Back to the on-screen lesson ·
Regions between two curves, the limits they produce, why the outer limits must be constants, reversing the order of integration when the inner integral will not go, and area and mass as double integrals.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to sketch a region, decide whether it is type one or type two, write the four limits of the corresponding iterated integral, evaluate it, and reverse the order of integration by re-describing the region. You will also be able to say why an outer limit can never contain a variable, and set up an area or a mass as a double integral.
The double integral over a rectangle, evaluated as an iterated integral from the inside out. Everything about that survives here. The one thing that changes is that the inner limits are allowed to be functions of the outer variable, and that single change is what lets a double integral reach every region you will ever want.
A region is type one when it lies between two graphs $y = g_1(x)$ and $y = g_2(x)$ over an interval of $x$ — vertical strips cross it cleanly. It is type two when it lies between $x = h_1(y)$ and $x = h_2(y)$ over an interval of $y$ — horizontal strips cross it cleanly. Many regions are both, some are neither and must be cut into pieces. Reversing the order of integration means describing the same region the other way and rewriting the integral to match.
For a type one region $D = \{(x,y) : a \le x \le b,\; g_1(x) \le y \le g_2(x)\}$,
$$\int\!\!\int_D f\,dA = \int_a^b\!\!\int_{g_1(x)}^{g_2(x)} f(x,y)\,dy\,dx.$$
For a type two region $D = \{(x,y) : c \le y \le d,\; h_1(y) \le x \le h_2(y)\}$,
$$\int\!\!\int_D f\,dA = \int_c^d\!\!\int_{h_1(y)}^{h_2(y)} f(x,y)\,dx\,dy.$$
Why the outer limits must be constants. The outer integral is the last thing done, and after it the answer is a number. If an outer limit still mentioned a variable, that variable would survive into the answer — so an outer limit containing $x$ or $y$ is not a hard case, it is a mistake, and it is the fastest self-check available on a set-up.
Reading the limits off a sketch. Draw the region. For $dy\,dx$: the outer limits are the leftmost and rightmost $x$ anywhere in the region; then draw a vertical line at a typical $x$ and read where it enters and leaves — those are the inner limits, and they are functions of $x$. For $dx\,dy$ do the same with horizontal lines.
Reversing the order. The two descriptions above are of the same region, so the two integrals are equal. Swapping is never done by moving the limits around algebraically; it is done by re-reading the sketch. You reverse when the inner integral in one order is impossible or unpleasant, and that happens often enough that it is a standard technique rather than a curiosity.
Two quantities that are double integrals. The area of $D$ is $\int\int_D 1\,dA$, and the mass of a plate with density $\rho(x,y)$ is $\int\int_D \rho\,dA$. Both are set up exactly like any other double integral; only the integrand differs.
Another way: picture
Put a pencil down vertically on the region and slide it from left to right. Where the pencil first touches and last leaves gives the outer limits; where each position of the pencil enters and leaves the region gives the inner limits. Turning the pencil horizontal and sliding it upwards instead gives the other order — same region, same total, different strips.
Another way: steps
To set up a double integral over a general region:
Consider $\int_0^1\!\int_x^1 \sin(y^2)\,dy\,dx$. The inner integral asks for an antiderivative of $\sin(y^2)$, which no elementary function provides, so as written the problem is not merely hard but closed.
Sketch the region: $0 \le x \le 1$ and $x \le y \le 1$ is the triangle with corners $(0,0)$, $(0,1)$ and $(1,1)$. Now read it the other way. The extreme values of $y$ are $0$ and $1$. At a fixed $y$, the horizontal line enters at $x = 0$ and leaves at $x = y$. So the integral is
$$\int_0^1\!\!\int_0^y \sin(y^2)\,dx\,dy = \int_0^1 y\sin(y^2)\,dy = \tfrac12\left(1 - \cos 1\right).$$
The inner integral became trivial — the integrand has no $x$ in it, so it is just the width of the strip — and the factor $y$ that appeared is exactly what the substitution $u = y^2$ needs. This is the whole pattern: reversing supplies the missing factor, because the width of the strip is a function of the outer variable.
It is worth noticing that nothing about the integrand was manipulated. The entire move was a re-description of the region.
Leaving a variable in an outer limit. $\int_0^x\!\int_0^y f\,dy\,dx$ is not an integral of anything; the answer would still contain $x$. Check the outer limits are constants before doing any work.
Swapping the limits without re-reading the region. Turning $\int_0^1\!\int_0^x f\,dy\,dx$ into $\int_0^1\!\int_0^y f\,dx\,dy$ by exchanging the letters describes a different triangle and gives a different answer. Only the sketch decides.
Using one order on a region that needs two integrals. A region that a vertical strip leaves and re-enters is not type one, and forcing one integral onto it silently includes territory that is not in the region.
Forgetting that the strips must fill the region exactly. The inner limits describe where the strip is inside $D$ — not the whole line, and not the bounding box.
There is a strong pull towards treating $\int_a^b\!\int_{g_1}^{g_2} \ldots dy\,dx$ as a piece of notation that can be rearranged — exchange the differentials, exchange the limits, done. It cannot. The four limits are a description of a region, and the same region described the other way round has genuinely different formulas in it, often involving inverse functions that the original description never mentioned.
The reliable procedure is always the same three moves: sketch the region the current limits describe, describe that region the other way, write the new integral. Any shortcut that skips the sketch is guessing, and the guess is usually a region with the same bounding box and a different area.
Find the area between $y = x^2$ and $y = x$ for $0 \le x \le 1$. As a double integral it is $\int_0^1\!\int_{x^2}^{x} 1\,dy\,dx$.
Area is the double integral of one.
The inner integral is $\left[y\right]_{x^2}^{x} = x - x^2$, which is the height of the strip at $x$.
The inner integral of $1$ is always the strip's width.
Then $\int_0^1 (x - x^2)\,dx = \tfrac12 - \tfrac13 = \tfrac16$ — the familiar area-between-curves formula from Calculus II, which this shows is not a separate rule but a double integral with the inner integral already done.
The old formula was a special case all along.
The region under $y = x^2$ from $x = 0$ to $x = 2$ is type one: $0 \le x \le 2$, $0 \le y \le x^2$.
Vertical strips run from the axis up to the parabola.
As a type two region it is $0 \le y \le 4$ with $\sqrt{y} \le x \le 2$, because a horizontal line at height $y$ enters at the parabola and leaves at the right-hand edge.
Horizontal strips need the parabola solved for $x$.
So $\int_0^2\!\int_0^{x^2} f\,dy\,dx = \int_0^4\!\int_{\sqrt{y}}^{2} f\,dx\,dy$. Both describe the same region, and which one to use is decided entirely by which inner integral you can actually do.
The region is the same; the work is not.
The region is $0 \le y \le 2$ with $y \le x \le 2$: the triangle with corners $(0,0)$, $(2,0)$ and $(2,2)$.
Sketch first, always.
Read it the other way: $x$ runs from $0$ to $2$, and at a fixed $x$ the vertical strip runs from $y = 0$ up to $y = x$.
So the reversed integral is $\int_0^2\!\int_0^{x} f(x,y)\,dy\,dx$. The quickest check on any reversal is to test a corner: $(2,0)$ satisfies both descriptions and $(0,2)$ satisfies neither, as it should.
The region $D$ lies between $y = 0$ and $y = 5x$, for $0 \le x \le 7$. Give the limits of $\int\int_D f\,dy\,dx$.
| Value | |
|---|---|
| The outer lower limit for $x$ | |
| The outer upper limit for $x$ | |
| The inner lower limit at $x = 2$ | |
| The inner upper limit at $x = 2$ |
Put the stages of reversing the order in $\int_{0}^{8}\!\int_{0}^{x} f\,dy\,dx$ into order.
Number the steps in order (write the number in the box):
Evaluate $\displaystyle\int_{0}^{3}\!\int_{0}^{x} x\,dy\,dx$ over the triangle $0 \le y \le x \le 3$.
Answer:
In $\displaystyle\int_{0}^{8}\!\int_{x}^{8} e^{y^{2}}\,dy\,dx$ the order must be reversed. Why?
The region $D$ lies between $y = 0$ and $y = 2x$, for $0 \le x \le 6$. Give the limits of $\int\int_D f\,dy\,dx$.
| Value | |
|---|---|
| The outer lower limit for $x$ | |
| The outer upper limit for $x$ | |
| The inner lower limit at $x = 1$ | |
| The inner upper limit at $x = 1$ |
A plate covers the triangle $0 \le y \le x \le 6$ and has density $7y$ at the point $(x,y)$. What is its mass?
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Put the stages of reversing the order in $\int_{0}^{4}\!\int_{0}^{x} f\,dy\,dx$ into order.
Number the steps in order (write the number in the box):
You can read the limits of a double integral off a sketch, evaluate it inside out, and reverse the order when the inner integral will not go. Next: the regions that are circles rather than triangles, where a different coordinate system does the work.
10. Your turn: reverse the order in $\int_0^2\!\int_{y}^{2} f(x,y)\,dx\,dy$, step 3