This deck builds the double integral from Riemann sums up to volume under a surface. It covers the properties of the integral, Fubini's theorem in either order, the volume between two surfaces as top minus bottom, average value, and setting up Type I and Type II regions. It targets the classic traps: forgetting top minus bottom, blindly swapping the limits over a region that is not a rectangle, and confusing a signed integral with a true volume.
Subject: Calculus III · 106 slides · symbolic lesson
Open the interactive version of this deck · Homework for this lesson
Objectives
By the end of this deck you can:
1. Read a double integral as a limit of Riemann sums and as the volume under a surface.
2. Use linearity and additivity to break integrals apart.
3. Apply Fubini's theorem to iterate in either order over a rectangle.
4. Set up limits over Type I and Type II regions, and reverse the order of integration.
5. Find the volume between two surfaces and the average value of a function over a region.
Warm-up
Discussion prompt
Before we open Week 9 - Double Integrals & Volume: without looking back, what was the main idea of Week 8 - Iterated Integrals & Area, 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 partial, or iterated, integration: integrating one variable while holding the other constant, evaluating iterated integrals with constant and with variable limits, finding plane area as the double integral of 1, setting limits from a sketch, and reversing the order of integration. It targets the traps of letting the outer limits contain the outer variable, reversing the order without redrawing the region, and mismatching the differentials.
Concept
A single definite integral measures the area under one curve over an interval.
\[ \int_a^b f(x)\,dx = \text{(area under the curve above } [a,b] \text{)} \]
We built it by slicing the interval into tiny pieces, forming rectangles, and adding them up. The double integral copies this idea one dimension higher.
Counterexample
Discussion prompt
A single definite integral measures the area under one curve over an interval.
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:
We built it by slicing the interval into tiny pieces, forming rectangles, and adding them up. The double integral copies this idea one dimension higher.
Intuition
Instead of a curve over an interval, picture a surface hovering over a flat region in the plane.
The single integral filled in the area between a curve and a line. The double integral fills in the volume between a surface and the ground.
Same recipe: chop the base into tiny pieces, build a column over each piece, add the columns, then let the pieces shrink to zero.
Analogy
Discussion prompt
Explain From area to volume 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:
Instead of a curve over an interval, picture a surface hovering over a flat region in the plane.
Concept
Let the base be a region in the plane. Cover it with a fine grid, cutting it into many small rectangular cells.
\[ R \text{ is split into cells } R_1, R_2, \dots, R_n \]
Each cell has a small area. We will build one column of height given by the surface over each cell.
Explain it
Discussion prompt
Explain Partition the region into cells 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:
Let the base be a region in the plane. Cover it with a fine grid, cutting it into many small rectangular cells.
Concept
In each cell, pick any sample point and read off the surface height there.
\[ \text{sample point } (x_i^{*}, y_i^{*}) \in R_i, \qquad \text{cell area } \Delta A_i \]
The column over that cell has base area given by the cell and height given by the function value at the sample point.
\[ \text{column volume} \approx f(x_i^{*}, y_i^{*})\,\Delta A_i \]
Socratic
Discussion prompt
In each cell, pick any sample point and read off the surface height there.
Suppose that were not true. What is the first thing in Week 9 - Double Integrals & Volume that would stop working?
Hint: Follow it one step downstream. The answer is whatever was quietly relying on it.
Answer:
The column over that cell has base area given by the cell and height given by the function value at the sample point.
Concept
Add up all the columns. That total estimates the volume under the surface.
\[ S_n = \sum_{i=1}^{n} f(x_i^{*}, y_i^{*})\,\Delta A_i \]
The finer the grid, the closer this sum gets to the true volume.
Concept
Now let every cell shrink toward zero, so the number of cells grows without bound. If the sums approach a single number, that number is the double integral.
\[ \iint_R f(x,y)\,dA = \lim_{\lVert P\rVert \to 0} \sum_{i=1}^{n} f(x_i^{*}, y_i^{*})\,\Delta A_i \]
Here the norm of the partition is the size of the largest cell. Sending it to zero forces every cell to vanish.
Concept
dA — The differential of area, the limiting form of a small cell area. Over rectangular grids it becomes the product of a small change in x and a small change in y.
\[ dA = dx\,dy = dy\,dx \]
The double sign on the integral reminds you there are two directions to sweep through.
Concept
If the function is continuous on a closed, bounded region, the limit of the Riemann sums exists and does not depend on how you choose the sample points.
That is the safety net for everything ahead: for the smooth functions in this course, the double integral is a well-defined number.
Concept
When the surface never dips below the ground over the region, the double integral is exactly the volume of the solid trapped between them.
\[ f(x,y) \ge 0 \text{ on } R \;\Longrightarrow\; V = \iint_R f(x,y)\,dA \]
The surface is the roof, the region is the floor, and the integral measures the space in between.
Picture it
Figure (svg): A surface above a grid of columns, each column rising from a cell of the base to meet the surface
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:
Think of pouring the solid full of thin square posts, each standing on one grid cell and rising to touch the surface.
Intuition
Think of pouring the solid full of thin square posts, each standing on one grid cell and rising to touch the surface.
Figure (svg): A surface above a grid of columns, each column rising from a cell of the base to meet the surface
Shrinking the posts until they are infinitely thin turns the staircase of columns into the smooth solid. That limit is the integral.
Concept
If part of the surface drops below the ground, that part contributes a negative amount. The double integral is a signed volume.
\[ f < 0 \text{ there} \;\Longrightarrow\; \text{that region subtracts from the total} \]
So the raw integral can be smaller than the true amount of material, and can even be zero when the parts above and below cancel.
Intuition
Treat height above the ground as a deposit and height below the ground as a withdrawal. The integral reports the net balance, not the total money moved.
Keep this in mind: true volume of material is never negative, but the signed integral can be. We will hit this as a trap later.
Ranking
Put in order
Put the moves of Estimate a double integral with a Riemann sum 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. A two-by-two grid on a two-by-two square gives four cells, each one unit wide and one unit tall, so each cell area is one.
Worked example
Estimate the integral of the surface below over the square region using a two-by-two grid and midpoint sample points.
\[ f(x,y) = x^2 + y^2, \qquad R = [0,2]\times[0,2] \]
Cut R into four unit cells
Why: A two-by-two grid on a two-by-two square gives four cells, each one unit wide and one unit tall, so each cell area is one.
List the four midpoints
Why: The center of each unit cell sits at a half-integer in each coordinate.
\[ (0.5,0.5),\;(1.5,0.5),\;(0.5,1.5),\;(1.5,1.5) \]
Evaluate the height at each midpoint
Why: Plug each point into x squared plus y squared.
| point | value of f |
|---|---|
| (0.5, 0.5) | 0.5 |
| (1.5, 0.5) | 2.5 |
| (0.5, 1.5) | 2.5 |
| (1.5, 1.5) | 4.5 |
Multiply each height by the cell area and add
Why: Each cell area is one, so the estimate is just the sum of the four heights.
\[ S_4 = (0.5 + 2.5 + 2.5 + 4.5)(1) = 10 \]
Verify against the exact value
Why: Splitting the integral, the exact answer is thirty-two thirds, about 10.67, so the coarse estimate of ten is close and a bit low, exactly as expected for a bowl-shaped surface sampled at centers.
\[ \iint_R (x^2+y^2)\,dA = \tfrac{32}{3} \approx 10.67 \]
Picture it
Animation
Shows: Each line of the worked example "Estimate a double integral with a Riemann sum", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Splitting the integral, the exact answer is thirty-two thirds, about 10.67, so the coarse estimate of ten is close and a bit low, exactly as expected for a bowl-shaped surface sampled at centers.
Concept
You can pull constants out and split sums, exactly like single integrals.
\[ \iint_R \big(c\,f + g\big)\,dA = c\iint_R f\,dA + \iint_R g\,dA \]
This lets you break a messy integrand into simple pieces you already know how to integrate.
Concept
If you cut the region into two non-overlapping pieces, the integral over the whole equals the sum over the pieces.
\[ R = R_1 \cup R_2, \quad R_1 \cap R_2 = \varnothing \;\Longrightarrow\; \iint_R f\,dA = \iint_{R_1} f\,dA + \iint_{R_2} f\,dA \]
This is how we handle awkward shapes: split them into friendly pieces and add.
Concept
Integrating the constant function one over a region measures the area of that region.
\[ \iint_R 1\,dA = \text{area of } R \]
Intuitively, a flat roof at height one over the region encloses a solid whose volume number equals the floor area number.
Concept
If one surface sits below another everywhere on the region, its integral is no larger.
\[ f \le g \text{ on } R \;\Longrightarrow\; \iint_R f\,dA \le \iint_R g\,dA \]
A quick corollary: the integral lies between the smallest value times the area and the largest value times the area. Great for sanity checks.
Concept
Computing the limit of Riemann sums directly is painful. Fubini's theorem lets you compute a double integral as two ordinary single integrals, one after the other.
\[ \iint_R f\,dA = \int_a^b \!\! \int_c^d f(x,y)\,dy\,dx = \int_c^d \!\! \int_a^b f(x,y)\,dx\,dy \]
Over a rectangle you may integrate in either order and get the same number. Pick the order that is easier.
Intuition
Slice the solid with a plane at a fixed x. That slice is a flat region whose area you get from one single integral in y.
\[ A(x) = \int_c^d f(x,y)\,dy \]
Now sweep the slice across the region, adding up the slice areas. That second integral in x accumulates the whole volume.
\[ V = \int_a^b A(x)\,dx \]
Socratic
Discussion prompt
Slice the solid with a plane at a fixed x. That slice is a flat region whose area you get from one single integral in y.
Suppose that were not true. What is the first thing in Week 9 - Double Integrals & Volume that would stop working?
Hint: Follow it one step downstream. The answer is whatever was quietly relying on it.
Concept
Work from the inside out. The inner integral treats the outer variable as a frozen constant.
\[ \int_a^b \underbrace{\left( \int_c^d f(x,y)\,dy \right)}_{\text{a function of } x}\,dx \]
The inner integral collapses the y direction and leaves a plain single-variable function of x. The outer integral then finishes the job.
Concept
The outer limits are always plain constants. Only the inner limits may depend on the outer variable.
\[ \int_a^b \!\! \int_{g_1(x)}^{g_2(x)} f\,dy\,dx \quad\text{(inner limits may hold } x\text{)} \]
If the outer variable ever appears in the outer limits, the setup is broken. The final answer must be a pure number, not an expression with a leftover variable.
Step zero
Discussion prompt
The inner integral is a function of x — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Hold x fixed and integrate in y
Answer:
Worked example
Do only the inner integral of the surface from the estimate example, over the same square.
\[ \int_0^2 (x^2 + y^2)\,dy \]
Hold x fixed and integrate in y
Why: The x squared term is a constant with respect to y, so it integrates to x squared times y.
\[ = \left[\, x^2 y + \frac{y^3}{3} \,\right]_{y=0}^{y=2} \]
Evaluate at the y limits
Why: Substitute y equals two and subtract the value at y equals zero.
\[ = 2x^2 + \frac{8}{3} \]
Verify it is a function of x only
Why: There is no y left anywhere, confirming the inner integral collapsed the y direction as it should. This expression is ready for the outer integral.
Picture it
Animation
Shows: Each line of the worked example "The inner integral is a function of x", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: There is no y left anywhere, confirming the inner integral collapsed the y direction as it should. This expression is ready for the outer integral.
Estimation
Predict first
Finish the previous integral over the square in the order dy then dx.
Commit before you compute: what does Rectangle, integrating dy then dx 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 earlier estimate
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. The exact value thirty-two thirds is about 10.67, just above the coarse midpoint estimate of ten from before, which is the expected direction for this bowl surface.
Worked example
Finish the previous integral over the square in the order dy then dx.
\[ \iint_R (x^2+y^2)\,dA = \int_0^2 \!\! \int_0^2 (x^2+y^2)\,dy\,dx \]
Insert the inner result
Why: We already found the inner integral in y equals two x squared plus eight thirds.
\[ = \int_0^2 \left( 2x^2 + \frac{8}{3} \right) dx \]
Integrate in x
Why: The power rule on two x squared gives two thirds x cubed; the constant integrates to eight thirds x.
\[ = \left[\, \frac{2x^3}{3} + \frac{8}{3}x \,\right]_0^2 = \frac{16}{3} + \frac{16}{3} = \frac{32}{3} \]
Verify against the earlier estimate
Why: The exact value thirty-two thirds is about 10.67, just above the coarse midpoint estimate of ten from before, which is the expected direction for this bowl surface.
Picture it
Animation
Shows: Each line of the worked example "Rectangle, integrating dy then dx", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The exact value thirty-two thirds is about 10.67, just above the coarse midpoint estimate of ten from before, which is the expected direction for this bowl surface.
Missing information
Discussion prompt
Now integrate the same surface in the reversed order dx then dy. Fubini promises the same answer.
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:
Now y squared is the constant, so it integrates to y squared times x, and x squared integrates to x cubed over three.
Worked example
Now integrate the same surface in the reversed order dx then dy. Fubini promises the same answer.
\[ \int_0^2 \!\! \int_0^2 (x^2+y^2)\,dx\,dy \]
Do the inner integral in x
Why: Now y squared is the constant, so it integrates to y squared times x, and x squared integrates to x cubed over three.
\[ \int_0^2 (x^2+y^2)\,dx = \left[\, \frac{x^3}{3} + y^2 x \,\right]_0^2 = \frac{8}{3} + 2y^2 \]
Do the outer integral in y
Why: Integrate eight thirds plus two y squared over y from zero to two.
\[ \int_0^2 \left( \frac{8}{3} + 2y^2 \right) dy = \left[\, \frac{8}{3}y + \frac{2y^3}{3} \,\right]_0^2 = \frac{16}{3} + \frac{16}{3} = \frac{32}{3} \]
Verify both orders agree
Why: Both orders give thirty-two thirds, confirming Fubini's theorem on this rectangle.
Picture it
Animation
Shows: Each line of the worked example "Same rectangle, the other order", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Both orders give thirty-two thirds, confirming Fubini's theorem on this rectangle.
Concept
Over a rectangle, if the integrand factors into a part in x times a part in y, the double integral factors into a product of two single integrals.
\[ \iint_R g(x)\,h(y)\,dA = \left( \int_a^b g(x)\,dx \right)\!\left( \int_c^d h(y)\,dy \right) \]
This only works over a rectangle and only when the integrand truly separates. It can save a lot of writing.
Step zero
Discussion prompt
Using the separable shortcut — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Split into a product of single integrals
Answer:
Worked example
Integrate a product integrand over a rectangle.
\[ \iint_R x\,y\,dA, \qquad R = [1,3]\times[0,2] \]
Split into a product of single integrals
Why: The integrand is x times y, a function of x times a function of y, and the region is a rectangle, so the shortcut applies.
\[ = \left( \int_1^3 x\,dx \right)\!\left( \int_0^2 y\,dy \right) \]
Evaluate each factor
Why: The first is nine halves minus one half, which is four; the second is two.
\[ = \left[ \frac{x^2}{2} \right]_1^3 \left[ \frac{y^2}{2} \right]_0^2 = (4)(2) = 8 \]
Verify by iterating instead
Why: The inner integral in y is x times y squared over two evaluated to two, giving two x; then the integral of two x from one to three is nine minus one, which is eight. Same answer.
Picture it
Animation
Shows: Each line of the worked example "Using the separable shortcut", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The inner integral in y is x times y squared over two evaluated to two, giving two x; then the integral of two x from one to three is nine minus one, which is eight. Same answer.
Trap
Over the triangle with the corners at the origin, at one on the x-axis, and at the point where x and y both equal one, a student sets up the area and then just relabels dy dx as dx dy, keeping the same numbers.
\[ \text{start: } \int_0^1 \!\! \int_0^x 1\,dy\,dx \;\; \xrightarrow{\text{wrong}} \;\; \int_0^1 \!\! \int_0^x 1\,dx\,dy \]
The inner integral in x runs to x, and x also sits in the outer limit. The inner integral leaves a leftover x, so the answer still contains x and cannot be a number.
\[ \int_0^x 1\,dx = x \;\Rightarrow\; \int_0^1 x\,dy = xy\Big|_0^1 = x \;\; (\text{not a number}) \]
Redraw the region and re-derive the limits. For the reversed order, fix y and ask how far x ranges: from the line x equals y across to x equals one.
\[ 0 \le y \le 1, \quad y \le x \le 1 \;\Rightarrow\; \int_0^1 \!\! \int_y^1 1\,dx\,dy \]
Now the outer limits are constants and the answer is a clean number, matching the original.
\[ \int_0^1 (1 - y)\,dy = 1 - \tfrac{1}{2} = \tfrac{1}{2} = \int_0^1 \!\! \int_0^x 1\,dy\,dx \]
Concept
A Type I region is described by letting x run between two constants, while y is trapped between a bottom curve and a top curve that depend on x.
\[ R = \{\,(x,y) : a \le x \le b,\; g_1(x) \le y \le g_2(x)\,\} \]
Read it as: pick an x, then y sweeps from the lower boundary up to the upper boundary.
Picture it
Figure (svg): A region with a vertical strip; the strip runs from a lower boundary curve up to an upper boundary curve, and x runs left to right
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:
Imagine sweeping a thin vertical strip across the region from left to right.
Intuition
Imagine sweeping a thin vertical strip across the region from left to right.
Figure (svg): A region with a vertical strip; the strip runs from a lower boundary curve up to an upper boundary curve, and x runs left to right
The inner integral in y adds up the strip from its bottom to its top. The outer integral in x slides the strip across, from x equals a to x equals b.
Concept
The order is dy on the inside and dx on the outside, matching the strip picture.
\[ \iint_R f\,dA = \int_a^b \!\! \int_{g_1(x)}^{g_2(x)} f(x,y)\,dy\,dx \]
The inner limits are the boundary curves; the outer limits are the constant range of x.
Estimation
Predict first
Integrate over the triangle where x runs from zero to two and y runs from the x-axis up to the line y equals x.
Commit before you compute: what does Evaluate over a Type I triangle 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 size is reasonable
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. The integrand ranges from zero to four over a triangle of area two, so a value of four sits comfortably below the maximum height times the area, which is eight.
Worked example
Integrate over the triangle where x runs from zero to two and y runs from the x-axis up to the line y equals x.
\[ \iint_R (x+y)\,dA, \quad 0 \le x \le 2,\; 0 \le y \le x \]
Write the iterated integral
Why: Type I means dy inside with the curve limits, dx outside with the constant limits.
\[ = \int_0^2 \!\! \int_0^x (x+y)\,dy\,dx \]
Do the inner integral in y
Why: Integrate x plus y in y to get x times y plus y squared over two, then evaluate from zero to x.
\[ \int_0^x (x+y)\,dy = x^2 + \frac{x^2}{2} = \frac{3x^2}{2} \]
Do the outer integral in x
Why: Integrate three halves x squared from zero to two.
\[ \int_0^2 \frac{3x^2}{2}\,dx = \frac{3}{2}\cdot\frac{8}{3} = 4 \]
Verify the size is reasonable
Why: The integrand ranges from zero to four over a triangle of area two, so a value of four sits comfortably below the maximum height times the area, which is eight. The answer four is plausible.
Concept
A Type II region flips the roles: y runs between two constants, while x is trapped between a left curve and a right curve that depend on y.
\[ R = \{\,(x,y) : c \le y \le d,\; h_1(y) \le x \le h_2(y)\,\} \]
Read it as: pick a y, then x sweeps from the left boundary across to the right boundary.
Intuition
Now sweep a thin horizontal strip up the region from bottom to top.
Figure (svg): A region with a horizontal strip running from a left boundary curve to a right boundary curve, and y running bottom to top
The inner integral in x adds up the strip from its left edge to its right edge. The outer integral in y slides the strip upward.
Concept
Here the order flips: dx on the inside and dy on the outside.
\[ \iint_R f\,dA = \int_c^d \!\! \int_{h_1(y)}^{h_2(y)} f(x,y)\,dx\,dy \]
The inner limits are the left and right boundary curves written as functions of y; the outer limits are the constant range of y.
Ranking
Put in order
Put the moves of Evaluate over a Type II triangle 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. Type II means dx inside with the curve limits, dy outside with the constant limits.
Worked example
Integrate over the triangle where y runs from zero to one and x runs from the y-axis to the line x equals y.
\[ \iint_R 6xy\,dA, \quad 0 \le y \le 1,\; 0 \le x \le y \]
Write the iterated integral
Why: Type II means dx inside with the curve limits, dy outside with the constant limits.
\[ = \int_0^1 \!\! \int_0^y 6xy\,dx\,dy \]
Do the inner integral in x
Why: Treat y as constant; integrate six x y in x to get three x squared y, then evaluate from zero to y.
\[ \int_0^y 6xy\,dx = 3x^2 y \Big|_0^y = 3y^3 \]
Do the outer integral in y
Why: Integrate three y cubed from zero to one.
\[ \int_0^1 3y^3\,dy = \frac{3}{4} \]
Verify with the reversed order
Why: As a Type I region the same triangle is x from zero to one, y from x to one; integrating six x y that way also gives three quarters, confirming the result.
Concept
Many regions can be described either way. Choose the order whose boundary curves are simpler and whose inner integral you can actually do.
Sketch first. If the top and bottom boundaries are single clean curves in x, go Type I. If the left and right boundaries are single clean curves in y, go Type II.
Sometimes one order is impossible in closed form and the other is easy. When stuck, try reversing the order.
Worked example
Set up the area of the triangle with corners at the origin, at two on the x-axis, and at four on the y-axis, as a Type I integral.
Figure (svg): A triangle with vertices at the origin, at (2,0), and at (0,4), with the slanted side connecting (2,0) to (0,4)
Find the slanted boundary line
Why: The line through the point two on the x-axis and four on the y-axis has intercept form, which rearranges to the top boundary for y.
\[ \frac{x}{2} + \frac{y}{4} = 1 \;\Longrightarrow\; y = 4 - 2x \]
Read off the Type I limits
Why: For each x from zero to two, y runs from the x-axis up to the slanted line.
\[ \text{Area} = \int_0^2 \!\! \int_0^{\,4-2x} 1\,dy\,dx \]
Evaluate
Why: The inner integral gives four minus two x; integrate that from zero to two.
\[ = \int_0^2 (4 - 2x)\,dx = \big[\,4x - x^2\,\big]_0^2 = 4 \]
Verify with the triangle area formula
Why: One half base times height is one half times two times four, which equals four, matching the integral.
Notation
Annotate
From Set up limits from a sketch — read this one piece at a time. What is each part doing?
On: \( \frac{x}{2} + \frac{y}{4} = 1 \;\Longrightarrow\; y = 4 - 2x \)
Step zero
Discussion prompt
Reversing the order of integration — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Describe the region from the given limits
Answer:
Worked example
Rewrite this integral with the order reversed, then evaluate.
\[ \int_0^1 \!\! \int_x^1 2y\,dy\,dx \]
Describe the region from the given limits
Why: Here x runs from zero to one, and for each x, y runs from the line y equals x up to y equals one. That is the triangle above the diagonal.
\[ 0 \le x \le 1, \quad x \le y \le 1 \]
Re-read the region with y on the outside
Why: For a fixed y between zero and one, x runs from the y-axis across to the diagonal x equals y.
\[ 0 \le y \le 1, \quad 0 \le x \le y \]
Evaluate the reversed integral
Why: The inner integral in x of two y is two y times y; integrating two y squared from zero to one gives two thirds.
\[ \int_0^1 \!\! \int_0^y 2y\,dx\,dy = \int_0^1 2y^2\,dy = \frac{2}{3} \]
Verify against the original order
Why: The original inner integral gives one minus x squared; integrating that from zero to one gives one minus one third, which is two thirds. The two orders match.
Picture it
Animation
Shows: Each line of the worked example "Reversing the order of integration", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The original inner integral gives one minus x squared; integrating that from zero to one gives one minus one third, which is two thirds. The two orders match.
Trap
Given the integral over the region below the diagonal, a student just moves the numbers straight across when switching the order, keeping the same limit values.
\[ \int_0^1 \!\! \int_0^x f\,dy\,dx \;\; \xrightarrow{\text{wrong}} \;\; \int_0^1 \!\! \int_0^x f\,dx\,dy \]
The variable x now appears in both an inner limit and an outer role, which is impossible. Copying limits is not reversing the order.
Sketch the region, then re-derive the new limits from the picture. Below the diagonal means y from zero to x, which becomes x from y to one.
\[ \int_0^1 \!\! \int_0^x f\,dy\,dx = \int_0^1 \!\! \int_y^1 f\,dx\,dy \]
Always redraw. The new limits come from the geometry, never from shuffling symbols.
Break the constraint
Discussion prompt
The rule this trap just fixed:
Always redraw. The new limits come from the geometry, never from shuffling symbols.
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
When the surface stays above the ground, the volume under it over any region is just the double integral of the height.
\[ V = \iint_R f(x,y)\,dA, \qquad f \ge 0 \text{ on } R \]
The only new work over a general region is finding the correct Type I or Type II limits.
Hypothesis
Predict first
Volume under a plane over a square 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: Confirm the surface stays above the floor
Why: On the square, the smallest value of four minus x minus y is at the far corner, giving four minus one minus one, which is two, still positive. So the integral is a genuine volume.
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 volume of the solid under the plane below and above the unit square.
\[ z = 4 - x - y, \qquad R = [0,1]\times[0,1] \]
Confirm the surface stays above the floor
Why: On the square, the smallest value of four minus x minus y is at the far corner, giving four minus one minus one, which is two, still positive. So the integral is a genuine volume.
Set up and do the inner integral in y
Why: Integrate four minus x minus y in y from zero to one, treating x as constant.
\[ \int_0^1 (4 - x - y)\,dy = 4 - x - \tfrac{1}{2} = \tfrac{7}{2} - x \]
Do the outer integral in x
Why: Integrate seven halves minus x from zero to one.
\[ \int_0^1 \left( \tfrac{7}{2} - x \right) dx = \tfrac{7}{2} - \tfrac{1}{2} = 3 \]
Verify with the center-height shortcut
Why: A plane over a square has average height equal to its value at the center. At the point one half, one half the height is four minus one, which is three, times the area one gives three. It matches.
Concept
When a solid is capped by an upper surface and floored by a lower surface, its height at each point is the top value minus the bottom value.
\[ V = \iint_R \big[\, z_{\text{top}}(x,y) - z_{\text{bottom}}(x,y) \,\big]\,dA \]
The integrand is the gap between the two surfaces, measured straight up.
Intuition
Picture a single vertical toothpick standing in the solid. It starts on the lower surface and ends on the upper surface.
Figure (svg): Two stacked curves with a vertical segment between them, the top curve above the bottom curve, showing the height as the gap
The length of that toothpick is top minus bottom. Adding up all the toothpicks over the region is the volume.
Socratic
Discussion prompt
Picture a single vertical toothpick standing in the solid. It starts on the lower surface and ends on the upper surface.
Suppose that were not true. What is the first thing in Week 9 - Double Integrals & Volume that would stop working?
Hint: Follow it one step downstream. The answer is whatever was quietly relying on it.
Answer:
The length of that toothpick is top minus bottom. Adding up all the toothpicks over the region is the volume.
Estimation
Predict first
Find the volume between the upper plane and the lower plane over the unit square.
Commit before you compute: what does Volume between two surfaces 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 linearity
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. The integral of four is four, of two x is one, and of y is one half, over the unit square.
Worked example
Find the volume between the upper plane and the lower plane over the unit square.
\[ z_{\text{top}} = 4 - x, \quad z_{\text{bottom}} = x + y, \quad R = [0,1]\times[0,1] \]
Confirm the top really is on top
Why: At the corner one, one the top is three and the bottom is two, and the gap only grows toward the origin, so the top surface stays above the bottom on the whole square.
Form the height integrand
Why: Subtract bottom from top: four minus x minus the quantity x plus y.
\[ z_{\text{top}} - z_{\text{bottom}} = (4 - x) - (x + y) = 4 - 2x - y \]
Integrate over the square
Why: Inner integral in y of four minus two x minus y from zero to one gives seven halves minus two x; then integrate in x.
\[ \int_0^1 \!\! \int_0^1 (4 - 2x - y)\,dy\,dx = \int_0^1 \left( \tfrac{7}{2} - 2x \right) dx = \tfrac{7}{2} - 1 = \tfrac{5}{2} \]
Verify by linearity
Why: The integral of four is four, of two x is one, and of y is one half, over the unit square. Then four minus one minus one half equals five halves, matching.
Picture it
Animation
Shows: Each line of the worked example "Volume between two surfaces", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The integral of four is four, of two x is one, and of y is one half, over the unit square. Then four minus one minus one half equals five halves, matching.
Trap
For the volume between the top and bottom planes above, a student integrates only the top surface and calls it the volume.
\[ \iint_R (4 - x)\,dA = 4 - \tfrac{1}{2} = \tfrac{7}{2} \]
This measures the volume under the top surface down to the ground, not the volume of the slab between the two surfaces. It is too big.
The height of the solid is the gap between surfaces, so subtract the bottom before integrating.
\[ \iint_R \big[(4 - x) - (x + y)\big]\,dA = \tfrac{5}{2} \]
The difference between seven halves and five halves is exactly the volume under the bottom surface that was wrongly left in.
Concept
If you want true volume but the surface crosses below the ground, the plain double integral counts the below-ground part as negative and undercounts the material.
\[ \text{true volume} = \iint_R |f(x,y)|\,dA \;\ne\; \iint_R f(x,y)\,dA \text{ (in general)} \]
To get true volume, split the region where the surface changes sign, or integrate the absolute value.
Explain it
Discussion prompt
Explain Sign when the surface dips below the plane 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:
To get true volume, split the region where the surface changes sign, or integrate the absolute value.
Trap
Asked for the volume between the surface and the ground over the strip, a student integrates the raw function.
\[ f(x,y) = x - 1, \quad R = [0,2]\times[0,1] \]
\[ \iint_R (x - 1)\,dA = \left( \int_0^2 (x-1)\,dx \right)(1) = 0 \]
They report zero volume, which is nonsense for a real solid: the piece where x is below one sat under the ground and cancelled the piece above.
The surface is below the ground for x less than one and above it for x greater than one. Integrate the absolute value, splitting at x equals one.
\[ \iint_R |x-1|\,dA = \int_0^2 |x-1|\,dx = \tfrac{1}{2} + \tfrac{1}{2} = 1 \]
The true volume is one. The raw integral of zero was only the net signed value.
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
The average value of a function over a region is its integral divided by the area of the region.
\[ f_{\text{avg}} = \frac{1}{\text{area of } R}\iint_R f(x,y)\,dA \]
It is the single flat height that would enclose the same volume over the same floor.
Analogy
Discussion prompt
Explain Average value over a region 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 average value of a function over a region is its integral divided by the area of the region.
Intuition
Imagine the solid under the surface is made of sand. Level the sand flat over the region. The height it settles to is the average value.
That is why you divide by the area: total volume spread evenly over the floor gives the level height.
Counterexample
Discussion prompt
Imagine the solid under the surface is made of sand. Level the sand flat over the region. The height it settles to is the average value.
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.
Step zero
Discussion prompt
Average value over a rectangle — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Recall the integral
Answer:
Worked example
Find the average value of the bowl surface over the square from the estimate example.
\[ f(x,y) = x^2 + y^2, \quad R = [0,2]\times[0,2] \]
Recall the integral
Why: We already computed the double integral of x squared plus y squared over this square.
\[ \iint_R (x^2+y^2)\,dA = \frac{32}{3} \]
Divide by the area of the square
Why: The square has area four; divide the integral by four.
\[ f_{\text{avg}} = \frac{1}{4}\cdot\frac{32}{3} = \frac{8}{3} \]
Verify the average is between the extremes
Why: On this square the function ranges from zero at the origin to eight at the far corner. The average eight thirds, about 2.67, lies between zero and eight, as any average must.
Picture it
Animation
Shows: Each line of the worked example "Average value over a rectangle", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: On this square the function ranges from zero at the origin to eight at the far corner. The average eight thirds, about 2.67, lies between zero and eight, as any average must.
Ranking
Put in order
Put the moves of Average value over a triangle 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. Inner integral in y of x from zero to x gives x times x; then integrate x squared from zero to one.
Worked example
Find the average value of the function that returns x over the triangle where x runs from zero to one and y runs from the x-axis up to y equals x.
\[ f(x,y) = x, \quad 0 \le x \le 1,\; 0 \le y \le x \]
Integrate x over the region
Why: Inner integral in y of x from zero to x gives x times x; then integrate x squared from zero to one.
\[ \int_0^1 \!\! \int_0^x x\,dy\,dx = \int_0^1 x^2\,dx = \frac{1}{3} \]
Find the area of the triangle
Why: The triangle has base one and height one, so its area is one half.
\[ \text{area} = \int_0^1 \!\! \int_0^x 1\,dy\,dx = \int_0^1 x\,dx = \frac{1}{2} \]
Divide integral by area
Why: Average is one third divided by one half.
\[ f_{\text{avg}} = \frac{1/3}{1/2} = \frac{2}{3} \]
Verify the average is reasonable
Why: On this triangle x ranges from zero to one, and more of the region sits at larger x, so an average of two thirds pulled toward the high end is sensible.
Concept
For an L-shape or any region that is not a single clean Type I or Type II piece, cut it into pieces you can handle and add the results.
You can also compute over a big simple region and subtract a piece you do not want. Both are just additivity.
Worked example
Find the area of the L-shape: the two-by-two square with the top-right unit square removed.
Figure (svg): A two by two square with the top-right unit square cut out, forming an L-shape
Take the full square minus the missing corner
Why: Additivity lets us subtract the removed unit square from the full two-by-two square instead of integrating the L directly.
\[ \text{area}(L) = \iint_{\text{big}} 1\,dA - \iint_{\text{corner}} 1\,dA \]
Compute each area
Why: The full square has area four; the removed corner is a unit square with area one.
\[ = 4 - 1 = 3 \]
Verify by splitting into two rectangles instead
Why: The L splits into a two-by-one bottom strip of area two and a one-by-one block of area one, giving two plus one, which is three. Both methods agree.
Notation
Annotate
From Area of an L-shape by additivity — read this one piece at a time. What is each part doing?
On: \( \text{area}(L) = \iint_{\text{big}} 1\,dA - \iint_{\text{corner}} 1\,dA \)
Explain it to yourself
Discussion prompt
In Recipe: set up any double integral this move is made:
5. Sanity-check the size
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:
Compare the result against smallest-value-times-area and largest-value-times-area, or against a rough geometric estimate.
Pattern
1. Sketch the region
Why: Almost every mistake in double integrals is a limits mistake. A picture is the cure.
2. Choose Type I or Type II
Why: Decide whether vertical strips (dy inside) or horizontal strips (dx inside) give simpler boundary curves.
3. Write inner limits as curves, outer limits as constants
Why: The outer variable may never appear in the outer limits, or the answer will not be a number.
4. Integrate inside first, then outside
Why: The inner integral collapses one variable and leaves a single-variable function for the outer integral.
5. Sanity-check the size
Why: Compare the result against smallest-value-times-area and largest-value-times-area, or against a rough geometric estimate.
Explain it to yourself
Discussion prompt
In Recipe: volume between two surfaces this move is made:
3. Find the region of overlap and its limits
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:
The base region is usually where the two surfaces intersect or a given boundary; set Type I or Type II limits from it.
Pattern
1. Identify which surface is on top
Why: Test a point in the region; the larger z value is the top. If they cross, split the region where they meet.
2. Form the integrand as top minus bottom
Why: The height of the solid at each point is the gap between the surfaces, measured straight up.
3. Find the region of overlap and its limits
Why: The base region is usually where the two surfaces intersect or a given boundary; set Type I or Type II limits from it.
4. Integrate the gap over the region
Why: The result is always nonnegative when you subtract in the right order, which is a built-in check.
Real world
Discussion prompt
Outside this lesson: where does Week 9 - Double Integrals & Volume 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 Recipe: volume between two surfaces 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 builds the double integral from Riemann sums up to volume under a surface. It covers the properties of the integral, Fubini's theorem in either order, the volume between two surfaces as top minus bottom, average value, and setting up Type I and Type II regions. It targets the classic traps: forgetting top minus bottom, blindly swapping the limits over a region that is not a rectangle, and confusing a signed integral with a true volume.
Elimination
Eliminate the wrong options
What does the double integral of f over R represent?
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: With a nonnegative height, each column contributes height times base area, and the limit of these columns is exactly the volume of the solid trapped between the surface and the region below it.
Check
Suppose the surface never dips below the ground over the region.
\[ f(x,y) \ge 0 \text{ on } R \]
Check your understanding
What does the double integral of f over R represent?
Answer: A
Why: With a nonnegative height, each column contributes height times base area, and the limit of these columns is exactly the volume of the solid trapped between the surface and the region below it.
Check
Evaluate the double integral over the rectangle.
\[ \int_0^1 \!\! \int_0^2 6x^2 y\,dy\,dx \]
Check your understanding
What is the value of this integral?
Answer: A
Why: The inner integral in y of six x squared y from zero to two is six x squared times two, which is twelve x squared. Integrating twelve x squared in x from zero to one gives four.
Prediction
Predict first
Which reversed integral describes the same region?
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: The integral with y from 0 to 1 and x from y to 1
Why: The region is below the diagonal: y runs from zero up to x. For a fixed y, x runs from the line x equals y across to x equals one, and y ranges from zero to one, giving y outer, x from y to one.
Check
Consider the region for this integral, then reverse the order to dx dy.
\[ \int_0^1 \!\! \int_0^x f(x,y)\,dy\,dx \]
Check your understanding
Which reversed integral describes the same region?
Answer: A
Why: The region is below the diagonal: y runs from zero up to x. For a fixed y, x runs from the line x equals y across to x equals one, and y ranges from zero to one, giving y outer, x from y to one.
Check
Find the volume of the solid between the top plane and the bottom plane over the unit square.
\[ z_{\text{top}} = 4, \quad z_{\text{bottom}} = x + y, \quad R = [0,1]\times[0,1] \]
Check your understanding
What is the volume of this solid?
Answer: A
Why: The height is the top minus the bottom, four minus the quantity x plus y. Integrating four minus x minus y over the unit square gives four minus one half minus one half, which is three.
Check
Find the average value of the function over the square.
\[ f(x,y) = x + y, \quad R = [0,2]\times[0,2] \]
Check your understanding
What is the average value of f over R?
Answer: A
Why: The integral of x plus y over the square is eight, and the area of the square is four, so the average is eight divided by four, which is two.
Elimination
Eliminate the wrong options
What is the value of the raw double integral?
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: The integral of x minus one over x from zero to two is zero, times the y-length one, so the signed integral is zero: the part below the ground for x under one cancels the part above for x over one.
Check
Evaluate the raw double integral over the strip.
\[ \iint_R (x - 1)\,dA, \quad R = [0,2]\times[0,1] \]
Check your understanding
What is the value of the raw double integral?
Answer: A
Why: The integral of x minus one over x from zero to two is zero, times the y-length one, so the signed integral is zero: the part below the ground for x under one cancels the part above for x over one.
Connect it up
Draw it
One page, no notation unless you need it: draw how these connect — Recipe: set up any double integral · Recipe: volume between two surfaces · Start from what you know: the single integral · From area to volume · Partition the region into cells. Put an arrow wherever one of them is what makes another possible, and label the arrow with why.
Recap
The double integral is the limit of Riemann sums, and when the height is nonnegative it is the volume under the surface.
Fubini lets you iterate in either order over a rectangle; over a general region you must re-derive the limits from a sketch as Type I or Type II.
For volume between surfaces, integrate top minus bottom. For true volume when the surface dips below, use the absolute value. For average value, divide the integral by the area.
| Task | Set-up |
|---|---|
| Volume under a surface | integrate f over R (f nonnegative) |
| Volume between surfaces | integrate top minus bottom over R |
| Area of a region | integrate the constant one over R |
| Average value | integral over R divided by area of R |
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