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.

1 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.

2 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 rectangularpolar 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

\[ B = \iint _Q y \d A? \]

$A < B$ $A = B$ $A > B$

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*}

3 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 cylindricalspherical 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 $B$?

$A < B$ $A = B$ $A > B$

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.