Wichita State University Logo

Math 344: Calculus III

15.2 Double Integrals Over Regions

15.2.1 Double Integrals Over General Regions

Integrating functions of one variable did not require special setup or techniques because every integration region is just an interval or a finite sum of intervals. Intervals are the one-dimensional analogue to rectangles, so the ease their computation makes sense. Integration of two-dimensional regions takes more care, though, since we can easily have non-rectangular regions over which we would like to integrate. We integrate a function $f(x,y)$ over a bounded general region $D$ by assuming that it may be contained in a rectangle $R$ (this is what makes it bounded) and define a new function $F(x,y)$ in the rectangle so that

$$ \def\ihat{\mathbf{\hat{\mmlToken{mi}[mathvariant="bold"]{ı}}}} \def\jhat{\mathbf{\hat{\mmlToken{mi}[mathvariant="bold"]{ȷ}}}} \def\khat{\mathbf{\hat{k}}} \newcommand{\pzi}[2][]{\dfrac{\partial #1}{\partial #2}} F(x,y) = \left\{ \begin{array}{ccl} f(x,y) & \quad & (x,y)\in D \\ 0 & \quad & (x,y)\in R\setminus D \end{array}\right. $$
where $(x,y)\in R\setminus D$ means that $(x,y)$ is in $R$ but not in $D$.

Two side-by-side plots of a general region D plotted on a 2D coordinate plane. On the left side, just the region is plotted. On the right side, the region is plotted enclosed in a rectangle R.

If the function $F$ is integrable over the region $R$, we define the following:

Definition

The double integral of $f$ over $D$ is

$$ \iint_D f(x,y)\,dA = \iint_R F(x,y)\,dA $$
where

$$ F(x,y) = \left\{ \begin{array}{ccl} f(x,y) & \quad & (x,y)\in D \\ 0 & \quad & (x,y)\in R\setminus D \end{array}\right. $$

By setting things up like this, we get to use the previous definition of integrating over rectangle that we had from last section. This means that our integral $\iint_R F(x,y)\,dA$ is meaningfully defined and $\iint_D f(x,y)\,dA$ can be interpreted to be the volume of a solid which is bounded above by the surface $z = f(x,y)$ and below by $D$.

Two side-by-side plots of a general region D plotted on a 3D coordinate plane with a surface above. On the right side, the integration region is plotted enclosed in a rectangle R.

It is possible to find the volume between $D$ and this surface $z = f(x,y)$ so long as the curve bounding $D$ is"nice." Defining this precisely and proving the condition we choose valid is (way) beyond the scope of this class, but suppose at minimum the boundary is piecewise smooth , that is, the boundary may be written as a finite collection of smooth (differentiable) curves.

15.2.2 Types of Regions

We commonly separate the planar regions we wish to integrate into types, so that we may set up iterated integrals to evaluate them. We call these regions type I and type II . A type I region is one whose boundaries may be expressed in terms of a function of $x$ and a type II region is one whose boundaries may be expressed as a function of $y$.

Definition

  • A type I planar region is given by

    $$ D = \left\{ (x,y) : x\in [a,b], g_1(x) \leq y \leq g_2(x) \right\} $$ where $g_1$ and $g_2$ are continuous functions on $[a,b]$.
  • A type II planar region is given by

    $$ D = \left\{ (x,y) : y\in [c,d], h_1(y) \leq x \leq h_2(y) \right\} $$ where $h_1$ and $h_2$ are continuous functions on $[c,d]$.

Type I Regions

Three side-by-side examples of a type I integration regions. All of these functions are bounded above and below by continuous functions of y.

The regions depicted here are type I , so they are contained within an interval $[a,b]$ in the $x$-direction and by two functions of $x$, $g_1(x)$ and $g_2(x)$ in the $y$-direction. For these types of regions, we compute the double integral of $f$ over $D$ by writing it as an iterated integral

$$ \iint_D f(x,y)\,dA = \int_a^b \int_{g_1(x)}^{g_2(x)} f(x,y)\,dy\,dx $$
Note that here we integrate with respect to $y$ first. When we evaluate the integral with respect to $y$, we will use the functions $g_1$ and $g_2$ as our limits and then evaluate the second integral with respect to $x$, yielding the numerical value for definite integral.

Type II Regions

Three side-by-side examples of a type II integration regions. All of these functions are bounded on the left and right by continuous functions of x.

For type II regions, we see that the lower and upper limits are given by an interval of $y$-values $[c,d]$ and the left and right boundaries are defined by function of $x$, $h_1(x)$ and $h_2(x)$. Therefore, the iterated integrals for these regions have the form

$$ \iint_D f(x,y)\,dA = \int_c^d \int_{h_1(y)}^{h_2(y)} f(x,y)\,dx\,dy $$
where we integrate with respect to $x$ first.

Video Summary

15.2.2 Type I Examples and Exercises

Example 1

Evaluate the double integral

$$ \iint_D \dfrac{y}{x^2 + 1}\,dA $$
where $D = \left\{(x,y) : x\in [0,4],0\leq y \leq \sqrt{x}\right\}$.

Solution

The first step for this problem is the sketch the region to make sure that we have a clear understanding of the integration domain.

A region bounded on the left and right by 0 and 4, below by y = 0 and above by y = sqrt(x).

The arrowed line in the middle indicates the direction of the interior integral with respect to $y$. We compute this first, then "slide" that line from left to right over $[0,4]$ as we integrate with respect to $x$. Therefore, the iterated integral has the form

$$ \iint_D \dfrac{y}{x^2 + 1}\,dA = \int_0^4 \int_{y\,=\,0}^{y\,=\,\sqrt{x}} \dfrac{y}{x^2 + 1}\,dy\,dx $$
We now evaluate

$$ \begin{align*} \int_0^4 \int_{y\,=\,0}^{y\,=\,\sqrt{x}} \dfrac{y}{x^2 + 1}\,dy\,dx &= \int_0^4 \left[ \int_{y\,=\,0}^{y\,=\,\sqrt{x}} \dfrac{y}{x^2 + 1}\,dy\right]\,dx \\ \\ &= \int_0^4 \left[ \left.\dfrac{y^2}{2(x^2+1)}\right|_{\, y\,=\,0}^{\, y\,=\,\sqrt{x}}\right]\,dx \\ \\ &= \int_0^4 \left[ \dfrac{\left(\sqrt{x}\right)^2}{2(x^2+1)} - \dfrac{(0)^2}{2(x^2+1)} \right]\,dx \\ \\ &= \int_0^4 \dfrac{x}{2(x^2+1)}\, dx \\ \\ &= \left|\begin{array}{cl} u = x^2+1 & 4\rightarrow (4)^2 + 1 = 17 \\ du = 2x\,dx & 0 \rightarrow (0)^2+1 = 1 \end{array}\right| \\ \\ &= \dfrac{1}{4} \int_1^{17} \dfrac{du}{u} \\ \\ &= \left.\left(\dfrac{1}{4}\ln |u|\right)\right|_1^{17} \\ \\ &= \dfrac{1}{4}\ln 17 \end{align*} $$

Exercise 1

Evaluate the iterated integral

$$ \int_0^\frac{\pi}{2} \int_0^x x\sin y\,dy\,dx $$

Check Your Work $$ \int_0^\frac{\pi}{2} \int_0^x x\sin y\,dy\,dx = \dfrac{\pi^2}{8} - \dfrac{\pi}{2} + 1 $$

Video Solution

15.2.3 Type II Regions Examples and Exercises

Example 2

Evaluate the iterated integral

$$ \int_0^1 \int_0^{e^y} \sqrt{1 + e^y}\,dx\,dy $$

Solution

Since our problem is already presented as an iterated integral, we do not need to sketch the region prior to beginning, but it is generally good practice to do so.

A region bounded on the left by x = 0, on the right by x = exp(y), and above and below 0 and 1.

Evaluating the integral:

$$ \begin{align*} \int_0^1 \int_0^{e^y} \sqrt{1 + e^y}\,dx\,dy &= \int_0^1 \left[\int_0^{e^y} \sqrt{1 + e^y}\,dx\right]\,dy \\ \\ &= \int_0^1 \left[ \left. x\sqrt{1 + e^y}\right|_{\,x\,=\,0}^{\,x\,=\,e^y} \right]\,dy \\ \\ &= \int_0^1 e^y\sqrt{1 + e^y}\,dy \\ \\ &= \left| \begin{array}{cl} u = 1 + e^y & 1\rightarrow 1 + e^1 = 1 + e \\ du = e^y\,dy & 0\rightarrow 1 + e^0 = 2 \end{array} \right| \\ \\ &= \int_2^{1+e} \sqrt{u}\,du \\ \\ &= \left.\dfrac{2}{3}u^\frac{3}{2}\right|_{\,2}^{\,1+e} \\ \\ &= \dfrac{2}{3}\left((1+e)^\frac{3}{2} - 2^\frac{3}{2}\right) \\ \\ &= \dfrac{2}{3}\left((1+e)^\frac{3}{2} - 2\sqrt{2}\right) \end{align*} $$

Exercise 2

Evaluate the double integral

$$ \iint_D y^2\,dA $$
where $D$ is the triangular region with vertices $(0,1)$, $(1,2)$, and $(4,1)$.

Check Your Work $$ \int_1^2 \int_{x = y-1}^{x=7-3y} y^2\,dx\,dy = \frac{11}{3} $$

Follow Along The first task for this problem is to sketch the region, so that we may find the bounding functions and interval. The region $D$ looks like A region bounded on the left by x = y-1, on the right by x = 7-3y, and above and below 1 and 2.
which is a type II region. The equations for the bounding lines are shown in the figure, and may be determined by finding the equations for the lines that pass through $(0,1)$, $(1,2)$ and $(1,2)$,$(4,1)$, respectively. The slope between $(0,1)$ and $(1,2)$ is $1$, and by the point-slope formula the equation of this line is
$$ \begin{align*} y - y_0 &= m(x-x_0) \\ \\ y - 1 &= 1(x-0) \\ \\ x &= y-1 \end{align*} $$
The slope between $(1,2)$ and $(4,1)$ is $-\frac{1}{3}$, and by the point-slope formula
$$ \begin{align*} y - y_0 &= m(x-x_0) \\ \\ y - 2 &= -\frac{1}{3}(x-1) \\ \\ -3y + 6 &= x - 1 \\ \\ x &= 7 - 3y \end{align*} $$
Our iterated integral has the form
$$ \int_1^2 \int_{x = y-1}^{x=7-3y} y^2\,dx\,dy $$
Evaluating the integral
$$ \begin{align*} \int_1^2 \int_{x = y-1}^{x=7-3y} y^2\,dx\,dy &= \int_1^2 \left[\int_{x = y-1}^{x=7-3y} y^2\,dx\right]\,dy \\ \\ &= \int_1^2 \left[\left. xy^2\right|_{\,x = y-1}^{\,x=7-3y}\right]\,dy \\ \\ &= \int_1^2 \left[(7-3y)y^2 - (y-1)y^2 \right]\,dy \\ \\ &= \int_1^2 8y^2 -4y^3\, dy \\ \\ &= \left. \frac{8}{3}y^3 - y^4 \right|_{\,1}^{\,2} \\ \\ &= \left(\frac{64}{3} - 16\right) - \left(\frac{8}{3} - 1\right) \\ \\ &= \frac{11}{3} \end{align*} $$

15.2.4 Changing Integration Order

At the beginning of the section, we were careful to relate integrating over a general region to integration over rectangular regions. One of the things we gain from this is the ability to use Fubini's theorem. This means that we may integrate a general region with respect to $x$ and $y$ in either order. It is often the case that an integral is easier to compute in one direction than it is in the other. This is why we have been practicing making sketches for the region, so that we can use the sketches to change the integration order when it is convenient.

Example 3

Evaluate the iterated integral

$$ \int_0^4 \int_{y\,=\,\frac{1}{2}x}^{y\,=\,2} \cos\left(y^2\right)\,dy\,dx $$

Solution

Following the integration order as written here leaves with the unpleasant task of attempting to integrate $\cos\left(y^2\right)$. However, if we sketch the region we will see that this region is both type I and type II, so it is possible to use Fubini's theorem to change the order of integration. Two copies of the same integration region side by side, interpreted as a type I region on the left and a type II region on the right.
After sketching the region as given, we see that it is bounded in the $y$ direction by $[0,2]$ and in the $x$ direction by the curves $x = 0$ and $x = 2y$. This allows us to rewrite the iterated integral with a new order of integration

$$ \int_0^2 \int_{x\,=\,0}^{x\,=\,2y} \cos\left(y^2\right)\,dx\,dy $$
We now evaluate by integrating with respect to $x$ first and hoping that it will provide us with some help resolving the integral of $\cos\left(y^2\right)$.

$$ \begin{align*} \int_0^2 \int_{x\,=\,0}^{x\,=\,2y} \cos\left(y^2\right)\,dx\,dy &= \int_0^2 \left[ \int_{x\,=\,0}^{x\,=\,2y} \cos\left(y^2\right)\,dx\right]\,dy \\ \\ &= \int_0^2 \left[ \left. x\cos\left(y^2\right)\right|_{\,x\,=\,0}^{\,x\,=\,2y}\right]\,dy \\ \\ &= \int_0^2 2y\cos\left(y^2\right)\,dy \end{align*} $$
We see that yes, integrating with respect to $x$ first gives us a factor of $2y$ that will allow us to integrate the much easier $2y\cos\left(y^2\right)$ by substitution.

$$ \begin{align*} \int_0^2 2y\cos\left(y^2\right)\,dy &= \left|\begin{array}{cl} u = y^2 & 2\rightarrow (2)^2 = 4 \\ du = 2y\,dy & 0\rightarrow (0)^2 = 0 \end{array}\right| \\ \\ &= \int_0^4 \cos u\,du \\ \\ &= \cos(4) - 1 \end{align*} $$

Exercise 3

Evaluate the double integral

$$ \iint_D y^2 e^{xy}\,dA $$
where the region $D$ is bounded by the curves $y = x$, $y = 4$, and $x = 0$.

Check Your Work $$ \iint_D y^2 e^{xy}\,dA = \dfrac{e^{16} - 17}{2} $$

Video Solution

Creative Commons Logo - White


Your use of this self-initiated mediated course material is subject to our Creative Commons License .


Creative Commons Attribution-NonCommercial-ShareAlike 4.0

Creative Commons Logo - Black
Attribution
You must give appropriate credit, provide a link to the license, and indicate if changes were made. You may do so in any reasonable manner, but not in any way that suggests the licensor endorses you or your use.

Creative Commons Logo - Black
Noncommercial
You may not use the material for commercial purposes.

Creative Commons Logo - Black
Share Alike
You are free to share, copy and redistribute the material in any medium or format. If you adapt, remix, transform, or build upon the material, you must distribute your contributions under the same license as the original.