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\centerline
{
\bf Dr. Z's Math251 Handout \#15.4a (2nd ed.)
[Double Integrals in Polar Coordinates]
}

By Doron Zeilberger

{\bf Problem Type 15.4aa}: Evaluate the integral
$$
\int\int_{D} F(x,y) \, dA \quad,
$$
where $D$ is a region best described in polar coordinates,
$$
D=\{ \, (r,\theta) \,  \vert \, \alpha \leq \theta \leq \beta \, ,
\, h_1 (\theta ) \leq r \leq h_2 (\theta) \, \} \quad .
$$
{\bf Example Problem 15.4aa}: 
Evaluate the integral
$$
\int\int_{D} e^{-x^2-y^2} \, dA \quad,
$$
where $D$ is the region bounded by the semi-circle
$x=\sqrt{25-y^2}$ and the $y$-axis.
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{\bf Steps} & {\bf Example} \cr  &  \cr
{\bf 1.} Draw the region and express it, if possible and
convenient, as
$$
D=
$$
$$
\{ \, (r,\theta) \,  \vert \, \alpha \leq \theta \leq \beta \, ,
\, h_1 (\theta ) \leq r \leq h_2 (\theta) \, \} \quad .
$$
Of course, in many problems, the $h_1(\theta)$ and/or $h_2(\theta)$
may be plain numbers (i.e. not involve $\theta$).
&
{\bf 1.} This is a {\bf semi}-circle, i.e. {\bf half} a circle,
center origin, radius $5$, and since it is bounded by the
$y$-axis, and $x \geq 0$, it is the {\bf right} half 

[Had it been $x=-\sqrt{25-y^2}$ it would have been the left-half. 
Had it been $y=\sqrt{25-x^2}$ it would have been the upper-half.
Had it been $y=-\sqrt{25-x^2}$ it would have been the lower-half.]

Since it is the right-half, $\theta$ ranges from
$\theta=-\pi/2$ (the downwards direction) to
$\theta=\pi/2$ (the upwards direction). For each
ray $\theta=\theta_0$, $r$, the distance from the
origin, ranges from $r=0$ to $r=5$ (and indeed does
not depend on $\theta$ in this problem). So our region
phrased in {\bf polar coordinates} is:

$$
D=\{ \, (r,\theta) \,  \vert \,-\pi/2 \leq \theta \leq \pi/2 \, ,
\, 0 \leq r \leq 5 \, \} \quad .
$$

\cr

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{\bf 2.} Rewrite the area integral
$$
\int\int_{D} F(x,y) \, dA \quad,
$$
in {\bf polar} coordinates by replacing 

$x$ by $r \cos \theta$, $y$ by $r \sin \theta$, $dA$ by $r \, dr \, d\theta$.

[{\bf shortcut:} Whenever you see $x^2+y^2$ you can replace it by $r^2$.]

Write it as an iterated integral 
$$
\int\int_{D} F(x,y) \, dA =
\int_{\alpha}^{\beta} \int_{h_1(\theta)}^{h_2(\theta)} F
(r \cos \theta \, , \, r \sin \theta ) \, r \, dr \, d  \theta
\quad ,
$$
with the $\theta$-integral being at the {\bf outside} and
the $r$-integral being in the {\bf inside}.
&
{\bf 2.} 
$$
\int\int_{D} e^{-x^2-y^2} \, dA 
$$
$$
=\int_{-\pi/2}^{\pi/2} \int_{0}^{5} e^{-r^2}\, r dr \, d  \theta
\quad .
$$

\cr
\bigskip & \bigskip \cr

{\bf 3.} Evaluate this iterated integral by first doing the
inner-integral (possibly getting an expression in $\theta$,
or just a number), and then the outer integral.
&
{\bf 3.}  The inside integral is (do the change-of-variable $u=-r^2$):
$$
 \int_{0}^{5} e^{-r^2} \, r dr \, = \, (-1/2) e^{-r^2}\Bigl \vert_{0}^{5}=
(1-e^{-25})/2 \quad,
$$
and the whole double-integral is
$$
\int_{-\pi/2}^{\pi/2} \int_{0}^{5} e^{-r^2}\, r dr \, d  \theta
$$
$$
=\int_{-\pi/2}^{\pi/2} \left [ \int_{0}^{5} e^{-r^2}\, r dr \right ]\, d  \theta
$$
$$
=\int_{-\pi/2}^{\pi/2} [(1-e^{-25})/2]\ \,\, d  \theta
=(1-e^{-25})/2 \int_{-\pi/2}^{\pi/2}  d  \theta=
$$
$$
[(1-e^{-25})/2 ][\pi/2 - (-\pi/2)]=
\pi (1-e^{-25})/2 \quad .
$$
{\bf Ans.:} $\pi (1-e^{-25})/2$ .

\cr}

{\bf Problem Type 15.4ab}: 
Find the volume of the solid
above the surface $z=f(x,y)$ and below the 
surface $z=g(x,y)$ .


{\bf Example Problem 15.4ab}: 
Find the volume of the solid
above the cone $z=\sqrt{x^2+y^2}$ and below the 
sphere $x^2+y^2+z^2=2 \,\,$ .
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{\bf Steps} & {\bf Example} \cr  &  \cr
{\bf 1.} Find the ``floor'', let's call it $D$,  by setting
$f(x,y)=g(x,y)$ (or if convenient already convert
to polar coordinates). 

&
{\bf 1.} In polar coordinates, the two surfaces are
$z=r$ and $z=\sqrt{2-r^2}$. Setting them equal gives
$r=\sqrt{2-r^2}$. Squaring both sides gives
$r^2=2-r^2$, which gives $2r^2=2$, which gives
$r^2=1$ and so $r=\pm 1$. But $r$ is never negative, so
$r=-1$ is nonsense. Hence
the ``floor'', $D$, is the region bounded by the circle $r=1$,
or, if you wish, the disk $r \leq 1$.

So
$$
D=\{ (r, \theta) \vert
0 \leq r \leq 1 , 0 \leq \theta \leq 2 \pi \} \quad .
$$
\cr

\bigskip & \bigskip \cr
{\bf 2.} The volume is the area integral of TOP-BOTTOM
$$
\int\int_{D} [f(x,y)-g(x,y)] \, dA \quad .
$$
Set it up. Then convert it to polar-coordinates.
&
{\bf 2.} The bottom is $z=\sqrt{x^2+y^2}$, and
in polar $z=r$, and the top is $x^2+y^2+z^2=2$
which is $z=\sqrt{2-x^2-y^2}$ and in polar $z=\sqrt{2-r^2}$.
So the volume in polar coordinates is
$$
\int_{0}^{2 \pi} \int_{0}^{1} [\sqrt{2-r^2}-r] \, r \, dr \, d \theta
\quad .
$$
\cr \bigskip & \bigskip \cr

{\bf 3.} Evaluate the iterated integral. First do the
inner integral (w.r.t. to $r$) getting an expression
in $\theta$ (or just a number), and then do the outer integral.
&
{\bf 3.} 
The inner integral is
$$
 \int_{0}^{1} [\sqrt{2-r^2}-r] \, r \, dr=
 \int_{0}^{1} [r \sqrt{2-r^2}-r^2] \, dr
$$
$$
= \int_{0}^{1} r (2-r^2)^{1/2} \, dr
-
 \int_{0}^{1} r^2 \, dr
$$
$$
=-(1/3) (2-r^2)^{3/2} \Bigl \vert_{0}^{1}
\, - \, r^3/3 \Bigl \vert_{0}^{1}
$$
$$
=-(1/3) (2-r^2)^{3/2} \Bigl \vert_{0}^{1}\, - \, r^3/3 \Bigl \vert_{0}^{1}
$$
$$
=-(1/3) [(2-1^2)^{3/2} \, - \, (2-0^2)^{3/2}]-1/3
$$
$$
=[2^{3/2}-2]/3=(2\sqrt{2}-1)/3 \quad .
$$
The whole integral is thus:
$$
\int_{0}^{2 \pi} \int_{0}^{1} [\sqrt{2-r^2}-r] \, r \, dr \, d \theta
$$
$$
=\int_{0}^{2 \pi} \left [ \int_{0}^{1} [\sqrt{2-r^2}-r] \, r \, dr 
\right ] \, d \theta
$$
$$
=\int_{0}^{2 \pi}  (2\sqrt{2}-1)/3 \, d \theta
$$
$$
= 2\pi(2\sqrt{2}-1)/3  \quad .
$$
{\bf Ans.:} The volume is $2\pi(2\sqrt{2}-1)/3$.
\cr}

{\bf Problem Type 15.4ac}: 
Evaluate  the iterated integral
by converting to polar coordinates.
$$
\int_{a}^{b} \int_{f_1(y)}^{f_2(y)} F(x,y) \,dx \,dy
$$


{\bf Example Problem 15.4ac}: Evaluate  the iterated integral
by converting to polar coordinates.
$$
\int_{0}^{3} \int_{-\sqrt{9-y^2}}^{\sqrt{9-y^2}} x^2y \,dx \,dy
$$
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{\bf Steps} & {\bf Example} \cr  &  \cr
{\bf 1.} By looking at the limits of integration of
the outer and inner integral signs, figure out
the region $D$.
$$
D=\{ (x,y) \, \vert  \, a \leq y \leq b \, , \, f_1(y) 
\leq x \leq f_2(y) \, \} \quad .
$$
Draw this region, and express it in polar coordinates
$$
D=\{ (r,\theta) \, \vert  \, \alpha \leq \theta
\leq \beta  \, , \, g_1(\theta) 
\leq r \leq g_2(\theta) \, \}
$$

&
{\bf 1.} Our region is:
$$
D=\{ (x,y) \, \vert  \, 0 \leq y \leq 3 \, , \, 
-\sqrt{9-y^2} \leq x \leq \sqrt{9-y^2} \, \} \quad .
$$
Drawing it (do it!), we see that this is the upper-half
of the circle whose center is the origin and whose
radius is $3$. In polar coordiantes it is:
$$
D=\{ (r,\theta) \, \vert  \, 0 \leq \theta
\leq \pi  \, , \, 0
\leq r \leq 3 \, \} \quad .
$$
\cr
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{\bf 2.} Write the iterated integral as an area integral,
then convert it to an iterated integral in polar coordinates.
Use the ``disctionary'' $x=r\cos \theta$
$y=r\sin \theta$ $\,dx\,dy=r \, dr \, d\theta$.
&
{\bf 2.} 
$$
\int_{0}^{3} \int_{-\sqrt{9-y^2}}^{\sqrt{9-y^2}} x^2 y \,dx \,dy
$$
$$
=
\int_{0}^{\pi} \int_{0}^{3} (r\cos \theta)^2 (r \sin \theta) r  \,dr \,d\theta
$$
$$
=\int_{0}^{\pi} \int_{0}^{3} r^4 \sin \theta \cos^2 \theta  \,dr \,d\theta
\quad .
$$
\cr
\smallskip & \smallskip \cr

{\bf 3.} Evaluate that iterated integral by doing the inner
integral first, and then the outer integral.

&
{\bf 3.} The inner integral is
$$
\int_{0}^{3} r^4 \sin \theta \cos^2 \theta  \,dr=
\sin \theta \cos^2 \theta \int_{0}^{3} r^4   \,dr
$$
$$
=\sin \theta \cos^2 \theta \,
\left [ {{r^5} \over {5}} \, \Bigl \vert_{0}^{3} \right ] 
$$
$$
={{243} \over {5}} \sin \theta \cos^2 \theta \quad .
$$
The outer integral is:
$$
\int_{0}^{\pi} \int_{0}^{3} r^4 \sin \theta \cos^2 \theta  \,dr \,d\theta
$$
$$
=\int_{0}^{\pi} 
\left [ \int_{0}^{3} r^4 \sin \theta \cos^2 \theta  \,dr \right ] \,d\theta
$$
$$
=\int_{0}^{\pi} {{243} \over {5}}  \cos^2 \theta \sin \theta \,d\theta
={{243} \over {5}}
\int_{0}^{\pi}  \cos^2  \theta \sin \theta \,d\theta
$$
$$
=
{{243} \over {5}} \cdot \, \left [
{{-\cos^3  \theta} \over {3}} \Bigl \vert_{0}^{\pi} \right ]
=
{{81} \over {5}} \cdot \, \,( -\cos^3 ( \pi)  - -\cos^3 (0) )
={{162} \over {5}}\quad .
$$
{\bf Ans.:} ${{162} \over {5}}$.

\cr}

{\bf A Problem from a Previous Final}

Use polar coordinates to compute the double
integral
$$
\int \, \int_D xy \, dA \quad ,
$$
where 
$$
D= \{ (x,y) \, \vert \, x^2+y^2 \leq 4 \, , \,
x \geq 0 \, , \, y \geq 0 \, \} \quad .
$$
\bigskip
(Hint: recall the trig identity $\sin 2 \theta = 2 \sin \theta \cos \theta$)

{\bf Ans.:} $2$.


{\bf Another Problem from a Previous Final}

Evaluate the iterated integral by converting it to polar coordinates
$$
\int_0^1 \, \int_0^{\sqrt{1-x^2}} e^{x^2+y^2} \, dy \, dx 
\quad .
$$

{\bf Ans.}: $\pi(e-1)/4$.

\end



