Triple Integral of a cored apple Use a triple integral in spherical coordinates to ﬁnd the volume, $V$, of a cored apple, which consists of a sphere of radius $2$, $x^2 + y^2 + z^2 = 4$, and a cylindrical hole of radius one, $x^2 + y^2 = 1$. In other words, ﬁnd the volume of the sphere with the cylinder removed.
I know that $\theta$ goes from $0$ to $2\pi$ since the sphere is complete. What I don't understand is how to:
1) Convert rectangular coordinates to spherical coordinates. My textbook gives a terrible explanation.
2) Find the bounds of the triple integral. As I said, $\theta$ goes from $0$ to $2\pi$. I have no clue as to how to find $p$ or $\varphi$. The center being removed is also throwing me.
Could some explain as to how to go about this? I'm not necessarily looking for answer but a means to solve this problem. We have an exam next week and I would really love to be able to understand it.
 A: Here, I evaluate the forth part of the region so after evaluating the triple integral  you should multiply the result to $4$. According to Fig below, we have a rectangle triangular  $OCD$. So $\sin(\phi_0)=1/2$ and so $\phi_0=\pi/6$. It means that $\phi\in [\pi/6,\pi/2]$. As you found out the rang of $\theta$ is $[0,\pi/2]$. But about the $\rho$:
You see that our region is starting to shape from the intersection till we meet $z=0$. For finding the range of $\rho$, make an arrow arbitrary to cut the cylinder at $A$ and next cut the sphere at $B$. What is $\rho$ at $A$ and what is at $B$? We know that in a spherical coordinates: $$x=\rho\sin(\phi)\cos(\theta)\\y=\rho\sin(\phi)\sin(\theta)\\z=\rho\cos(\phi)$$ then the equation of our cylinder $x^2+y^2=1$  be converted to $\rho^2\sin^2(\phi)=1$ or $\rho=\frac{1}{\sin(\phi)}$. This is the first coordinate $\rho$ at $A$. Clearly, $\rho$ at $B$ is $2$. Hence you have: $$\int_0^{\pi/2}d\theta\int_{\pi/6}^{\pi/2}\sin(\phi)d\phi\int_{\frac{1}{\sin(\phi)}}^{2}\rho^2d\rho$$.

A: This is the sneaky solution.  No integration needed. Consider the case where the radius of the cylindrical hole is zero. The length of the hole is then twice the radius of the sphere and the volume of the remainder is $\frac{4 \pi}{3}$ $(\frac{L}{2})^3=\frac{\pi L^3}{6}$ where $L$ is the length of the hole. With 
$y=\frac{1}{2}$ we get $L=\sqrt{3}$ and $V=\frac{pi 3^{3/2}}{6} =2.72$ approx. 
A: I would use cylindrical coordinates and not spherical coordinates.
$x = r\cos\theta\\
y = r\sin\theta\\
z = z$
What does our region look like under this parameterization?
$x^2 + y^2 = 1$ becomes $r^2 = 1$ and $x^2 + y^2 + z^2 = 4$ becomes $r^2 + z^2 = 4$ 
We also need the "jacobain."
The jacobian measures how much volume it taken up when we nudge $r, \theta, z$ each by a small amount.
$dx\ dy\ dz = \left|\begin{array}{}\frac {\partial x}{\partial r}&\frac{\partial x}{\partial \theta}&\frac{\partial x}{\partial z}\\
\frac {\partial y}{\partial r}&\frac{\partial y}{\partial \theta}&\frac{\partial y}{\partial z}\\
\frac {\partial z}{\partial r}&\frac{\partial z}{\partial \theta}&\frac{\partial z}{\partial z}\end{array}\right|\ dz\ dr\ d\theta = 
\left|\begin{array}{}\cos\theta&-r\sin\theta&0\\
\sin\theta&r\cos\theta&0\\0&0&1\end{array}\right| = r\ dz\ dr\ d\theta$
$\int_0^{2\pi}\int_1^2\int_{-\sqrt{4-r^2}}^{\sqrt{4-r^2}} r \ dz\ dr\ d\theta$
If you wanted to use spherical
$x = \rho\cos\theta\sin\phi\\
y = \rho\sin\theta\sin\phi\\
z = \rho\cos\phi$
The cylinder 
$\rho^2\sin\phi^2 = 1\\\rho = csc\phi$
The sphere is $\rho = 1$
And we have a the circles where the two intersect.
$\csc\phi = 1\\
\rho = \frac {\pi}{6}, \frac {5\pi}{6}$
The jacobian:
$dx\ dy\ dz = \left|\begin{array}{}\cos\theta\sin\phi&-\rho\sin\theta\sin\phi&\rho\cos\theta\cos\phi\\
\sin\theta\sin\phi&\rho\cos\theta\sin\phi&\rho\sin\theta\cos\phi\\
\cos\phi&0&\rho\sin\phi\end{array}\right| = \rho^2\sin\phi \ d\rho\ d\phi\ d\theta$
$\int_0^{2\pi}\int_{\frac{\pi}{6}}^{\frac{5\pi}{6}}\int_{\sec\phi}^2 \rho^2\sin\phi\ d\rho\ d\phi\ d\theta$
