Computations and interpretations

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We practice more computations and think about what integrals mean.

In this section we will continue to set-up (and sometimes compute) double and triple integrals and think about what these mean.

Triangles

Let $T$ be the triangle with vertices $(0,0)$ and $(\pi ,0)$ and $(\pi ,\pi )$ . Then we can write $T$ as $T = \{ (x,y) : 0 \leq y \leq \pi \text { and } \answer [given]{y} \leq x \leq \answer [given]{\pi } \},$ but we could also write $T = \{ (x,y) : 0 \leq x \leq \pi \text { and } \answer [given]{0} \leq y \leq \answer [given]{x} \}.$ Which of these might be a better choice to compute $\iint _T 2 \sin (x^2) \d A?$

Based on these two different descriptions of $T$ , we can evaluate the double integral $\iint _T 2 \sin (x^2) \d A$ through two rather different looking iterated integrals. Write with me, $\iint _T 2 \sin (x^2) \d A =\begin {cases} &= \int _{\answer {0}}^{\answer {\pi }} \int _{y}^\pi 2 \sin (x^2) \d x \d y \\ &= \int _{0}^\pi \int _{\answer {0}}^{\answer {x}} 2 \sin (x^2) \d y \d x. \end {cases}$ If we were to use the first description of $T$ , we might have trouble finding an $x$ -antiderivative of $\sin (x^2)$ , so let’s try the second description. In that case, $\begin{align*} \iint_T 2 \sin (x^2) \d A &= \int_{0}^\pi 2 x \sin (x^2) \d x\\ &= \answer[given]{1 - \cos (\pi^2)}. \end{align*}$

It’s important to do a self-check to see if our purported value for an integral is at all plausible.

The region $T$ is a triangle with base $\answer {\pi }$ and height $\answer {\pi }$ , so the area of the region $T$ is $\answer {\pi ^2/2}$ which is about $5$ square units. In other words, $\iint _T 1 \d A = \answer {\pi ^2/2}$ which also means that $\iint _T 2 \d A = \answer {\pi ^2}.$ We are claiming that $\iint _T 2 \sin (x^2) \d A$ equals $1 - \cos (\pi ^2)$ , which is about $1.9$ .

When $0 \leq x \leq \pi$ , the value of $\sin (x^2)$ is sometimes positive, sometimes negative, but at least we know that $\answer {-2} \leq 2 \sin (x^2) \leq \answer {2},$ and this inequality then implies that $\left | \iint _T 2 \sin (x^2) \d A \right | \leq \left | \iint _T 2 \d A \right | = \answer {\pi ^2}.$ So $1.9$ is certainly in the ballpark of plausibility.

Polar coordinates

Evaluate the integral $\iint _Q \left (x + y\right ) \d A$ where $Q$ is the quarter circle, $Q = \{ (x,y) : \text {$x \geq 0$, $y \geq 0$, $x^2 + y^2 \leq 1$} \}.$

We use coordinates. $\begin{align*} \iint_Q \left(x + y\right) \d A &= \int_{\answer[given]{0}}^{\answer[given]{\pi/2}} \int_{0}^{1} (r \cos \theta + r \sin \theta) r \d r \d \theta \\ &= \int_{\answer[given]{0}}^{\answer[given]{\pi/2}} \eval{\answer[given]{ (r^3/3) \cos \theta + (r^3/3) \sin \theta }}_0^1\d \theta \\ &= \int_{\answer[given]{0}}^{\answer[given]{\pi/2}}\answer[given]{\frac{(\sin \theta + \cos \theta)}{3}}\d\theta\\ &= \eval{\answer[given]{\frac{\sin \theta - \cos \theta}{3}}}_{0}^{\pi/2} \\ &= \answer[given]{2/3}. \end{align*}$

Again consider the region $Q = \{ (x,y) : x \geq 0,\hspace {1ex} y \geq 0,\hspace {1ex} x^2 + y^2 \leq 1 \}.$ How does $A = \iint _Q x \d A$ compare to $B = \iint _Q y \d A?$

The region $Q$ is symmetric across the line $x = y$ . As a consequence of this, we might have computed that $A = \iint _Q x \d A = 1/3,$ and likewise $B = 1/3$ . Because of this, and the fact that the integral of a sum is the sum of integrals, we could have deduced $\begin{align*} A + B &= (1/3) + (1/3) \\ &= \iint_Q x \d A + \iint_Q y \d A \\ &= \iint_Q (x+y) \d A = 2/3. \end{align*}$

Spheres and hemispheres

Let $B$ be the region $B = \{ (x,y,z) : x^2 + y^2 + z^2 \leq 1\}$ .

Explain why $\iiint _B z^3 \d V = \answer [given]{0}$ .

This integral vanishes because integral over the northern hemisphere of $B$ will cancel the contribution from the southern hemisphere of $B$ .

Let $H$ be the region $H = \{ (x,y,z) : x^2 + y^2 + z^2 \leq 1 \text { and } z \geq 0 \}.$

Show that $\iiint _H z^3 \d V = \pi /12$ .

Unlike the previous example, this does not vanish.

We use

coordinates. $\begin{align*} \iiint_H z^3 \d V &= \int_{0}^{2\pi} \int_{0}^{\pi/2} \int_{0}^1 (\rho \cos \varphi)^3 \rho^2 \sin \varphi \d \rho \d \varphi \d \theta \\ &= \int_{0}^{2\pi} \int_{0}^{\pi/2} \int_{0}^1 \rho^5 \cos^3 \varphi \sin \varphi \d \rho \d \varphi \d \theta \\ &= \int_{0}^{2\pi} \int_{0}^{\pi/2} \eval{\answer[given]{\frac{\rho^6 \cos^3 \varphi \sin \varphi}{6}}}_0^1 \d \varphi \d \theta \\ &= \int_{0}^{2\pi} \int_{0}^{\pi/2} \answer[given]{\frac{\cos^3 \varphi \sin \varphi}{6}} \d \varphi \d \theta \\ &= \int_{0}^{2\pi} \eval{\answer[given]{\frac{-\cos^4 \varphi}{24}}}_{0}^{\pi/2} \d \theta \\ &= \int_{0}^{2\pi} \answer[given]{\frac{1}{24}} \d \theta \\ &= \answer[given]{\frac{\pi}{12}}. \end{align*}$

Again let $H$ be the region $H = \{ (x,y,z): \text {$x^2 + y^2 + z^2 \leq 1$ and $z \geq 0 $}\}.$ Set $A = \iiint _H z^3 \d V$ and $B = \iiint _H z^{10} \d V$ . How does $A$ relate to $B$ ?

Indeed, $A > B$ because, for points $(x,y,z) \in H$ , we have $-1 \leq z \leq 1$ and $z^3 \geq z^{10}$ . Moreover, except when $z = \pm 1$ , it is the case that $z^3 > z^{10}$ . By comparing the integrands, we can gain insight into the relative sizes of the integrals.

This same kind of thinking can lend insight into the question of what happens when $\iiint _H z^N \d V$ when $N$ is very large.

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Triangles

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Spheres and hemispheres

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