# 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.

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Think about my approach below. – Babak S. Oct 31 '12 at 9:41
does the volume of this shape not need to be multiplied by 8 because there are 8 octants and the graph above in only in the first octant? – user108829 Nov 14 '13 at 17:55

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: $$\int_0^{\pi/2}d\theta\int_{\pi/6}^{\pi/2}\sin(\phi)d\phi\int_{\frac{1}{\sin(\phi)}}^{2}\rho^2d\rho$$.