Integrals with trivial integrands

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We study integrals over general regions with and integrand equaling one.

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 $\R ^3$ .

Double integrals and area

We start with Fubini’s Theorem. While we will state it generally, in this section the integrand, $F$ , will always be $1$ .

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 $\iint _R F(x,y)\d A = \int _c^d\int _{g_3(y)}^{g_4(y)} F(x,y)\d x\d y.$

When the integrand of a double integral is $1$ , like this $\iint _R\d A$ this integral computes the area of the region. This is because you are finding the “signed volume” between the curve and the plane $z=1$ . In our first example, we will use this “hammer to swat a fly,” meaning, we are essentially using a much more difficult method than necessary! We apologize in advance, and include the example because it is illustrative of the method of using double integrals to compute areas.

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

In the previous example, we integrated with respect to $y$ first. Set-up an integral that computes the area of $R$ that integrates with respect to $x$ first. Start by finding overall 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: $\iint _R \d A = \int _{\answer {1}}^{\answer {2}}\int _{\answer {1}}^{\answer {5-2y}} \d x \d y$

In our example above, we used an integral to find the area of a triangle. Whenever you learn a new technique, you should always “try it out” on a computation where you know the answer through a different method. With our next two examples, we’ll work with regions where calculus really helps out.

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, $\begin{align*} \int_{\answer[given]{-6}}^{\answer[given]{6}}\int_{\answer[given]{y^2/3}}^{\answer[given]{12}}\d x \d y &= \int_{\answer[given]{-6}}^{\answer[given]{6}} \eval{\answer[given]{x}}_{\answer[given]{y^2/3}}^{\answer[given]{12}} \d y\\ &= \int_{\answer[given]{-6}}^{\answer[given]{6}} \left(\answer[given]{12-y^2/3}\right) \d y\\ &= \eval{\answer[given]{12y-y^3/9}}_{\answer[given]{-6}}^{\answer[given]{6}}\\ &= \answer[given]{96}. \end{align*}$

In the previous example, we integrated with respect to $x$ first. Set-up an integral that computes the area of $R$ that integrates with respect to $y$ first. Start by finding overall 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: $\iint _R \d A = \int _{\answer {0}}^{\answer {12}}\int _{\answer {-\sqrt {3x}}}^{\answer {\sqrt {3x}}} \d y \d x$

And now for one more example of using a double integral to compute the area of a region.

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 {$0\leq x\leq 1$ and $x^4\leq y\leq \sqrt {x}$}\}$ The area of this region is given by, write with me, $\begin{align*} \int_{\answer[given]{0}}^{\answer[given]{1}}\int_{\answer[given]{x^4}}^{\answer[given]{\sqrt{x}}}\d y \d x &= \int_{\answer[given]{0}}^{\answer[given]{1}} \eval{\answer[given]{y}}_{\answer[given]{x^4}}^{\answer[given]{\sqrt{x}}} \d x\\ &= \int_{\answer[given]{0}}^{\answer[given]{1}} \left(\answer[given]{\sqrt{x}-x^4}\right) \d x\\ &= \eval{\answer[given]{2x^{3/2}/3 - x^5/5}}_{\answer[given]{0}}^{\answer[given]{1}}\\ &= \answer[given]{7/15}. \end{align*}$

In the previous example, we integrated with respect to $y$ first. Set-up an integral that computes the area of $R$ that integrates with respect to $x$ first. Start by finding overall bounds for our variables. In this case: $\begin{align*} 0\le &x \le 1\\ \answer{0}\le x^4\le &y \le \sqrt{x}\le \answer{1} \end{align*}$ Now, write $x$ in terms of $y$ . Since $y$ is bounded by $x$ on both sides, we’ll do this in two steps. Write with me: $\begin{align*} \answer{0}\le &x^4\le y\\ 0 \le &x \le \answer{y^{1/4}} \end{align*}$ and $\begin{align*} y\le &\sqrt{x}\le \answer{1}\\ \answer{y^2} \le &x \le 1 \end{align*}$ We may now write our desired integral: $\iint _R \d A = \int _{\answer {0}}^{\answer {1}}\int _{\answer {y^2}}^{\answer {y^{1/4}}} \d x \d y$

Triple integrals and volume

We start by again(!) introducing another 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.

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, $\int _{\answer [given]{0}}^{\answer [given]{4}} \int _{\answer [given]{0}}^{\answer [given]{-3x/4+3}} \int _{\answer [given]{0}}^{\answer [given]{\frac {12-3x-4y}{6}}}\d z \d y\d x$ $\begin{align*} &=\int_{\answer[given]{0}}^{\answer[given]{4}} \int_{\answer[given]{0}}^{\answer[given]{-3x/4+3}} \answer[given]{\frac{12-3x-4y}{6}} \d y \d x\\ &= \int_{\answer[given]{0}}^{\answer[given]{4}} \eval{\answer[given]{2y-\frac{xy}{2}-\frac{y^2}{3}}}_{\answer[given]{0}}^{\answer[given]{-3x/4+3}} \d x\\ &= \int_{\answer[given]{0}}^{\answer[given]{4}} \left(\answer[given]{\frac{3x^2}{16} -\frac{3x}{2}+3}\right) \d x\\ &= \eval{\answer[given]{x^3/16-3x^2/4+3x}}_{\answer[given]{0}}^{\answer[given]{4}} \d x\\ &=\answer[given]{4}. \end{align*}$

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 $z$ . $\iiint _R \d V = \int _{\answer {0}}^{\answer {1}} \int _{\answer {-1+z}}^{\answer {1-z}} \int _{\answer {-\sqrt {(1-z)^2-y^2}}}^{\answer {\sqrt {(1-z)^2-y^2}}} \d x \d y \d z$

When the integrand isn’t one

What if the integrand isn’t $1$ ? Computationally, you just do what you did before, but you have an additional antiderivative to compute. However, sometimes the nontrivial region can make a difficult antiderivative quite ease. Let’s see one more 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: $\begin{align*} \int_0^{\pi/2}\int_y^{\pi/2} \frac{\sin(x)}{x} \d x \d y &= \int_{\answer[given]{0}}^{\answer[given]{\pi/2}}\int_{\answer[given]{0}}^{\answer[given]{x}} \frac{\sin(x)}{x} \d y \d x\\ &= \int_{\answer[given]{0}}^{\answer[given]{\pi/2}} \answer[given]{\sin(x)} \d x\\ &=\answer[given]{1} \end{align*}$

You may be wondering, what do multiple integrals mean when the integrand is not $1$ ? Read on!