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\centerline
{
\bf Dr. Z's Math251 Handout \#15.8 [Triple Integrals in Cylindrical and
Spherical Coordinates]
}

By Doron Zeilberger

{\bf Problem Type 15.8a}: 
Evaluate
$$
\int\int\int_{E} F(x,y,z) \, dV \quad ,
$$
where $E$ is a solid region descibed in terms of
cylinders and other stuff.

{\bf Example Problem 15.8a}: Evaluate
$$
\int\int\int_{E} x^2 \, dV \quad ,
$$
where $E$ is the solid that lies within the cylinder
$x^2+y^2=1$, above the plane $z=0$, and below the cone
$z^2=4x^2+4y^2$.
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{\bf Steps} & {\bf Example} \cr  &  \cr
{\bf 1.} Chances are that you are supposed to use
{\it cylindrical coordinates}. Express $E$ in the
form
$$
E=\{ \,
(r, \theta, z) \, \vert \,
\alpha \leq \theta \leq \beta\, , \, 
$$
$$
h_1(\theta) \leq r \leq h_2 (\theta) \, , \,
u_1(r,\theta) \leq z \leq u_2(r, \theta )\} \quad .
$$
&
{\bf 1.} Since $x^2+y^2=r^2$, the cone $z^2=4x^2+4y^2$
can be writen $z^2=4r^2$. The cylinder $x^2+y^2=1$ is
really $r=1$, and this is the ``base''. 
$z^2=4r^2$ means that $z$ ranges between $-2r$ and $2r$.
{\bf but} we are also told that our solid is {\bf above}
the plane $z=0$, so $z$ ranges between $0$ and $2r$.
It turns out that
$$
E=\{ \,
(r, \theta, z) \, \vert \,
0 \leq \theta \leq 2 \pi \, , \, 
0 \leq r \leq 1 \, , \,
0 \leq z \leq 2r \} \quad .
$$

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{\bf 2.} Express the integrand $F(x,y,z)$ in cylindrical
coordinates,
using the ``dictionary'' 
$$
x=r \cos \theta \quad,\quad
y=r \sin \theta \quad,\quad
x^2+y^2=r^2 \quad.
$$
Also $dV=r dr d\theta dz$.
Then set up the volume integral as an iterated integral
$$
\int_{\alpha}^{\beta} \int_{h_1(\theta)}^{h_2(\theta)}
\int_{u_1(r,\theta)}^{u_2(r, \theta)} 
F(r \cos \theta ,r \sin \theta ,z) r \, dz \,dr \, d \theta \quad.
$$

&
{\bf 2.} 
$$
\int_{0}^{2\pi} \int_{0}^{1}
\int_{0}^{2r} (r \cos \theta)^2 r \, dz \,dr \, d \theta \quad.
$$

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{\bf 3.} Evalute the integral from the inside to the outside.
&
{\bf 3.} The inside integral is:
$$
\int_{0}^{2r} (r \cos \theta)^2 r \, dz =
r^3 \cos^2 \theta \int_{0}^{2r}  dz =
2 r^4 \cos^2 \theta  \quad .
$$
The middle integral is
$$
 \int_{0}^{1} \left [
\int_{-2r}^{2r} (r \cos \theta)^2 r \, dz \right ] \,dr
$$
$$
= \int_{0}^{1} 2 r^4 \cos^2 \theta \,dr
=\cos^2 \theta
\int_{0}^{1} 2 r^4 \,dr=
{{2} \over {5}} \cos^2 \theta \quad .
$$
The outer integral is
$$
\int_{0}^{2\pi} \left [ \int_{0}^{1}
\int_{-2r}^{2r} (r \cos \theta)^2 r \, dz \,dr \, 
\right ]
d \theta 
$$
$$
=\int_{0}^{2\pi} 
{{2} \over {5}} \cos^2 \theta \quad d \theta 
$$
$$
=
{{2} \over {5}} \int_{0}^{2\pi} 
{{1+\cos 2\theta} \over {2}} \quad d \theta 
$$
$$
=
{{2} \over {5}} \cdot
[{{\theta +(1/2)\sin 2\theta} \over {2}}] \Bigl \vert_{0}^{2 \pi}
={{2\pi} \over {5}} \quad.
$$
{\bf Ans.}: ${{2\pi} \over {5}}$.
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{\bf Problem Type 15.8b}: Evaluate
$$
\int\int\int_{E} F(x,y,z) \, dV \quad ,
$$
where $E$ is bounded by the $xz$-plane and the hemispheres
$y=\sqrt{r_1^2-x^2-z^2}$ and $y=\sqrt{r_2^2-x^2-z^2}$.

{\bf Example Problem 15.8b}: Evaluate
$$
\int\int\int_{E} x^2 \, dV \quad ,
$$
where $E$ is bounded by the $xz$-plane and the hemispheres
$y=\sqrt{1-x^2-z^2}$ and $y=\sqrt{4-x^2-z^2}$.
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{\bf Steps} & {\bf Example} \cr  &  \cr
{\bf 1.} This is best handled with spherical coordinates.
The hemispere $y=\sqrt{R^2-x^2-z^2}$  is half of the
sphere $x^2+y^2+z^2=R^2$ whose equation in spherical
coordinates is really simple: $\rho=R$.
Since $y>0$ (the square-root is always positive), the
range of $\theta$ is between $0$ and $\pi$.
$\rho$ is between $r_1$ and $r_2$ and $\phi$ has
its full range $0$ to $\pi$.
So
$$
E=\{ \, ( \rho, \theta, \phi ) \, \vert r_1 \leq \rho \leq r_2
\, , \, 0 \leq \theta \leq \pi \, , \, 0 \leq \phi \leq \pi \, \} \quad .
$$

&
{\bf 1.} Here the radii are $1$ and $2$ so
$$
E=\{ \, ( \rho, \theta, \phi ) \, \vert 1 \leq \rho \leq 2
\, , \, 0 \leq \theta \leq \pi \, , \, 0 \leq \phi \leq \pi \, \} \quad .
$$
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{\bf 2.} Using the `dictionary'
$$
x= \rho \sin \phi \cos \theta \, , \,
y= \rho \sin \phi \sin \theta \, , \,
$$
$$
z= \rho \cos \phi  \quad ; \quad
dV=\rho^2 \sin \phi \, d \rho \, d \theta \, d \phi \quad .
$$
Convert the volume integral into an iterated spherical
integral (using the description of $E$ in step 1).
$$
\int_{0}^{\pi}\int_{0}^{\pi}\int_{r_1}^{r_2} 
F(\rho \sin \phi \cos \theta,
\rho \sin \phi \sin \theta,
\rho \cos \phi) \rho^2 \sin \phi \, d \rho \, d \theta \, d \phi \quad .
$$


&
{\bf 2.} 
$$
\int_{0}^{\pi}\int_{0}^{\pi}\int_{1}^{2} 
(\rho \sin \phi \cos \theta)^2
\rho^2 \sin \phi \, d \rho \, d \theta \, d \phi
$$
$$
=\int_{0}^{\pi}\int_{0}^{\pi}\int_{1}^{2} 
\rho^4 \sin^3 \phi \cos^2 \theta \, d \rho \, d \theta \, d \phi \quad .
$$
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{\bf 3.} If you are lucky and all the limits of integration
are numbers (do not involve $\rho$, $\phi$ or $\theta$) and
the integrand is a product of functions of a single variable,
then the iterated integral is simply a product of three
simple integrals. Express the big integral like that,
 and evaluate each single integral separately. Then
multiply them together.

{\bf Warning:} This is only possible if {\it all} the
limits of integration are numbers and the integrand
is {\it completely} separable as a product of functions
of a single variable. 
&
{\bf 3.} 
$$
\int_{0}^{\pi}\int_{0}^{\pi}\int_{1}^{2} 
\rho^4 \sin^3 \phi \cos^2 \theta \, d \rho \, d \theta \,  d \phi
$$
$$
=\int_{0}^{\pi} \sin^3 \phi \, d \phi
\int_{0}^{\pi}  \cos^2 \theta  \, d \theta 
\int_{1}^{2} \rho^4   \, d \rho \quad .
$$

The first integral is ($u=\cos \phi$)
$$
\int_{0}^{\pi} \sin^3 \phi \, d \phi=
\int_{0}^{\pi} \sin^2 \phi \, d (-\cos \phi) =
$$
$$
\int_{0}^{\pi} (1- \cos^2 \phi) \, d (-\cos \phi) 
=-\int_{1}^{-1} (1- u^2) \, du =
$$
$$
-u+{{u^3} \over {3}} \Bigl \vert_{1}^{-1}
={{4} \over {3}} \quad .
$$
The second integral is
$$
\int_{0}^{\pi}  \cos^2 \theta  \, d \theta =
\int_{0}^{\pi}  {{1+ \cos 2 \theta} \over {2}}  \, d \theta =
$$
$$
={{\theta} \over {2}} + {{\sin 2 \theta} \over {4}} \Bigl \vert_{0}^{\pi}=
{{\pi} \over {2}} \quad .
$$
The third integral is
$$
\int_{1}^{2} \rho^4   \, d \rho = {{\rho^{5}} \over {5}} \Bigl \vert_{1}^{2}=
{{2^{5}-1^5} \over {5}}={{31} \over {5}} \quad .
$$
Multipluing these three single-integrals, we get that the
original triple integral equals
$$
{{4} \over {3}} \cdot {{\pi} \over {2}} \cdot {{31} \over {5}}=
{{62 \pi } \over {15}} \quad.
$$
{\bf Ans.}: ${{62 \pi } \over {15}}$.

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