The surface area of the part of the sphere $x^2 + y^2 + z^2 = a^2 $ that lies within the cylinder $x^2 + y^2 = ax $ and above the $xy$-plane.
When determining boundaries, how do you get $ r = a \cos \theta $ and why is $ -\pi/2 \le \theta \le \pi/2 $ ?
Thank you!
The equation of the cylinder is a circle of radius $a/2$ centered at $(a/2,0)$ in the $xy$ plane. This is a circle tangent to the origin, and in polar coordinates, its equation is $r = a \cos \theta$. (When you convert, you get $r^2=ar\cos\theta$, then divide by $r$.) As $\theta$ goes from $-\pi$ to $\pi$, the circle is drawn twice, so we need only $-\pi/2 \leq \theta \leq \pi/2.$
$x^2+y^2=r^2$ and $x=r \cos (\theta)$
So,
$$x^2+y^2=ax$$
Means,
$$r^2=ar\cos (\theta)$$
Or when we are not the origin,
$$r=a \cos (\theta)$$
As to why the $-\frac{\pi}{2}$ to $\frac{\pi}{2}$, this should be apparent from this picture.
Note that the cylinder equation can be transformed as follows
$$
\begin{gathered}
x^{\,2} + y^{\,2} = a\,x\quad \Rightarrow \quad x^{\,2} - 2\frac{a}
{2}x + \left( {\frac{a}
{2}} \right)^{\,2} + y^{\,2} = \left( {\frac{a}
{2}} \right)^{\,2} \quad \Rightarrow \hfill \\
\left( {x - \frac{a}
{2}} \right)^{\,2} + y^{\,2} = \left( {\frac{a}
{2}} \right)^{\,2} \hfill \\
\end{gathered}
$$
So the base circle of the cylinder is centered at $C=(a/2,0)$ and has radius $a/2$,
which is half of that of the sphere.
Then the ray from the origin to a point $P$ on the cylinder base circle, making an angle $\theta$ with the $x$ axis
will correspond to a ray from $C$ to $P$ making an angle $2 \theta$ with the $x$ axis.
So $\theta$ will range from $-\pi /2$ to $\pi/2$, where it is tangent to the base circle, and
$$
\begin{gathered}
\left| {OP} \right| = \frac{a}
{2}\sqrt {\left( {1 + \cos \left( {2\theta } \right)} \right)^{\,2} + \sin ^{\,2} \left( {2\theta } \right)} = \frac{a}
{2}\sqrt {1 + \cos ^{\,2} \left( {2\theta } \right) + 2\cos \left( {2\theta } \right) + \sin ^{\,2} \left( {2\theta } \right)} = \hfill \\
= \frac{a}
{2}\sqrt {2 + 2\cos \left( {2\theta } \right)} = \frac{a}
{2}\sqrt {2 + 2\cos ^{\,2} \theta - 2\sin ^{\,2} \theta } = a\cos \theta \hfill \\
\end{gathered}
$$