Math 234 — Third Midterm Review
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About double & triple integrals
The conceptual question.
To compute the total mass of the region divide it into small pieces
(“partition the region $\cR$”). Call the volume of one small piece
$\Delta V$. The mass of that piece will be approximately “density times
volume,” i.e.
\[
\mu(x, y, z) \Delta V
\]
where $(x,y,z)$ is a sample point in the small piece. To get the total mass
in the region $\cR$ we add over all the small pieces and get
\[
\text{Total mass} \approx \sum_{\text{all pieces}} \mu(x, y, z) \Delta V.
\]
As we choose finer partitions this approximation should get better, and in the limit we get
\[
\text{Total mass} = \iiint_\cR \mu(x, y, z) \; dV.
\]
To compute the total kinetic energy of the region we do almost the same thing: we divide
it into small pieces (“partition the region $\cR$”). Call the
volume of one small piece $\Delta V$. The kinetic energy of that piece will be
approximately “one half mass times velocity squared,” i.e.
\[
\frac12 \underbrace{\mu(x, y, z) \Delta V}_{\text{mass of small piece}} v(x, y, z)^2,
\]
where $(x,y,z)$ is a sample point in the small piece. To get the total kinetic energy
in the region $\cR$ we add over all the small pieces and get
\[
\text{Total K.E.} \approx \sum_{\text{all pieces}} \frac12 \mu(x, y, z) v(v, y, z)^2 \Delta V.
\]
As we choose finer partitions this approximation should get better, and in the limit we get
\[
\text{Total K.E.} = \iiint_\cR \frac12 \mu(x, y, z) v(x, y, z)^2 \; dV.
\]
Polar, Spherical and Cylindrical Coordinates.
The region for $I_1$ is the region that is between the circles with radius
$a$ and$b$. In polar coordinates it is given by
\[
0\leq \theta \leq 2\pi, \quad a\leq r\leq b
\]
Using $x=r\cos \theta$, $y=r\sin\theta$, we get
\begin{align*}
I_1 &= \int_0^{2\pi} \int_{r=a}^b (r\cos \theta)^2\, r dr d\theta \\
&= \int_0^{2\pi} \int_{r=a}^b \cos^2 \theta \, r^3 dr d\theta \\
&= \int_0^{2\pi} \cos^2 \theta \bigl[\tfrac14r^4\bigr]_a^b d\theta \\
&= \frac14 \bigl(b^4-a^4\bigr)\int_0^{2\pi} \cos^2 \theta d\theta \\
&= \frac14 \bigl(b^4-a^4\bigr) \pi.
\end{align*}
(use the double angle formula for the last integral).
The region for $I_2$ is contained in the region for $I_1$. It consists of
all points between the circles with radii $a$ and $b$ that lie above the graph
of $y=c|x|$. This graph is v-shaped and consists of two half lines. The region
is therefore given in polar coordinates by
\[
a\leq r\leq b, \quad \alpha\leq\theta\leq\beta,
\]
where the angles $\alpha$ and $\beta$ are determined by the slopes of the graph of $y=c|x|$.
We have $\tan\alpha = c$, so $\alpha = \arctan c$; the other angle is $\beta = \pi-\alpha$.
Thus we get
\begin{align*}
I_2 &= \int_\alpha^\beta \int_{r=a}^b (r\cos \theta)^2\, r dr d\theta \\
&= \int_\alpha^\beta \int_{r=a}^b \cos^2 \theta \, r^3 dr d\theta \\
&= \int_\alpha^\beta \cos^2 \theta \bigl[ \frac14r^4 \bigr]_a^b d\theta \\
&= \frac14 \bigl(b^4-a^4\bigr)\int_\alpha^\beta \cos^2 \theta d\theta \\
&= \frac14 \bigl(b^4-a^4\bigr) \int_\alpha^\beta \frac12 \{1+\cos 2\theta\} d\theta\\
&= \frac14 \bigl(b^4-a^4\bigr) \frac12 \{\beta-\alpha+\frac12\sin 2\beta-\frac12\sin2\alpha\}
\end{align*}
which one could simplify.
The region for $J_1$ is three
dimensional. It is the region contained between the two spheres with radius $a$
and $b$. The best choice of coordinates is spherical coordinates. In these
coordinates the region is described by
\[
a\leq \rho\leq b,\quad
0\leq \phi\leq \pi, \quad
0\leq \theta\leq 2\pi.
\]
The coordinate $x$ is $x=\rho\sin\phi\cos\theta$, and
the integral therefore is
\begin{align*}
J_1 &= \int_{\theta=0}^{2\pi}\int_{\phi=0}^\pi \int_{\rho=a}^b
(\rho\sin\phi\cos \theta)^2\, \rho^2 \sin\phi d\rho d\phi d\theta \\
&= \int_{\theta=0}^{2\pi}\int_{\phi=0}^\pi \int_{\rho=a}^b
\sin^3\phi\, \cos^2 \theta\, \rho^4 d\rho d\phi d\theta \\
&= \tfrac15\bigl(b^5-a^5\bigr) \int_{\theta=0}^{2\pi}\int_{\phi=0}^\pi
\sin^3\phi\, \cos^2 \theta\, d\phi d\theta \\
&= \tfrac15\bigl(b^5-a^5\bigr) \int_{\theta=0}^{2\pi} \cos^2\theta \left\{\int_{\phi=0}^\pi
\sin^3\phi\, d\phi\right\} d\theta
\end{align*}
At this point we have to know these integrals,
\[
\int_0^\pi \sin^3\phi\,d\phi \stackrel{u=-\cos \phi}= \int_{-1}^1 (1-u^2)du
=\frac43
\]
and
\[
\int_0^{2\pi} \cos^2\theta\, d\theta = \pi. \qquad\text{(double angle trick)}
\]
We get
\[
J_1 = \tfrac15\bigl(b^5-a^5\bigr) \frac43 \pi
\]
The region for $J_2$ consists of
all points between the spheres with radii $a$ and $b$ that also lie above the
cone $z=\sqrt{x^2+y^2}$. The opening angle of the cone is $\pi/4 = 45^\circ$,
so in spherical coordinates the region is described by
\[
a\leq \rho \leq b, \qquad
0\leq \phi\leq \frac\pi4, \qquad
0\leq \theta \leq 2\pi.
\]
The integral is therefore almost the same as $J_1$ except one of the integration
bounds are different:
\begin{align*}
J_2 &= \int_{\theta=0}^{2\pi}\int_{\phi=0}^{\pi/4} \int_{\rho=a}^b
(\rho\sin\phi\cos \theta)^2\, \rho^2 \sin\phi d\rho d\phi d\theta \\
&= \int_{\theta=0}^{2\pi}\int_{\phi=0}^{\pi/4} \int_{\rho=a}^b
\sin^3\phi\, \cos^2 \theta\, \rho^4 d\rho d\phi d\theta \\
&= \tfrac15\bigl(b^5-a^5\bigr) \int_{\theta=0}^{2\pi}\int_{\phi=0}^{\pi/4}
\sin^3\phi\, \cos^2 \theta\, d\phi d\theta \\
&= \tfrac15\bigl(b^5-a^5\bigr) \int_{\theta=0}^{2\pi} \cos^2\theta \left\{\int_{\phi=0}^{\pi/4}
\sin^3\phi\, d\phi\right\} d\theta
\end{align*}
Again using $u=-\cos\phi$, we have
\[
\int_0^{\pi/4} \sin^3\phi\,d\phi = \int_{-1}^{\sqrt2/2} (1-u^2)du
=\left[u-\tfrac13u^3\right]_{-1}^{\sqrt2/2}
=\tfrac5{12}\sqrt2 +\tfrac23
\]
so we get
\begin{align*}
J_2
&= \tfrac15\bigl(b^5-a^5\bigr) \int_{\theta=0}^{2\pi}
\cos^2\theta \left\{\int_{\phi=0}^{\pi/4} \sin^3\phi\, d\phi\right\} d\theta \\
&= \tfrac15\bigl(b^5-a^5\bigr)\; \bigl(\tfrac5{12}\sqrt2 +\tfrac23\bigr) \int_{\theta=0}^{2\pi}
\cos^2\theta d\theta \\
&= \tfrac15\bigl(b^5-a^5\bigr)\; \bigl(\tfrac5{12}\sqrt2 +\tfrac23\bigr) \pi.
\end{align*}
Switching the order of integration.
The domain of integration in the integral
\[
I = \int_0^{\infty} \int_0^x e^{-x}\;dy\, dx
=\iint_\cR e^{-x}\;dA
\]
is given by
\[
0\leq y\leq x, \qquad 0\leq x\leq \infty
\]
These inequalities describe the region in terms of vertical slices: for
each $x$ the $y$ values corresponding to points in $\cR$ satisfy $0 \leq y
\leq x$. If you describe the region in terms of horizontal slices, then
for any fixed $y$ we check which $x$ values correspond to points in $\cR$,
and we find $y\leq x \le \infty$. So the other description of $\cR$ is
\[
0\le y\le \infty, \qquad y\leq x\le \infty.
\]
This gives us two ways of computing the integral. First,
\begin{align*}
I
&= \int_0^{\infty} \int_0^x e^{-x}\;dy\, dx\\
&= \int_0^\infty e^{-x}\left\{ \int_0^xdy\right\} dx \quad\text{($e^{-x}$ is constant for
the inner integral)}\\
&= \int_0^\infty xe^{-x}\;dx \quad\text{(integrate by parts)}\\
&= \left[(x-1)e^{-x}\right]_0^\infty\\
&=1.
\end{align*}
The other way is to first integrate with respect to $x$:
\begin{align*}
I
&= \int_0^{\infty} \int_y^\infty e^{-x}\;dx\, dy\\
&= \int_0^{\infty} \left[-e^{-x}\right]_y^\infty dy\\
&= \int_0^{\infty} e^{-y}\;dy\\
&= 1.
\end{align*}