Integrals with trivial integrands

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We study integrals over general regions by integrating $1$ .

Now we will integrate over regions that are more complex than rectangles and boxes. In particular we will find areas of regions bounded by curves in $\R ^2$ , and volumes of regions bounded by surfaces in $\R ^3$ . To do this we will use integrals whose integrand is $1$ . The integrand is the “thing” you are integrating: $\iint _R \underbrace {F(x,y)}_{\text {integrand}} \d A$ Think about this: The integral $\int _a^b \d x = \int _a^b 1 \d x = b-a$ computes the length between $a$ and $b$ . This is because the area between the constant curve of height one and the $x$ -axis is just the length of the curve! In a similar way: $\iint _R \d A = \iint _R 1 \d A = \text {Area of $R$}$ Here you imagine a surface of constant height $1$ above a region $R$ :

The volume under the plane of constant height $1$ above the region $R$ is numerically equal to the area of $R$ . In an entirely similar way, if $R$ is a subset of $\R ^3$ , then $\iiint _R \d V = \iiint _R 1 \d V= \text {Volume of $R$}$ In this section we will focus on integrals with a trivial integrand, meaning it is $1$ , and hence the section is called “Integrals with trivial integrands.”

Double integrals and area

We start with a third version of Fubini’s Theorem.

Fubini’s Theorem Let $R$ be a closed, bounded region in the $(x,y)$ -plane and let $F(x,y)$ be a continuous function on $R$ .

If $R=\{(x,y):\text {$a\leq x\leq b$ and $g_1(x)\leq y\leq g_2(x)$}\}$

where $g_1$ and $g_2$ are continuous functions on $[a,b]$ , then $\iint _R F(x,y) \d A = \int _a^b\int _{g_1(x)}^{g_2(x)} F(x,y)\d y\d x.$
If $R=\{(x,y):\text {$c\leq y\leq d$ and $g_3(y)\leq x\leq g_4(y)$}\}$

where $g_3$ and $g_4$ are continuous functions on $[c,d]$ , then $\iint _R F(x,y)\d A = \int _c^d\int _{g_3(y)}^{g_4(y)} F(x,y)\d x\d y.$

It is important to note that when using Fubini’s Theorem, we must always have numbers as the limits of the outer-most integral and curves (note constant curves are numbers) as the limits of the inner-most integral: $\int _{\text {number}}^{\text {number}}\int _{\text {curve}}^{\text {curve}} F(x,y) \d A$ Note, if we set $g_1$ and $g_2$ (or $g_3$ and $g_4$ ) to be constant functions, we recover the version of Fubini’s Theorem from the previous section.

Whenever you learn a new technique, you should always “try it out” on a computation where you know the answer through a different method. So let’s used an integral to find the area of a triangle.

Set-up and evaluate an iterated integral that will compute the area of the region below via a double integral:

Our region above can be defined by: $R=\left \{(x,y):\text {$1\leq x\leq \answer [given]{3}$ and $1\leq y\leq \answer [given]{-x/2+5/2}$}\right \}$ The area of this region is given by, write with me, $\begin{align*} \int_{\answer[given]{1}}^{\answer[given]{3}}\int_{\answer[given]{1}}^{\answer[given]{-x/2+5/2}}\d y \d x &= \int_{\answer[given]{1}}^{\answer[given]{3}} \eval{\answer[given]{y}}_{\answer[given]{1}}^{\answer[given]{-x/2+5/2}} \d x\\ &= \int_{\answer[given]{1}}^{\answer[given]{3}} \left(\answer[given]{-x/2+3/2}\right) \d x\\ &= \eval{\answer[given]{-x^2/4+3x/2}}_{\answer[given]{1}}^{\answer[given]{3}}\\ &= \answer[given]{1}. \end{align*}$

Set-up and evaluate an iterated integral that will compute the area of the region below via a double integral:

Our region above can be defined by: $R=\{(x,y):\text {$-6\leq y\leq 6$ and $y^2/3\leq x\leq 12$}\}$ The area of this region is given by, write with me, $\begin{align*} \int_{\answer[given]{-6}}^{\answer[given]{6}}\int_{\answer[given]{y^2/3}}^{\answer[given]{12}}\d x \d y &= \int_{\answer[given]{-6}}^{\answer[given]{6}} \eval{\answer[given]{x}}_{\answer[given]{y^2/3}}^{\answer[given]{12}} \d y\\ &= \int_{\answer[given]{-6}}^{\answer[given]{6}} \left(\answer[given]{12-y^2/3}\right) \d y\\ &= \eval{\answer[given]{12y-y^3/9}}_{\answer[given]{-6}}^{\answer[given]{6}}\\ &= \answer[given]{96}. \end{align*}$

Changing the order of integration

We can change the order of integration. This can make a big difference in the difficulty of the integral in question. This can be somewhat challenging. If you have a graph of the region in question, this can help quite a bit. If not, then a strategy is to

Find absolute bounds for each variable.
Solve inequalities that will allow you to integrate: $\int _{\text {number}}^{\text {number}}\int _{\text {curve}}^{\text {curve}} F(x,y) \d A$

Let’s see an example.

Consider: $\int _1^3 \int _1^{-x/2 +5/2}\d y \d x$ Set-up an iterated integral that integrates with respect to $x$ and then integrates with respect to $y$ .

Start by finding absolute bounds for our variables. In this case: $\begin{align*} 1\le &x \le 3\\ 1\le &y \le -x/2 + 5/2 \le \answer{2} \end{align*}$ Now, write $x$ in terms of $y$ . Write with me: $\begin{align*} \answer{-3/2}\le &y -5/2 \le -x/2 \le \answer{-1/2}\\ \answer{1/2}\le &x/2 \le 5/2-y \le \answer{3/2}\\ 1\le & x\le \answer{5-2y} \le 3 \end{align*}$ We may now write our desired integral: $\iint _R \d A = \int _{\answer {1}}^{\answer {2}}\int _{\answer {1}}^{\answer {5-2y}} \d x \d y$

Let’s see another example.

Consider: $\int _{-6}^6 \int _{y^2/3}^{12} \d x \d y$ Set-up an iterated integral that computes the area of $R$ that integrates with respect to $y$ and then with respect to $x$ .

Start by finding absolute bounds for our variables. In this case: $\begin{align*} \answer{0}\le y^2/3\le &x \le 12\\ -6\le &y \le 6 \end{align*}$ Now, write $y$ in terms of $x$ . Write with me: $\begin{align*} \answer{0}\le &y^2 \le 3x \le \answer{36}\\ \answer{-6}\le -\sqrt{3x}\le &y \le \sqrt{3x} \le \answer{6}\\ \end{align*}$ We may now write our desired integral: $\iint _R \d A = \int _{\answer {0}}^{\answer {12}}\int _{\answer {-\sqrt {3x}}}^{\answer {\sqrt {3x}}} \d y \d x$

Sometimes when you switch the order of integration, you will need to write the sum of iterated integrals. You can recognize this by the fact that you won’t be able to find a single curve to bound your inner integral. This will be more clear with an example.

Consider the region:

Set-up iterated integrals that compute the area of $R$ .

First we’ll integrate with respect to $y$ and then $x$ . Here $\answer [given]{1}\le x \le \answer [given]{5}$ and $\answer [given]{(x+3)/4}\le y \le \answer [given]{(9-x)/2}$ so our integral is: $\int _{\answer [given]{1}}^{\answer [given]{5}} \int _{\answer [given]{(x+3)/4}}^{\answer [given]{(9-x)/2}}\d y \d x$

Now let’s integrate with respect to $x$ and then $y$ . If we give absolute bounds to our inequalites above, we see something strange: $1 \le \answer [given]{(x+3)/4}\le y \le \answer [given]{(9-x)/2} \le 4$ This is actually two inequalities (note we can see this from the graph):

$\begin{align*} 1 &\le \answer[given]{(x+3)/4}\le y && \text{and}\\ y &\le \answer[given]{(9-x)/2} \le 4 \end{align*}$ Solving for $x$ in these inequalities we find: $\begin{align*} 1 &\le x \le 4y-3 && \text{and}\\ 1 &\le x \le 9-2y \end{align*}$ Here $x$ is bounded by two different curves! When this happens, it is probably best to look at the graph of the situation:

Thus we see that the area can be found by the sum: $\int _1^2 \int _{\answer [given]{1}}^{\answer [given]{4y-3}}\d x \d y +\int _2^4 \int _{\answer [given]{1}}^{\answer [given]{9-2y}}\d x \d y$

Why change the order of integration?

You may ask yourself, “Why change the order of integration?” Sometimes changing the region can make a difficult (impossible) antiderivative easier (not impossible). Let’s see an example.

Compute: $\int _0^{\pi /2}\int _y^{\pi /2} \frac {\sin (x)}{x} \d x \d y$

Here you don’t stand a chance if you try to antidifferentiate with respect to $x$ . So the trick is to immediately swap the order of antidifferentation. Starting by writing absolute bounds for $x$ and $y$ : $0\le \answer [given]{y}\le \answer [given]{x}\le \pi /2$ Now we see that: $\begin{align*} \int_0^{\pi/2}\int_y^{\pi/2} \frac{\sin(x)}{x} \d x \d y &= \int_{\answer[given]{0}}^{\answer[given]{\pi/2}}\int_{\answer[given]{0}}^{\answer[given]{x}} \frac{\sin(x)}{x} \d y \d x\\ &= \int_{\answer[given]{0}}^{\answer[given]{\pi/2}} \answer[given]{\sin(x)} \d x\\ &=\answer[given]{1} \end{align*}$

This author does not know how to solve the problem above without changing the order of integration.

Triple integrals and volume

We start by introducing a fourth version of Fubini’s Theorem.

Fubini Let $R$ be a closed, bounded region in $\R ^3$ and let $F(x,y,z)$ be a continuous function on $R$ . If $\begin{align*} R=\{(x,y,z):&\text{$a\leq x\leq b$, $g_1(x)\leq y\leq g_2(x)$,}\\ &\text{and $H_1(x,y) \leq z \leq H_2(x,y)$}\} \end{align*}$ where $g_1$ and $g_2$ are continuous functions on $[a,b]$ , and $H_1$ and $H_2$ are continuous on the region $S =\{(x,y):\text {$a\leq x\leq b$ and $g_1(x)\leq y\leq g_2(x)$}\}$ then $\iiint _R F(x,y,z) \d V = \int _a^b\int _{g_1(x)}^{g_2(x)}\int _{H_1(x,y)}^{H_2(x,y)} F(x,y,z)\d z\d y\d x.$ There are six reorderings total with three variables. We will spare the young mathematician the details, and trust that you will sort it out.

Again note that when using Fubini’s Theorem, we must always have numbers as the limits of the outer-most integral, curves as the limits of the middle integral, and surfaces as the limits of the inner-most integral: $\int _{\text {number}}^{\text {number}}\int _{\text {curve}}^{\text {curve}}\int _{\text {surface}}^{\text {surface}} F(x,y,z) \d V$

Now with Fubini’s help, we will use triple integrals to compute volumes.

Set-up and evaluate an iterated integral that will compute the volume of the region bounded by:

The plane $x=0$ .
The plane $y=0$ .
The plane $z=0$ .
The plane $3x + 4y + 6z = 12$ .

Our region above produces a tetrahedron, a triangular-based pyramid. It intersects the $x$ -axis at $(4,0,0)$ , the $y$ -axis at $(0,3,0)$ , and the $z$ -axis at $(0,0,2)$ . The region can be defined by: $\begin{align*} R=\Big\{(x,y,z):&0\leq x\leq 4, \\ &0\leq y\leq -3x/4+3, \\ &0\leq z \leq \frac{12-3x-4y}{6}\Big\} \end{align*}$ The volume of this region is given by, write with me, $\int _{\answer [given]{0}}^{\answer [given]{4}} \int _{\answer [given]{0}}^{\answer [given]{-3x/4+3}} \int _{\answer [given]{0}}^{\answer [given]{\frac {12-3x-4y}{6}}}\d z \d y\d x$ $\begin{align*} &=\int_{\answer[given]{0}}^{\answer[given]{4}} \int_{\answer[given]{0}}^{\answer[given]{-3x/4+3}} \answer[given]{\frac{12-3x-4y}{6}} \d y \d x\\ &= \int_{\answer[given]{0}}^{\answer[given]{4}} \eval{\answer[given]{2y-\frac{xy}{2}-\frac{y^2}{3}}}_{\answer[given]{0}}^{\answer[given]{-3x/4+3}} \d x\\ &= \int_{\answer[given]{0}}^{\answer[given]{4}} \left(\answer[given]{\frac{3x^2}{16} -\frac{3x}{2}+3}\right) \d x\\ &= \eval{\answer[given]{x^3/16-3x^2/4+3x}}_{\answer[given]{0}}^{\answer[given]{4}} \\ &=\answer[given]{4}. \end{align*}$

In the previous example, we integrated with respect to $z$ , then $y$ , then $x$ . Set-up an integral that computes the volume of $R$ that integrates with respect to $x$ , then $z$ , then $y$ . Start by finding overall bounds for our variables. In this case: $\begin{align*} 0\le &x \le 4\\ 0\le &y \le -3x/4+3 \le \answer{3}\\ 0\le &z \le \frac{12-3x-4y}{6}\le \answer{2} \end{align*}$ At this point we see that our bounds for $y$ are $\answer {0}$ to $\answer {3}$ . Now we will find our bounds for $x$ . We must find an expression for $x$ in terms of $y$ and $z$ . Write with me: $\begin{align*} 0\le z \le &\frac{12-3x-4y}{6}\le 2\\ 0\le 6z \le &\answer{12-3x-4y} \le 12\\ -12+4y\le \answer{6z+4y-12}\le &-3x \le 4y\\ \frac{12-4y}{3}\ge \answer{\frac{12-6z-4y}{3}}\ge &x \ge -4y/3 \end{align*}$ However, we know that $y$ is nonnegative, so $-4y/3\le \answer {0}$ , and $x$ is bounded below by $\answer {0}$ . So $x$ runs from $0$ to $\answer {\frac {12-6z-4y}{3}}$ .

Finally we must write $z$ in terms of $y$ . Unfortunately, from our inequalities above, there is no direct way to get this. We must think about what our solid looks like. Recall that the plane bounding the solid is $3x + 4y + 6z = 12$ . If $x=0$ , then our plane is the line $\answer {4y+6z}=12$ . Hence $0\le z \le \answer {2-2y/3}$ We may now write our desired integral: $\iiint _R \d V = \int _{\answer {0}}^{\answer {3}}\int _{\answer {0}}^{\answer {-2y/3+2}} \int _{\answer {0}}^{\answer {\frac {12-4y-6z}{3}}} \d x \d z \d y$

Set-up an iterated integral that will compute the volume of the region bounded by the cone below:

First note that $\begin{align*} R=\{(x,y,z):&\text{$-1\leq x\leq 1$, $-\sqrt{1-x^2}\leq y\leq \sqrt{1-x^2}$}, \\ &\text{and $0\leq z \leq 1-\sqrt{x^2+y^2}$}\} \end{align*}$ Hence, the volume of this region is given by $\int _{\answer [given]{-1}}^{\answer [given]{1}} \int _{\answer [given]{-\sqrt {1-x^2}}}^{\answer [given]{\sqrt {1-x^2}}} \int _{\answer [given]{0}}^{\answer [given]{1-\sqrt {x^2+y^2}}}\d z \d y\d x.$

In the previous example, we integrated with respect to $z$ , then $y$ , then $x$ . Set-up an iterated integral that computes the volume of $R$ that integrates with respect to $x$ , then $y$ , then $z$ . $\iiint _R \d V = \int _{\answer {0}}^{\answer {1}} \int _{\answer {-1+z}}^{\answer {1-z}} \int _{\answer {-\sqrt {(1-z)^2-y^2}}}^{\answer {\sqrt {(1-z)^2-y^2}}} \d x \d y \d z$