This 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.
Subject: Calculus III · 107 slides · symbolic lesson
Open the interactive version of this deck · Homework for this lesson
Objectives
By the end of this deck you can:
1. Integrate a two-variable expression with respect to one variable while holding the other constant.
2. Evaluate an iterated integral from the inside out, with both constant and variable limits.
3. Set up the limits so that the outer limits are always constants.
4. Write the area of a plane region as a double iterated integral of 1.
5. Read limits from a sketch using vertical or horizontal strips, and reverse the order of integration correctly.
Warm-up
Discussion prompt
Before we open Week 8 - Iterated Integrals & Area: without looking back, what was the main idea of Week 7 - Gradient, Extrema & Lagrange Multipliers, 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 the gradient and directional derivatives, steepest ascent, tangent planes and normal lines, relative and absolute extrema with the Second Partials Test, and constrained optimization by Lagrange multipliers. It targets the classic traps: using a direction vector that is not a unit vector, misreading the D-test, forgetting the boundary, and dropping the constraint equation.
Concept
A double integral is not solved all at once. You integrate with respect to one variable first, then with respect to the other.
The inner integral treats the other variable as a frozen constant, exactly the way a partial derivative freezes one variable.
\[ \int_{a}^{b}\!\!\int_{c}^{d} f(x,y)\,dy\,dx \]
Counterexample
Discussion prompt
A double integral is not solved all at once. You integrate with respect to one variable first, then with respect to the other.
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 inner integral treats the other variable as a frozen constant, exactly the way a partial derivative freezes one variable.
Intuition
You already know how to hold a variable constant: that is what a partial derivative does. Partial integration simply undoes that.
If differentiating the following with respect to x (treating y as a constant) gives the integrand, then that is the partial antiderivative.
\[ \frac{\partial}{\partial x}\left(x^2 y\right) = 2xy \quad\Longrightarrow\quad \int 2xy \,dx = x^2 y + C(y) \]
The constant of integration is not a number here: it can be any function of the frozen variable.
Analogy
Discussion prompt
Explain It is a partial derivative run in reverse 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:
You already know how to hold a variable constant: that is what a partial derivative does. Partial integration simply undoes that.
Concept
To integrate with respect to x, treat every y as if it were a fixed number and integrate the x-powers as usual.
partial integration — Integrating an expression in two variables with respect to one of them while the other is held constant. The result still depends on the held variable.
Explain it
Discussion prompt
Explain Partial integration with respect to x 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 integrate with respect to x, treat every y as if it were a fixed number and integrate the x-powers as usual.
Ranking
Put in order
Put the moves of Worked example: integrate x squared times y over x 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. y does not change as x varies, so it is a constant multiplier for this integration.
Worked example
Evaluate the following, integrating with respect to x while y stays constant.
\[ \int_{1}^{2} x^2 y \, dx \]
Pull the constant y outside
Why: y does not change as x varies, so it is a constant multiplier for this integration.
\[ y\int_{1}^{2} x^2 \, dx \]
Antidifferentiate x squared
Why: The power rule for x: raise the exponent by one and divide by the new exponent.
\[ y\left[\frac{x^3}{3}\right]_{1}^{2} \]
Substitute the x-limits
Why: Evaluate at the top limit minus the bottom limit; y rides along untouched.
\[ y\left(\frac{8}{3}-\frac{1}{3}\right) = \frac{7y}{3} \]
Verify by differentiating back
Why: The partial derivative with respect to x of the antiderivative must return the integrand.
\[ \frac{\partial}{\partial x}\left(\frac{x^3 y}{3}\right) = x^2 y \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "integrate x squared times y over x", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The partial derivative with respect to x of the antiderivative must return the integrand.
Concept
The same idea works the other way: to integrate with respect to y, freeze x and integrate the y-powers.
Whatever variable you are not integrating over behaves like a plain constant coefficient.
Step zero
Discussion prompt
Worked example: integrate six x squared y over y — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Pull the constant part outside
Answer:
Worked example
Evaluate the following, integrating with respect to y while x stays constant.
\[ \int_{0}^{1} 6x^2 y \, dy \]
Pull the constant part outside
Why: The factor six x squared does not depend on y, so it is a constant for this integration.
\[ 6x^2\int_{0}^{1} y \, dy \]
Antidifferentiate y
Why: The power rule gives one half y squared.
\[ 6x^2\left[\frac{y^2}{2}\right]_{0}^{1} \]
Substitute the y-limits
Why: At the top y equals one, at the bottom y equals zero.
\[ 6x^2\left(\frac{1}{2}-0\right) = 3x^2 \]
Verify by differentiating back
Why: The partial derivative with respect to y of three x squared y squared should return the integrand.
\[ \frac{\partial}{\partial y}\left(3x^2 y^2\right) = 6x^2 y \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "integrate six x squared y over y", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The partial derivative with respect to y of three x squared y squared should return the integrand.
Concept
Stacking two integral signs means: do the inner integral first, then feed its result into the outer integral.
\[ \int_{a}^{b}\left[\int_{c}^{d} f(x,y)\,dy\right]dx \]
The bracket is usually dropped, but it is always there in spirit. The innermost differential is paired with the innermost integral sign.
Intuition
Think of the two integrals as nested boxes. You cannot open the outer box until the inner box is completely resolved into a single expression.
After the inner integral is done, its variable is gone: only the outer variable survives, and then the outer integral finishes the job into a number.
Concept
The order of the differentials tells you which limits belong to which variable. Read from the inside out.
\[ \int_{\text{outer }x}\;\int_{\text{inner }y}\; f\,\underbrace{dy}_{\text{inner}}\;\underbrace{dx}_{\text{outer}} \]
The differential closest to the integrand is the first one you integrate, and it is paired with the limits closest to the integrand.
Estimation
Predict first
Evaluate the iterated integral over a rectangle.
Commit before you compute: what does Worked example: constant limits, x then simplify 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 reversing the order
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. Over a rectangle with constant limits, integrating in the opposite order must give the same number.
Worked example
Evaluate the iterated integral over a rectangle.
\[ \int_{0}^{1}\!\!\int_{0}^{2} \left(x + 3y^2\right)\,dx\,dy \]
Integrate the inner integral with respect to x
Why: The differential dx is innermost, so x is the inner variable; treat y as constant.
\[ \int_{0}^{2}\left(x+3y^2\right)dx = \left[\frac{x^2}{2}+3y^2 x\right]_{0}^{2} \]
Substitute the x-limits
Why: At x equals two the bracket gives two plus six y squared; at x equals zero it gives zero.
\[ = \frac{4}{2} + 3y^2(2) = 2 + 6y^2 \]
Integrate the result with respect to y
Why: The inner variable is gone; now finish the outer integral over y from zero to one.
\[ \int_{0}^{1}\left(2+6y^2\right)dy = \left[2y + 2y^3\right]_{0}^{1} \]
Substitute the y-limits
Why: At y equals one we get two plus two; at y equals zero we get zero.
\[ = 2 + 2 = 4 \]
Verify by reversing the order
Why: Over a rectangle with constant limits, integrating in the opposite order must give the same number.
\[ \int_{0}^{2}\!\!\int_{0}^{1}\!\left(x+3y^2\right)dy\,dx = \int_{0}^{2}(x+1)\,dx = \left[\tfrac{x^2}{2}+x\right]_0^2 = 4 \;\checkmark \]
Concept
When every limit is a plain number, the region is a rectangle. But most regions are not rectangles.
For a non-rectangular region, the inner limits are allowed to depend on the outer variable. That is how the shape bends.
\[ \int_{a}^{b}\!\!\int_{g_1(x)}^{g_2(x)} f(x,y)\,dy\,dx \]
Intuition
Picture a single vertical strip standing at some x. Its bottom and top can slide up or down as you move the strip left or right.
That is what a variable inner limit records: the floor and ceiling of the strip, written as functions of the outer variable.
Figure (svg): A curved region with one vertical strip whose top follows an upper curve and whose bottom follows a lower curve.
Missing information
Discussion prompt
Evaluate the iterated integral where the inner limit depends on x.
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:
The dy is innermost; treat x as constant and integrate y from zero up to x.
Worked example
Evaluate the iterated integral where the inner limit depends on x.
\[ \int_{0}^{1}\!\!\int_{0}^{x} \left(x + y\right)\,dy\,dx \]
Integrate the inner integral with respect to y
Why: The dy is innermost; treat x as constant and integrate y from zero up to x.
\[ \int_{0}^{x}(x+y)\,dy = \left[xy + \frac{y^2}{2}\right]_{0}^{x} \]
Substitute the variable limit y equals x
Why: The upper limit is x itself, so replace every y with x; the lower limit zero contributes nothing.
\[ = x\cdot x + \frac{x^2}{2} = x^2 + \frac{x^2}{2} = \frac{3x^2}{2} \]
Integrate the result with respect to x
Why: Now a clean single-variable integral over x from zero to one remains.
\[ \int_{0}^{1}\frac{3x^2}{2}\,dx = \frac{3}{2}\left[\frac{x^3}{3}\right]_{0}^{1} = \frac{3}{2}\cdot\frac{1}{3} \]
Verify the arithmetic
Why: Multiply the fractions and confirm the final value is a single number.
\[ \frac{3}{2}\cdot\frac{1}{3} = \frac{1}{2} \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "variable inner limit, y then x", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Multiply the fractions and confirm the final value is a single number.
Concept
The inner limits may contain the outer variable. The outer limits may not contain any variable at all.
The reason is simple: after both integrals are done, the answer must be a plain number. If a variable survived in the outer limits, the answer would still depend on it.
\[ \int_{\text{constants}}^{\text{constants}}\!\!\int_{\text{may use outer var}}^{\text{may use outer var}} f\,dy\,dx \]
Intuition
The inner integral is done first, so the outer variable still exists when the inner limits are applied. That is why the inner limits may use it.
The outer integral is done last, when no other variable is left to lean on. So its limits have nothing to depend on but numbers.
Trap
It is tempting to write both limits in terms of x when the region looks triangular.
\[ \int_{0}^{x}\!\!\int_{0}^{x} f(x,y)\,dy\,dx \quad\text{(wrong)} \]
This is wrong. The outer sign carries dx, yet its upper limit is x. After finishing, the answer would still contain x, which is not a number.
\[ \text{result} = F(x) \quad\text{not a number} \]
The right setup puts constants on the outer sign and lets only the inner limit use x.
\[ \int_{0}^{1}\!\!\int_{0}^{x} f(x,y)\,dy\,dx \quad\text{(right)} \]
Now the outer limits zero and one are numbers, so the final value is a genuine number.
\[ \text{result} = \text{a real number} \;\checkmark \]
Fill the middle
Fill in the blanks
From Worked example: x outer, variable inner in x — finish the line. Write what belongs on the right of the equals sign before you look.
\int_\left[\frac{y^3}{3}\right]_{1}^{2}^___ y^2\,dy = ___
Why: Producing the right-hand side unprompted is the difference between recognising this line and being able to use it. The dx is innermost; treat y as constant and integrate two x from zero up to y.
Worked example
Evaluate the iterated integral with y as the outer variable and an inner limit that depends on y.
\[ \int_{1}^{2}\!\!\int_{0}^{y} 2x \, dx\,dy \]
Integrate the inner integral with respect to x
Why: The dx is innermost; treat y as constant and integrate two x from zero up to y.
\[ \int_{0}^{y} 2x\,dx = \left[x^2\right]_{0}^{y} = y^2 \]
Integrate the result with respect to y
Why: The inner variable is gone; a single integral in y from one to two remains.
\[ \int_{1}^{2} y^2\,dy = \left[\frac{y^3}{3}\right]_{1}^{2} \]
Substitute the y-limits
Why: At y equals two we get eight thirds; at y equals one we get one third.
\[ = \frac{8}{3} - \frac{1}{3} = \frac{7}{3} \]
Verify the inner result quickly
Why: Differentiating x squared with respect to x returns two x, confirming the inner antiderivative and the value seven thirds.
\[ \frac{d}{dx}\left(x^2\right) = 2x \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "x outer, variable inner in x", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Differentiating x squared with respect to x returns two x, confirming the inner antiderivative and the value seven thirds.
Concept
After the inner integral, its variable has been integrated away by the limits. What remains is a single-variable expression in the outer variable.
That leftover expression is exactly what the outer integral then works on, just an ordinary single-variable integral.
Step zero
Discussion prompt
Worked example: squared upper limit — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Integrate the inner integral with respect to y
Answer:
Worked example
Evaluate the iterated integral where the inner upper limit is x squared.
\[ \int_{0}^{1}\!\!\int_{0}^{x^2} 2y \, dy\,dx \]
Integrate the inner integral with respect to y
Why: The dy is innermost; the antiderivative of two y is y squared.
\[ \int_{0}^{x^2} 2y\,dy = \left[y^2\right]_{0}^{x^2} \]
Substitute the variable limit y equals x squared
Why: Square the upper limit: x squared squared is x to the fourth.
\[ = \left(x^2\right)^2 = x^4 \]
Integrate the result with respect to x
Why: Integrate x to the fourth over x from zero to one.
\[ \int_{0}^{1} x^4 \, dx = \left[\frac{x^5}{5}\right]_{0}^{1} = \frac{1}{5} \]
Verify the exponent step
Why: A common slip is to write x squared instead of x to the fourth; confirm the square of x squared really is x to the fourth.
\[ \left(x^2\right)^2 = x^{2\cdot 2} = x^4 \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "squared upper limit", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: A common slip is to write x squared instead of x to the fourth; confirm the square of x squared really is x to the fourth.
Trap
Here the inner differential is dx, yet the inner limits are written in terms of x. The variable of integration cannot appear in its own limits.
\[ \int_{0}^{1}\!\!\int_{0}^{x} f \, dx\,dy \quad\text{(wrong)} \]
Integrating over x from zero to x is nonsense: x is both the thing you sweep and the endpoint you sweep to.
Match the differential to its limits. If the inner limit uses x, then the inner variable must be y, so the inner differential is dy.
\[ \int_{0}^{1}\!\!\int_{0}^{x} f \, dy\,dx \quad\text{(right)} \]
Now x is the outer variable that the inner limit is allowed to reference, and dy sweeps the inner direction.
Break the constraint
Discussion prompt
The rule this trap just fixed:
Match the differential to its limits. If the inner limit uses x, then the inner variable must be y, so the inner differential is dy.
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
If you integrate the constant 1 over a region, each tiny piece contributes its own tiny area, and the total is the area of the region.
\[ \text{Area} = \iint_{R} 1 \, dA \]
Written as an iterated integral, the integrand is simply 1; all the information about the shape lives in the limits.
Intuition
Imagine covering the region with a grid of tiny rectangles. Each has a little area, the product of its width and height.
Adding up all those little areas and letting them shrink to zero gives the exact area of the region. That sum is what the double integral computes.
\[ dA = dy\,dx \quad\text{or}\quad dA = dx\,dy \]
Concept
In a vertical strip, x is held while y runs from the bottom curve to the top curve. So the inner integral is with respect to y, and dy is innermost.
Then x sweeps across the region from its smallest constant value to its largest.
\[ \text{Area} = \int_{a}^{b}\!\!\int_{g_1(x)}^{g_2(x)} dy\,dx \]
Picture it
Figure (svg): Several vertical strips filling a region bounded by a straight line above and a parabola below, sweeping from 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:
Build one thin vertical strip at position x: its height is the top curve minus the bottom curve. The inner integral of 1 over y gives exactly that height.
Intuition
Build one thin vertical strip at position x: its height is the top curve minus the bottom curve. The inner integral of 1 over y gives exactly that height.
Then slide the strip from the left edge to the right edge. The outer integral over x adds up all the strips.
Figure (svg): Several vertical strips filling a region bounded by a straight line above and a parabola below, sweeping from left to right.
Hypothesis
Predict first
Worked example: area between a line and a parabola, vertical strips 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: Find where the curves meet
Why: Set them equal to find the x-range of the region.
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 area of the region between the line and the parabola shown.
\[ y = x \quad\text{and}\quad y = x^2 \]
Find where the curves meet
Why: Set them equal to find the x-range of the region.
\[ x = x^2 \;\Rightarrow\; x - x^2 = 0 \;\Rightarrow\; x = 0,\, 1 \]
Identify top and bottom on this interval
Why: For x between zero and one, the line lies above the parabola, so the line is the top.
\[ \text{top: } y = x, \qquad \text{bottom: } y = x^2 \]
Set up the double integral of 1 with vertical strips
Why: Inner variable y runs from the bottom curve to the top curve; outer variable x runs over the constant interval.
\[ A = \int_{0}^{1}\!\!\int_{x^2}^{x} dy\,dx \]
Do the inner integral
Why: Integrating 1 over y gives the strip height, top minus bottom.
\[ \int_{x^2}^{x} dy = x - x^2 \]
Do the outer integral
Why: Integrate the strip height over x from zero to one.
\[ \int_{0}^{1}(x - x^2)\,dx = \left[\frac{x^2}{2} - \frac{x^3}{3}\right]_{0}^{1} = \frac{1}{2} - \frac{1}{3} = \frac{1}{6} \]
Verify with the single-variable area formula
Why: The area between two curves equals the integral of top minus bottom, which must match.
\[ \int_{0}^{1}\!\left(x - x^2\right)dx = \frac{1}{6} \;\checkmark \]
Concept
In a horizontal strip, y is held while x runs from the left curve to the right curve. So the inner integral is with respect to x, and dx is innermost.
Then y sweeps up the region from its smallest constant value to its largest.
\[ \text{Area} = \int_{c}^{d}\!\!\int_{h_1(y)}^{h_2(y)} dx\,dy \]
Intuition
Build one thin horizontal strip at height y: its width is the right curve minus the left curve. The inner integral of 1 over x gives that width.
Then slide the strip from the bottom edge to the top edge. The outer integral over y adds them all up.
Figure (svg): Several horizontal strips filling the same region, sweeping from the bottom to the top.
Ranking
Put in order
Put the moves of Worked example: same region, horizontal strips 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. For horizontal strips we need x as a function of y for the left and right edges.
Worked example
Recompute the area between the line and the parabola using horizontal strips, and confirm it matches.
\[ y = x \quad\text{and}\quad y = x^2 \]
Solve each boundary for x
Why: For horizontal strips we need x as a function of y for the left and right edges.
\[ y = x \Rightarrow x = y, \qquad y = x^2 \Rightarrow x = \sqrt{y} \]
Identify left and right on this interval
Why: For y between zero and one, the line x equals y is to the left and the square-root curve is to the right.
\[ \text{left: } x = y, \qquad \text{right: } x = \sqrt{y} \]
Set up the double integral of 1 with horizontal strips
Why: Inner variable x runs from the left curve to the right curve; outer variable y runs from zero to one.
\[ A = \int_{0}^{1}\!\!\int_{y}^{\sqrt{y}} dx\,dy \]
Do the inner integral
Why: Integrating 1 over x gives the strip width, right minus left.
\[ \int_{y}^{\sqrt{y}} dx = \sqrt{y} - y \]
Do the outer integral
Why: Integrate the width over y using the power rule on the square root.
\[ \int_{0}^{1}\!\left(y^{1/2} - y\right)dy = \left[\frac{2}{3}y^{3/2} - \frac{y^2}{2}\right]_{0}^{1} = \frac{2}{3} - \frac{1}{2} \]
Verify the two orders agree
Why: A correct region gives the same area either way; this must match the one sixth from vertical strips.
\[ \frac{2}{3} - \frac{1}{2} = \frac{4}{6} - \frac{3}{6} = \frac{1}{6} \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "same region, horizontal strips", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: A correct region gives the same area either way; this must match the one sixth from vertical strips.
Concept
Both orders give the same area, but one is often much cleaner. Choose the direction whose strips touch the same two boundary curves all the way across.
If a strip would change which curve bounds it partway across, split the region or switch strip direction to avoid that.
Estimation
Predict first
Find the area of the triangle bounded by the given lines using a double integral, then check with the geometry formula.
Commit before you compute: what does Worked example: area of a triangle by double integral 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 with one half base times height
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 triangle has base one along the x-axis and height two at x equals one; the geometry must agree.
Worked example
Find the area of the triangle bounded by the given lines using a double integral, then check with the geometry formula.
\[ y = 0, \quad y = 2x, \quad x = 1 \]
Describe the region with vertical strips
Why: At each x, the strip runs from the x-axis up to the line y equals two x; x goes from zero to one.
\[ A = \int_{0}^{1}\!\!\int_{0}^{2x} dy\,dx \]
Do the inner integral
Why: Integrating 1 over y gives the strip height two x.
\[ \int_{0}^{2x} dy = 2x \]
Do the outer integral
Why: Integrate two x over x from zero to one.
\[ \int_{0}^{1} 2x\,dx = \left[x^2\right]_{0}^{1} = 1 \]
Verify with one half base times height
Why: The triangle has base one along the x-axis and height two at x equals one; the geometry must agree.
\[ \frac{1}{2}(1)(2) = 1 \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "area of a triangle by double integral", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The triangle has base one along the x-axis and height two at x equals one; the geometry must agree.
Concept
Always sketch the region first. The picture, not the algebra, tells you the limits.
For vertical strips: the inner y-limits are the bottom and top curves; the outer x-limits are the leftmost and rightmost x-values the region reaches.
For horizontal strips: the inner x-limits are the left and right curves; the outer y-limits are the lowest and highest y-values the region reaches.
Trap
Given a vertical-strip integral, it is tempting to just swap the differentials and keep the same limits.
\[ \int_{0}^{1}\!\!\int_{0}^{x} f\,dy\,dx \;\xrightarrow{\;\text{swap}\;}\; \int_{0}^{1}\!\!\int_{0}^{x} f\,dx\,dy \quad\text{(wrong)} \]
This is wrong twice over: the outer sign now carries dy but has an x in its limit, and the region described is not the same triangle at all.
Redraw the region and re-read the limits for the new strip direction. The triangle is bounded by y equals zero, y equals x, and x equals one.
\[ \text{region: } 0 \le y \le 1,\quad y \le x \le 1 \]
For horizontal strips at height y, x runs from the line x equals y to the line x equals one, and y runs from zero to one.
\[ \int_{0}^{1}\!\!\int_{y}^{1} f\,dx\,dy \quad\text{(right)} \]
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
Reversing the order means describing the same region with strips in the other direction. The region never changes; only its description does.
You cannot simply exchange the differentials and keep the limits. You must go back to the sketch and re-read the new limits.
Intuition
Two iterated integrals describe the same region if and only if the set of points they sweep is identical. The limits are just two languages for one shape.
So the safe procedure is always: sketch the region from the given limits, then read off the limits for the other order from that same sketch.
Step zero
Discussion prompt
Worked example: reverse the order for a parabolic region — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Sketch the region from the given limits
Answer:
Worked example
Reverse the order of integration and evaluate, using the constant integrand 1 so we can check the area.
\[ \int_{0}^{2}\!\!\int_{0}^{x^2} 1 \, dy\,dx \]
Sketch the region from the given limits
Why: x runs from zero to two; for each x, y runs from zero up to x squared. That is the region under the parabola.
\[ 0 \le x \le 2, \qquad 0 \le y \le x^2 \]
Solve the boundary for x to get horizontal-strip limits
Why: The condition y less than or equal to x squared becomes x greater than or equal to the square root of y.
\[ y \le x^2 \;\Longrightarrow\; \sqrt{y} \le x \le 2 \]
Find the y-range
Why: The lowest y is zero and the highest is x squared at x equals two, namely four.
\[ 0 \le y \le 4 \]
Write the reversed integral and evaluate
Why: Inner x from the square root of y to two; outer y from zero to four.
\[ \int_{0}^{4}\!\!\int_{\sqrt{y}}^{2} dx\,dy = \int_{0}^{4}\left(2-\sqrt{y}\right)dy = \left[2y - \tfrac{2}{3}y^{3/2}\right]_{0}^{4} = 8 - \tfrac{16}{3} = \tfrac{8}{3} \]
Verify against the original order
Why: The original integral also gives the area, so the two must agree.
\[ \int_{0}^{2}\!\!\int_{0}^{x^2} dy\,dx = \int_{0}^{2} x^2\,dx = \left[\tfrac{x^3}{3}\right]_0^2 = \tfrac{8}{3} \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "reverse the order for a parabolic region", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The original integral also gives the area, so the two must agree.
Concept
Sometimes the inner integral in the given order has no elementary antiderivative, so you are stuck.
Reversing the order can turn that impossible inner integral into an easy one, without changing the value of the double integral.
Fill the middle
Fill in the blanks
From Worked example: reverse to make the inner integral doable — finish the line. Write what belongs on the right of the equals sign before you look.
\int_y\sin\!\left(y^2\right)^___\sin\!\left(y^2\right)dx = ___
Why: Producing the right-hand side unprompted is the difference between recognising this line and being able to use it. x runs from zero to one; for each x, y runs from x up to one.
Worked example
Evaluate the integral. The inner integral in y cannot be done in closed form as written.
\[ \int_{0}^{1}\!\!\int_{x}^{1} \sin\!\left(y^2\right)\,dy\,dx \]
Sketch the region
Why: x runs from zero to one; for each x, y runs from x up to one. That is the triangle above the line y equals x.
\[ 0 \le x \le 1, \qquad x \le y \le 1 \]
Re-read the region for horizontal strips
Why: For each y from zero to one, x runs from zero to the line x equals y.
\[ 0 \le y \le 1, \qquad 0 \le x \le y \]
Do the new inner integral in x
Why: The sine term is constant in x, so integrating over x just multiplies it by the strip width y.
\[ \int_{0}^{y}\sin\!\left(y^2\right)dx = y\sin\!\left(y^2\right) \]
Do the outer integral with a substitution
Why: Let u equal y squared so du equals two y dy; the y factor is exactly what the substitution needs.
\[ \int_{0}^{1} y\sin\!\left(y^2\right)dy = \left[-\tfrac{1}{2}\cos\!\left(y^2\right)\right]_{0}^{1} = \frac{1-\cos 1}{2} \]
Verify the antiderivative
Why: Differentiating the antiderivative must return the integrand y sine of y squared.
\[ \frac{d}{dy}\left(-\tfrac{1}{2}\cos y^2\right) = -\tfrac{1}{2}\cdot\left(-\sin y^2\right)\cdot 2y = y\sin\!\left(y^2\right) \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "reverse to make the inner integral doable", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Differentiating the antiderivative must return the integrand y sine of y squared.
Concept
First, sketch the region straight from the given limits. Second, choose the other strip direction. Third, read the new limits off the same sketch.
Never manipulate the limits algebraically without the picture; the sketch is what keeps the region fixed.
Missing information
Discussion prompt
Reverse the order of integration and evaluate with integrand 1 to check the area.
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:
x runs from zero to one; for each x, y runs from the square root of x up to one.
Worked example
Reverse the order of integration and evaluate with integrand 1 to check the area.
\[ \int_{0}^{1}\!\!\int_{\sqrt{x}}^{1} 1 \, dy\,dx \]
Sketch the region
Why: x runs from zero to one; for each x, y runs from the square root of x up to one.
\[ 0 \le x \le 1, \qquad \sqrt{x} \le y \le 1 \]
Solve the boundary for x
Why: The condition y greater than or equal to the square root of x becomes x less than or equal to y squared.
\[ y \ge \sqrt{x} \;\Longrightarrow\; 0 \le x \le y^2 \]
Write and evaluate the reversed integral
Why: Inner x from zero to y squared; outer y from zero to one.
\[ \int_{0}^{1}\!\!\int_{0}^{y^2} dx\,dy = \int_{0}^{1} y^2\,dy = \frac{1}{3} \]
Verify against the original order
Why: The original also computes the same area, so the answers must match.
\[ \int_{0}^{1}\!\left(1-\sqrt{x}\right)dx = \left[x - \tfrac{2}{3}x^{3/2}\right]_0^1 = 1 - \tfrac{2}{3} = \tfrac{1}{3} \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "reverse a square-root region", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The original also computes the same area, so the answers must match.
Concept
If the integrand factors into a function of x times a function of y, and the limits are all constants, the double integral splits into a product of two single integrals.
\[ \int_{a}^{b}\!\!\int_{c}^{d} g(x)h(y)\,dy\,dx = \left(\int_{a}^{b} g(x)\,dx\right)\!\left(\int_{c}^{d} h(y)\,dy\right) \]
Intuition
The split works because the inner integral of h over y is a constant that can be pulled outside the outer integral in x.
If a limit depended on the outer variable, that inner result would not be constant, and the clean product would fall apart.
Estimation
Predict first
Evaluate over the rectangle by splitting into a product.
Commit before you compute: what does Worked example: a separable integral 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 without splitting
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. Do the inner integral in x directly, then the outer in y, and confirm the same value.
Worked example
Evaluate over the rectangle by splitting into a product.
\[ \int_{0}^{1}\!\!\int_{0}^{2} xy \, dx\,dy \]
Separate into two single integrals
Why: The integrand is x times y and the limits are constants, so the double integral factors.
\[ \left(\int_{0}^{2} x\,dx\right)\!\left(\int_{0}^{1} y\,dy\right) \]
Evaluate each factor
Why: Each is a basic power-rule integral.
\[ \left[\tfrac{x^2}{2}\right]_0^2 \cdot \left[\tfrac{y^2}{2}\right]_0^1 = 2 \cdot \tfrac{1}{2} \]
Multiply
Why: The product of the two factor values gives the answer.
\[ 2 \cdot \tfrac{1}{2} = 1 \]
Verify without splitting
Why: Do the inner integral in x directly, then the outer in y, and confirm the same value.
\[ \int_{0}^{1}\!\left(\int_0^2 xy\,dx\right)dy = \int_{0}^{1} 2y\,dy = 1 \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "a separable integral", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Do the inner integral in x directly, then the outer in y, and confirm the same value.
Step zero
Discussion prompt
Worked example: area between two crossing parabolas — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Find the intersections
Answer:
Worked example
Find the area of the region enclosed between the two curves.
\[ y = x^2 \quad\text{and}\quad y = 2 - x^2 \]
Find the intersections
Why: Set the curves equal to find the x-range of the enclosed region.
\[ x^2 = 2 - x^2 \;\Rightarrow\; 2x^2 = 2 \;\Rightarrow\; x = \pm 1 \]
Identify top and bottom
Why: Between the intersections the downward parabola sits above the upward one.
\[ \text{top: } 2 - x^2, \qquad \text{bottom: } x^2 \]
Set up with vertical strips and do the inner integral
Why: The inner integral of 1 over y gives the strip height, top minus bottom.
\[ A = \int_{-1}^{1}\!\!\int_{x^2}^{2-x^2} dy\,dx = \int_{-1}^{1}\left(2 - 2x^2\right)dx \]
Do the outer integral
Why: Integrate the height over x from negative one to one.
\[ \left[2x - \tfrac{2}{3}x^3\right]_{-1}^{1} = \left(2 - \tfrac{2}{3}\right) - \left(-2 + \tfrac{2}{3}\right) = \tfrac{4}{3} + \tfrac{4}{3} = \tfrac{8}{3} \]
Verify using symmetry
Why: The region is symmetric about the y-axis, so twice the right half must give the same total.
\[ 2\int_{0}^{1}\!\left(2-2x^2\right)dx = 2\left[2x - \tfrac{2}{3}x^3\right]_0^1 = 2\cdot\tfrac{4}{3} = \tfrac{8}{3} \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "area between two crossing parabolas", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The region is symmetric about the y-axis, so twice the right half must give the same total.
Concept
The old single-variable area under a curve is just the special case where the bottom of every strip is the x-axis.
\[ \int_{a}^{b}\!\!\int_{0}^{f(x)} dy\,dx = \int_{a}^{b} f(x)\,dx \]
The inner integral of 1 collapses to the height of the curve, recovering the familiar area formula.
Ranking
Put in order
Put the moves of Worked example: one more variable-limit evaluation 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. Treat x as constant; the antiderivative of six x y in y is three x y squared.
Worked example
Evaluate the iterated integral with a variable inner limit and a non-constant integrand.
\[ \int_{0}^{1}\!\!\int_{0}^{x} 6xy \, dy\,dx \]
Integrate the inner integral with respect to y
Why: Treat x as constant; the antiderivative of six x y in y is three x y squared.
\[ \int_{0}^{x} 6xy\,dy = \left[3x y^2\right]_{0}^{x} \]
Substitute the variable limit y equals x
Why: Replace y with x in the antiderivative; the lower limit gives zero.
\[ = 3x\cdot x^2 = 3x^3 \]
Integrate the result with respect to x
Why: Integrate three x cubed over x from zero to one.
\[ \int_{0}^{1} 3x^3\,dx = \left[\frac{3x^4}{4}\right]_{0}^{1} = \frac{3}{4} \]
Verify the inner antiderivative
Why: Differentiating three x y squared with respect to y returns the integrand.
\[ \frac{\partial}{\partial y}\left(3x y^2\right) = 6xy \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "one more variable-limit evaluation", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Differentiating three x y squared with respect to y returns the integrand.
Pattern
1. Do the inner integral first
Why: Integrate with respect to the innermost differential, treating the outer variable as a constant.
2. Substitute the inner limits
Why: Plug the inner limits into the inner antiderivative; a variable inner limit is substituted just like a number.
3. Do the outer integral
Why: You now have a single-variable integral in the outer variable; integrate it over the constant outer limits.
4. Check the outer limits are constants
Why: If any variable survived into the answer, an outer limit was not a constant and the setup was wrong.
Pattern
1. Sketch the region and its boundary curves
Why: The picture, not the algebra, dictates the limits.
2. Choose strip direction
Why: Vertical strips give order dy dx; horizontal strips give order dx dy. Pick the one whose strips hit the same two curves all the way across.
3. Read the inner limits as the two curves the strip spans
Why: These are the bottom and top curves (vertical) or the left and right curves (horizontal), possibly depending on the outer variable.
4. Read the outer limits as the constant extent of the region
Why: The smallest and largest value of the outer variable the region reaches; these must be numbers.
Pattern
1. Sketch the region straight from the given limits
Why: Do not touch the limits algebraically until you can see the region.
2. Switch the strip direction
Why: If it was vertical strips, describe it with horizontal strips, and vice versa.
3. Re-read both sets of limits from the same sketch
Why: Solve each boundary curve for the new inner variable; find the new constant outer range.
4. Confirm the region is unchanged
Why: Both integrals must sweep exactly the same set of points; if not, a limit is wrong.
Real world
Discussion prompt
Outside this lesson: where does Week 8 - Iterated Integrals & Area actually turn up? Name one concrete situation — a job, a piece of software someone ships, a decision somebody has to make — and say which part of Pattern: reverse the order of integration 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 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
An area must come out positive. A negative area means the top and bottom curves were swapped or the limits were out of order.
If the region is symmetric, the two strip orders and any symmetry shortcut should all agree. Disagreement means a limit is wrong.
Concept
The differentials tell you the order. Innermost differential first, and its limits are the innermost pair.
\[ dy\,dx \;\Rightarrow\; \text{integrate } y \text{ first (vertical strips)} \]
\[ dx\,dy \;\Rightarrow\; \text{integrate } x \text{ first (horizontal strips)} \]
Intuition
You can run the whole process backward: given an iterated integral, read its limits as strips and rebuild the region on paper.
This is exactly the skill you use when reversing an order, and it is the single best way to catch a limit mistake before you integrate.
Concept
The symbol for the tiny piece of area is written as one object, but in an iterated integral it becomes the product of the two differentials.
\[ dA = dy\,dx = dx\,dy \]
Both orderings describe the same little rectangle; which one you use just decides which variable you integrate first.
Explain it
Discussion prompt
Explain The area element dA to a student a year behind you. No notation, no jargon they have not met — and it still has to be true.
Hint: If your explanation needs a symbol they have never seen, you are describing the notation rather than the idea.
Answer:
The symbol for the tiny piece of area is written as one object, but in an iterated integral it becomes the product of the two differentials.
Intuition
Every hard part of this topic becomes easy once the region is drawn. The sketch fixes the strips, the strips fix the limits, and the limits fix the integral.
The students who get these wrong almost always skipped the picture and pushed symbols around instead.
Analogy
Discussion prompt
Explain Always sketch first by analogy to something with no Calculus III in it at all — a queue, a recipe, a map, a bank balance, whatever fits. Then say where your analogy breaks.
Hint: An analogy that never breaks is not an analogy, it is the same idea wearing a hat. Find the seam — that is the part that is actually new.
Answer:
Every hard part of this topic becomes easy once the region is drawn. The sketch fixes the strips, the strips fix the limits, and the limits fix the integral.
Concept
When the integrand is 1, the double integral equals the area of the region. When the integrand is some other function, the same limits still describe the same region, but the value is no longer an area.
So keep two questions separate: what region are the limits describing, and what quantity is the integrand accumulating over it.
Concept
Setting up limits and integrating one variable at a time is the whole engine behind double integrals for volume, mass, and averages.
Master the region-to-limits skill here and the later topics become bookkeeping on top of the same idea.
Counterexample
Discussion prompt
Setting up limits and integrating one variable at a time is the whole engine behind double integrals for volume, mass, and averages.
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.
Elimination
Eliminate the wrong options
What is the value of this iterated 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: Inner integral over x gives one half plus y; integrating that over y from 0 to 1 gives one half plus one half, which equals 1.
Check
Work the inner integral first, then the outer, then choose.
\[ \int_{0}^{1}\!\!\int_{0}^{1} \left(x + y\right)\,dx\,dy \]
Check your understanding
What is the value of this iterated integral?
Answer: A
Why: Inner integral over x gives one half plus y; integrating that over y from 0 to 1 gives one half plus one half, which equals 1.
Check
The inner upper limit depends on x. Do the inner integral, substitute, then finish.
\[ \int_{0}^{1}\!\!\int_{0}^{x} 2y \, dy\,dx \]
Check your understanding
What is the value of this iterated integral?
Answer: A
Why: The inner integral of 2y from 0 to x is x squared; integrating x squared from 0 to 1 gives one third.
Prediction
Predict first
Which vertical-strip setup gives the area of this triangle?
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: Outer x from 0 to 1; inner y from 0 to x
Why: A vertical strip at x runs from the x-axis up to the diagonal y equals x, and x sweeps the constant range 0 to 1, so the inner y-limits are 0 and x.
Check
Consider the triangle with corners at the origin, at one on the x-axis, and at the point where x equals one and y equals one, bounded below by the x-axis, on the right by the vertical line, and above by the diagonal.
\[ \text{bounds: } y = 0,\quad x = 1,\quad y = x \]
Check your understanding
Which vertical-strip setup gives the area of this triangle?
Answer: A
Why: A vertical strip at x runs from the x-axis up to the diagonal y equals x, and x sweeps the constant range 0 to 1, so the inner y-limits are 0 and x.
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: Outer y from 0 to 1; inner x from y to 1
Why: The region is the set where 0 is at most y is at most x is at most 1; for a fixed y, x runs from the line x equals y up to x equals 1, and y runs from 0 to 1.
Check
Reverse the order of integration for the following region, which is the triangle below the diagonal.
\[ \int_{0}^{1}\!\!\int_{0}^{x} f(x,y)\,dy\,dx \]
Check your understanding
Which reversed integral describes the same region?
Answer: A
Why: The region is the set where 0 is at most y is at most x is at most 1; for a fixed y, x runs from the line x equals y up to x equals 1, and y runs from 0 to 1.
Check
Only one of these iterated integrals is set up legally. Look at the outer limits and at whether each differential matches its limits.
Check your understanding
Which iterated integral is set up correctly?
Answer: A
Why: The outer limits 0 and 2 are constants, and with order dy dx the inner variable is y while the inner limit uses only the outer variable x, so everything is consistent.
Elimination
Eliminate the wrong options
What is the result of this partial 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: Holding x constant, the antiderivative in y is 2x times y squared; evaluating from 0 to 2 gives 2x times 4, which is 8x.
Check
Integrate with respect to y, holding x constant, over the given limits.
\[ \int_{0}^{2} 4xy \, dy \]
Check your understanding
What is the result of this partial integral?
Answer: A
Why: Holding x constant, the antiderivative in y is 2x times y squared; evaluating from 0 to 2 gives 2x times 4, which is 8x.
Connect it up
Draw it
One page, no notation unless you need it: draw how these connect — Pattern: evaluate any iterated integral · Pattern: set up limits from a sketch · Pattern: reverse the order of integration · A double integral is done one variable at a time · It is a partial derivative run in reverse. Put an arrow wherever one of them is what makes another possible, and label the arrow with why.
Recap
You can integrate one variable at a time, holding the other constant, and evaluate an iterated integral from the inside out.
You know the golden rule: inner limits may depend on the outer variable, but the outer limits must be constants so the answer is a number.
You can write the area of a plane region as a double integral of 1, and set the limits from a sketch using vertical or horizontal strips.
You can reverse the order of integration by redrawing the region and re-reading the limits, which also rescues integrals whose inner antiderivative does not exist.
| Task | Key move |
|---|---|
| Evaluate | Inner integral first, then outer |
| Outer limits | Always constants |
| Area | Integrate 1 over the region |
| Reverse order | Redraw, then re-read the limits |
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