle.
\[ \theta = \arctan\frac{2}{2} = \frac{\pi}{4} \]
State the answer
Why: Report as rho, phi, theta.
\[ (\rho,\varphi,\theta) = \left(4,\ \tfrac{\pi}{4},\ \tfrac{\pi}{4}\right) \]
Verify by converting back
Why: Rebuild x, y, z from the spherical values; they must return the originals.
\[ x = 4\sin\tfrac{\pi}{4}\cos\tfrac{\pi}{4} = 2,\quad z = 4\cos\tfrac{\pi}{4} = 2\sqrt{2}\;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "a point in spherical form", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Rebuild x, y, z from the spherical values; they must return the originals.
Concept
Holding rho fixed means staying a fixed distance from the origin: that traces a sphere.
\[ \rho = c \quad\Rightarrow\quad \text{sphere of radius } c \]
This is why a ball centered at the origin is trivial in spherical: rho simply runs from 0 to the radius.
Concept
Holding phi fixed means keeping a fixed tilt from the z-axis: that sweeps out a cone with its tip at the origin.
\[ \varphi = c \quad\Rightarrow\quad \text{cone from the origin} \]
A narrow phi is a steep cone hugging the axis; phi at a right angle flattens the cone into the horizontal plane.
Concept
Holding theta fixed means facing one direction while rho and phi vary: that fills a vertical half-plane hinged on the z-axis.
\[ \theta = c \quad\Rightarrow\quad \text{vertical half-plane} \]
So the three constant-coordinate surfaces are a sphere, a cone, and a half-plane.
Concept
The best solids for spherical are the ones whose boundaries are spheres and cones, because then every limit is a constant.
rho runs from the inner radius to the outer radius; phi runs between two cone angles; theta sweeps around.
If the boundaries are not spheres or cones, spherical may not be the easy choice.
Concept
The spherical piece of volume carries a larger factor than the cylindrical one.
\[ dV = \rho^2 \sin\varphi \; d\rho\, d\varphi\, d\theta \]
There are two parts to remember: rho squared, and sin phi. Missing either one gives a wrong answer.
Fill the middle
Fill in the blanks
From Why rho squared sin phi — finish the line. Write what belongs on the right of the equals sign before you look.
d\rho \cdot \rho\, d\varphi \cdot \rho\sin\varphi\, d\theta = \rho^2 \sin\varphi\; d\rho\, d\varphi\, d\theta
Why: Producing the right-hand side unprompted is the difference between recognising this line and being able to use it. Volume of the little box is the product of its three edges.
Intuition
Build the tiny spherical piece from three edge lengths. One edge is the radial step d-rho.
A step in phi traces an arc on a circle of radius rho, so its length is rho times d-phi.
A step in theta traces an arc on a circle whose radius is only the horizontal distance rho sin phi, so its length is rho sin phi times d-theta.
Multiply the three edge lengths
Why: Volume of the little box is the product of its three edges.
\[ d\rho \cdot \rho\, d\varphi \cdot \rho\sin\varphi\, d\theta = \rho^2 \sin\varphi\; d\rho\, d\varphi\, d\theta \]
Explain it
Discussion prompt
Explain Why rho squared sin phi 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:
Build the tiny spherical piece from three edge lengths. One edge is the radial step d-rho.
Intuition
The sin phi factor tells a physical story. Near the equator, sin phi is large, so the theta-circles are wide and the pieces are fat.
Near the poles, sin phi shrinks toward zero, the theta-circles pinch to a point, and the pieces get thin.
That pinching is exactly why the volume element must carry sin phi.
Analogy
Discussion prompt
Explain The piece shrinks near the poles 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:
The sin phi factor tells a physical story. Near the equator, sin phi is large, so the theta-circles are wide and the pieces are fat.
Anomaly
Predict first
A student writes this, and it looks reasonable:
Treating the spherical element as the bare product of the three differentials.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: With no factor, every integral is trivial but meaningless.
Keep the full factor rho squared sin phi.
Why: With no factor, every integral is trivial but meaningless.
Trap
Treating the spherical element as the bare product of the three differentials.
\[ dV = d\rho\, d\varphi\, d\theta \qquad (\text{wrong}) \]
Compute the volume of a ball of radius 2 this way
Why: With no factor, every integral is trivial but meaningless.
\[ \int_0^{2\pi}\!\!\int_0^{\pi}\!\!\int_0^{2} d\rho\, d\varphi\, d\theta = 2\cdot\pi\cdot 2\pi = 4\pi^2 \]
That is not even the right kind of number for a volume of a ball.
Keep the full factor rho squared sin phi.
\[ dV = \rho^2\sin\varphi\; d\rho\, d\varphi\, d\theta \qquad (\text{right}) \]
Redo with the factor in place
Why: rho-squared integrates to 8 thirds, sin phi to 2, theta to two pi.
\[ \int_0^{2\pi}\!\!\int_0^{\pi}\!\!\int_0^{2} \rho^2\sin\varphi\, d\rho\, d\varphi\, d\theta = \tfrac{8}{3}\cdot 2\cdot 2\pi = \frac{32\pi}{3} \]
This matches four thirds pi radius cubed, the volume of a sphere of radius 2.
Notation
Annotate
From Trap: dropping rho squared sin phi — read this one piece at a time. What is each part doing?
On: \( dV = d\rho\, d\varphi\, d\theta \qquad (\text{wrong}) \)
Anomaly
Predict first
A student writes this, and it looks reasonable:
Assuming phi is the one that spins all the way around, so letting it run to two pi.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: The sine completes a full period and cancels itself out.
Phi only tilts from the top pole to the bottom pole, so it stops at pi. Theta is the one that spins to two pi.
Why: The sine completes a full period and cancels itself out.
Trap
Assuming phi is the one that spins all the way around, so letting it run to two pi.
\[ 0 \le \varphi \le 2\pi \qquad (\text{wrong}) \]
Integrate sin phi over that range
Why: The sine completes a full period and cancels itself out.
\[ \int_0^{2\pi} \sin\varphi\, d\varphi = 0 \]
The whole volume collapses to zero, an obvious sign the range is wrong.
Phi only tilts from the top pole to the bottom pole, so it stops at pi. Theta is the one that spins to two pi.
\[ 0 \le \varphi \le \pi, \qquad 0 \le \theta \le 2\pi \]
Integrate sin phi over the correct range
Why: From 0 to pi the sine stays positive and gives a clean total.
\[ \int_0^{\pi} \sin\varphi\, d\varphi = \big[-\cos\varphi\big]_0^{\pi} = 2 \]
A live sanity check: if a volume comes out zero, suspect the phi range first.
Notation
Annotate
From Trap: swapping phi and theta ranges — read this one piece at a time. What is each part doing?
On: \( \int_0^{\pi} \sin\varphi\, d\varphi = \big[-\cos\varphi\big]_0^{\pi} = 2 \)
Step zero
Discussion prompt
Worked example: volume of a sphere — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Set the limits
Answer:
Worked example
Find the volume of a ball of radius 2 using spherical coordinates.
Set the limits
Why: A full ball centered at the origin uses constant limits in every coordinate.
\[ 0 \le \rho \le 2,\quad 0 \le \varphi \le \pi,\quad 0 \le \theta \le 2\pi \]
Write the integral with the factor
Why: Volume is the triple integral of 1 times rho squared sin phi.
\[ V = \int_0^{2\pi}\!\!\int_0^{\pi}\!\!\int_0^{2} \rho^2\sin\varphi\; d\rho\, d\varphi\, d\theta \]
Separate the three integrals
Why: Every factor depends on only one variable, so the integral splits into a product.
\[ \left(\int_0^{2}\rho^2 d\rho\right)\!\left(\int_0^{\pi}\sin\varphi\, d\varphi\right)\!\left(\int_0^{2\pi} d\theta\right) \]
Evaluate each piece
Why: rho-squared gives 8 thirds, sin phi gives 2, theta gives two pi.
\[ \frac{8}{3}\cdot 2 \cdot 2\pi = \frac{32\pi}{3} \]
Verify with the sphere formula
Why: Four thirds pi radius cubed with radius 2 must match.
\[ \frac{4}{3}\pi (2)^3 = \frac{32\pi}{3} \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "volume of a sphere", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Four thirds pi radius cubed with radius 2 must match.
Missing information
Discussion prompt
Find the volume of the solid inside a sphere and above a cone, both meeting at the origin.
What do you need to know — or decide — before the first line can be written? List everything the problem has to hand you.
Hint: Anything you would have to invent to get started is a thing the problem must supply.
Answer:
rho from 0 to the sphere; phi from the axis down to the cone angle; theta a full turn.
Worked example
Find the volume of the solid inside a sphere and above a cone, both meeting at the origin.
\[ \rho \le 1, \qquad 0 \le \varphi \le \tfrac{\pi}{3} \]
The sphere caps it off; the cone forms the pointed bottom. This shape is a coordinate-surface match made for spherical.
Set the limits
Why: rho from 0 to the sphere; phi from the axis down to the cone angle; theta a full turn.
\[ 0 \le \rho \le 1,\quad 0 \le \varphi \le \tfrac{\pi}{3},\quad 0 \le \theta \le 2\pi \]
Evaluate the three integrals
Why: rho-squared gives one third; sin phi from 0 to pi over 3 gives one half; theta gives two pi.
\[ \tfrac{1}{3}\cdot\left(1 - \cos\tfrac{\pi}{3}\right)\cdot 2\pi = \tfrac{1}{3}\cdot\tfrac{1}{2}\cdot 2\pi = \frac{\pi}{3} \]
Verify with a solid-angle check
Why: The cone captures a fraction one minus cosine of the cone angle, all over 2, of the full ball; that fraction times four thirds pi gives the same result.
\[ \frac{1 - \cos(\pi/3)}{2}\cdot\frac{4}{3}\pi = \frac{0.5}{2}\cdot\frac{4}{3}\pi = \frac{\pi}{3} \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "the ice-cream-cone region", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The cone captures a fraction one minus cosine of the cone angle, all over 2, of the full ball; that fraction times four thirds pi gives the same result.
Estimation
Predict first
Evaluate the integral of the distance from the origin over the ball of radius 2.
Commit before you compute: what does Worked example: convert an integral to spherical 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 the power bookkeeping
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. One rho from the integrand plus two from the element makes rho cubed; missing the element factor would have left only rho and a wrong answer.
Worked example
Evaluate the integral of the distance from the origin over the ball of radius 2.
\[ \iiint_E \sqrt{x^2+y^2+z^2}\; dV, \qquad E:\; x^2+y^2+z^2 \le 4 \]
Rewrite the integrand
Why: The distance from the origin is exactly rho.
\[ \sqrt{x^2+y^2+z^2} = \rho \]
Attach the volume factor
Why: Integrand rho times element rho squared sin phi gives rho cubed sin phi.
\[ \int_0^{2\pi}\!\!\int_0^{\pi}\!\!\int_0^{2} \rho\cdot\rho^2\sin\varphi\; d\rho\, d\varphi\, d\theta = \int_0^{2\pi}\!\!\int_0^{\pi}\!\!\int_0^{2} \rho^3\sin\varphi\; d\rho\, d\varphi\, d\theta \]
Evaluate the three integrals
Why: rho-cubed gives 4; sin phi gives 2; theta gives two pi.
\[ 4\cdot 2\cdot 2\pi = 16\pi \]
Verify the power bookkeeping
Why: One rho from the integrand plus two from the element makes rho cubed; missing the element factor would have left only rho and a wrong answer.
\[ \rho^{1}\cdot\rho^{2} = \rho^{3} \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "convert an integral to spherical", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: One rho from the integrand plus two from the element makes rho cubed; missing the element factor would have left only rho and a wrong answer.
Step zero
Discussion prompt
Worked example: mass of a hemisphere — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Set the phi range for the upper half
Answer:
Worked example
A solid upper hemisphere of radius 1 has density equal to the distance from the origin. Find its mass.
\[ \delta = \rho, \qquad \rho \le 1,\; 0 \le \varphi \le \tfrac{\pi}{2} \]
Set the phi range for the upper half
Why: The upper half runs from the top pole down only to the equator, a right angle.
\[ 0 \le \varphi \le \frac{\pi}{2} \]
Write the mass integral
Why: Density rho times element rho squared sin phi gives rho cubed sin phi.
\[ m = \int_0^{2\pi}\!\!\int_0^{\pi/2}\!\!\int_0^{1} \rho^3\sin\varphi\; d\rho\, d\varphi\, d\theta \]
Evaluate the three integrals
Why: rho-cubed gives one quarter; sin phi from 0 to a right angle gives 1; theta gives two pi.
\[ \tfrac{1}{4}\cdot 1 \cdot 2\pi = \frac{\pi}{2} \]
Verify the phi integral
Why: The integral of sin phi from 0 to a right angle is 1 minus cosine of that angle, which is 1 minus 0.
\[ \int_0^{\pi/2}\sin\varphi\, d\varphi = \big[-\cos\varphi\big]_0^{\pi/2} = 0 - (-1) = 1 \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "mass of a hemisphere", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The integral of sin phi from 0 to a right angle is 1 minus cosine of that angle, which is 1 minus 0.
Ranking
Put in order
Put the moves of Worked example: centroid of a hemisphere 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 centroid height is the moment about the plane divided by the volume.
Worked example
Find the height of the centroid of a solid upper hemisphere of radius 2 (uniform density).
Recall the formula
Why: The centroid height is the moment about the plane divided by the volume.
\[ \bar{z} = \frac{1}{V}\iiint_E z\; dV \]
Find the volume
Why: A solid hemisphere is half of four thirds pi radius cubed.
\[ V = \frac{1}{2}\cdot\frac{4}{3}\pi (2)^3 = \frac{16\pi}{3} \]
Set up the moment with z equal to rho cos phi
Why: The integrand z times the element becomes rho cubed sin phi cos phi.
\[ \iiint_E z\, dV = \int_0^{2\pi}\!\!\int_0^{\pi/2}\!\!\int_0^{2} \rho^3\sin\varphi\cos\varphi\; d\rho\, d\varphi\, d\theta \]
Evaluate the moment
Why: rho-cubed gives 4; sin phi cos phi from 0 to a right angle gives one half; theta gives two pi.
\[ 4\cdot\tfrac{1}{2}\cdot 2\pi = 4\pi \]
Verify against the known centroid
Why: Divide the moment by the volume; the standard result is three eighths of the radius.
\[ \bar{z} = \frac{4\pi}{16\pi/3} = \frac{3}{4} = \frac{3(2)}{8} \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "centroid of a hemisphere", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Divide the moment by the volume; the standard result is three eighths of the radius.
Concept
The integrand often tells you the coordinate system before the region does.
If the integrand depends on the distance from the z-axis, that distance is r, so cylindrical is natural.
If the integrand depends on the distance from the origin, that distance is rho, so spherical is natural.
Counterexample
Discussion prompt
The integrand often tells you the coordinate system before the region does.
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:
If the integrand depends on the distance from the z-axis, that distance is r, so cylindrical is natural.
Concept
Both handle round solids, but they shine on different shapes.
| shape | best system |
|---|---|
| cylinder, can, tube | cylindrical |
| paraboloid over a disk | cylindrical |
| ball or spherical shell | spherical |
| region between a sphere and a cone | spherical |
Rule of thumb: flat top and bottom favor cylindrical; curved distance-from-origin boundaries favor spherical.
Discrimination
Sort into buckets
Sort these by best system, from memory, without looking back at Choosing cylindrical versus spherical. Telling them apart on the spot is the skill; the table is only where the answer happens to be written down.
Anomaly
Predict first
A student writes this, and it looks reasonable:
Forcing the ball of radius 2 into cylindrical coordinates.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: The z limits carry a square root, so the r integral needs a substitution to finish.
Use spherical, where every boundary of the ball is a constant.
Why: The z limits carry a square root, so the r integral needs a substitution to finish.
Trap
Forcing the ball of radius 2 into cylindrical coordinates.
\[ V = \int_0^{2\pi}\!\!\int_0^{2}\!\!\int_{-\sqrt{4-r^2}}^{\sqrt{4-r^2}} r\, dz\, dr\, d\theta \]
See the pain coming
Why: The z limits carry a square root, so the r integral needs a substitution to finish.
\[ \int_0^{2} 2r\sqrt{4-r^2}\, dr \quad\text{needs}\quad u = 4 - r^2 \]
Use spherical, where every boundary of the ball is a constant.
\[ V = \int_0^{2\pi}\!\!\int_0^{\pi}\!\!\int_0^{2} \rho^2\sin\varphi\; d\rho\, d\varphi\, d\theta \]
Finish in one line
Why: No square roots, no substitution; the answer is immediate.
\[ \tfrac{8}{3}\cdot 2\cdot 2\pi = \frac{32\pi}{3} \]
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
Three systems, three volume elements. The factor is the price of bending straight coordinates into round ones.
| system | volume element | extra factor |
|---|---|---|
| rectangular | dx dy dz | none |
| cylindrical | r dz dr d(theta) | r |
| spherical | rho-squared sin(phi) d(rho) d(phi) d(theta) | rho-squared sin(phi) |
The two round systems always cost a factor. If you ever write one without it, you have made the classic error.
Comparison
Comparison matrix
From The volume-factor cheat sheet: refill the volume element column from what you know. The rest of the table is as it appeared.
| system | volume element | extra factor |
|---|---|---|
| rectangular | dx dy dz | none |
| cylindrical | r dz dr d(theta) | r |
| spherical | rho-squared sin(phi) d(rho) d(phi) d(theta) | rho-squared sin(phi) |
Pattern
1. Rewrite surfaces and the integrand in rho, phi, theta
Why: Distance from origin is rho; distance from the z-axis is rho sin phi; height is rho cos phi.
2. Set rho limits from inner radius to outer radius
Why: For a solid ball, rho starts at 0.
3. Set phi limits between cone angles, and theta around
Why: phi lives in 0 to pi; theta lives in 0 to two pi.
4. Attach rho squared sin phi and integrate
Why: The element is rho squared sin phi d-rho d-phi d-theta. Both parts are required.
Pattern
Boundaries are boxes or planes
Why: Stay in rectangular; nothing round to gain from.
Axis of symmetry: cylinder, cone, or paraboloid
Why: Use cylindrical, and remember the factor r.
Boundaries are spheres and cones about the origin
Why: Use spherical, and remember rho squared sin phi.
Check the integrand too
Why: Distance from the axis points to cylindrical; distance from the origin points to spherical.
Real world
Discussion prompt
Outside this lesson: where does Week 12 - Cylindrical & Spherical Integrals 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 The coordinate decision guide 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 covers triple integrals in cylindrical and spherical coordinates: what each coordinate means, how to convert between systems, and the all-important volume factors r and rho-squared sin-phi. It targets the classic traps of dropping the volume factor, swapping phi and theta, and choosing the harder coordinate system.
Check
Evaluate the volume of a ball of radius 3 using spherical coordinates.
\[ V = \int_0^{2\pi}\!\!\int_0^{\pi}\!\!\int_0^{3} \rho^2\sin\varphi\; d\rho\, d\varphi\, d\theta \]
Check your understanding
What is the value of this integral?
Answer: A
Why: The rho-squared integral gives 9, the sin phi integral over 0 to pi gives 2, and the theta integral gives two pi; the product is 9 times 2 times two pi, which is 36 pi, matching four thirds pi times 3 cubed.
Prediction
Predict first
For a full solid, what is the correct range of the angle phi (measured from the positive z-axis)?
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: 0 to (pi)
Why: Phi tilts from the north pole at 0 down to the south pole at pi, so it covers 0 to pi. The full spin belongs to theta, which is the coordinate that runs 0 to two pi.
Check
Recall which angle sweeps how far in the math convention used here.
Check your understanding
For a full solid, what is the correct range of the angle phi (measured from the positive z-axis)?
Answer: A
Why: Phi tilts from the north pole at 0 down to the south pole at pi, so it covers 0 to pi. The full spin belongs to theta, which is the coordinate that runs 0 to two pi.
Elimination
Eliminate the wrong options
Which coordinate system makes this triple integral easiest to set up?
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: Both boundaries are single coordinate surfaces in spherical: the sphere is rho equal to a constant and the cone is phi equal to a constant, so every limit is a constant.
Check
A solid is the region inside a sphere and above a cone, both centered at the origin.
Check your understanding
Which coordinate system makes this triple integral easiest to set up?
Answer: A
Why: Both boundaries are single coordinate surfaces in spherical: the sphere is rho equal to a constant and the cone is phi equal to a constant, so every limit is a constant.
Check
Convert the integral of x-squared plus y-squared over the solid cylinder of radius 2 and height 3 to cylindrical form.
\[ \iiint_E (x^2+y^2)\, dV, \qquad E:\; r \le 2,\; 0 \le z \le 3 \]
Check your understanding
Which cylindrical integral is correct?
Answer: B
Why: The integrand x-squared plus y-squared becomes r-squared, and the volume element contributes another factor of r, so the full integrand is r-squared times r, which is r-cubed.
Connect it up
Draw it
One page, no notation unless you need it: draw how these connect — The cylindrical recipe · The spherical recipe · The coordinate decision guide · A triple integral adds up a solid · Why leave rectangular coordinates. Put an arrow wherever one of them is what makes another possible, and label the arrow with why.
Recap
You can now switch a triple integral into cylindrical or spherical coordinates and carry the correct volume factor each time.
Cylindrical is polar plus height; its element is r dz dr d-theta.
\[ dV = r\, dz\, dr\, d\theta \]
Spherical uses distance and two angles; its element is rho squared sin phi d-rho d-phi d-theta.
\[ dV = \rho^2\sin\varphi\; d\rho\, d\varphi\, d\theta \]
Remember the four traps: dropping the r, dropping rho squared sin phi, swapping phi and theta, and forcing the harder system. If a volume ever comes out zero or negative, check the phi range and the factor first.
| if the solid is | reach for | and never forget |
|---|---|---|
| a cylinder, cone, or paraboloid | cylindrical | the factor r |
| a ball, shell, or sphere-cone region | spherical | rho-squared sin(phi) |
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