Triple integral involving spherical coordinates I am extremely confused on this question and I am wondering if anyone can help me.
The question says:
Consider the Integral:
$$\int_0^8 \int_{-\sqrt{64-y^2}}^{\sqrt{64-y^2}}\int_{\sqrt{x^2+y^2}}^{ 8} 1\, dx\, dy \, dz\, .$$

I am completely lost on how to solve this problem. Any help would really be appreciated.
 A: Here's answers...
Consider the integral given. It runs on 0 to 8 for the outermost bound, then the next bound runs $\pm\sqrt{64-y^2}.$ That indicates the semicircle portion of the origin-centered circle of radius 8 that has positive y-coordinate. And the last part give the semicircle a 3-d shape.
So what is this?
It's a quarter of a sphere. Because the final limit goes to 8, the quarter-circle is definitely bounded by z=8 and x=0.
So your limits are 0 to 8 (for the radius), 0 to $\frac{\pi}{2},$ (for $\theta$), and 0 to $\pi$ (for $\phi$). Or at least I'm fairly sure.. I could have screwed up. Again.
Part (c)'s answer is obvious: We can already calculate the volume of this solid with formulas we know.
That means that the integral's result (and the volume of the solid given by the triple integral) is $\frac{\frac{4}{3}\pi(8)^3}{4}=\boxed{\frac{512\pi}{3}}.$
A: You have to convert the equations of the surfaces to spherical coordinates first.
So $z=\sqrt{x^2+y^2}$ becomes $\phi=\frac{\pi}{4}$ (the side surface of the cone) and $\rho=0$ (the point at the origin.) 
$z=8$ becomes $\rho=8\sec{\phi}$
The region in the $xy$ plane appears to be the half disk where $y>0$, so your $\theta$ limits should be from $0$ to $\pi$.
The limits for $\phi$ should be from $0$ (the positive z direction) to $\frac{\pi}{4}$ (the side of the cone.)
The limits for $\rho$ should be from the inside surface $\rho=0$ to the outside surface $\rho=8\sec{\phi}$
A: As we see in 
$$\int_0^8 \int_{-\sqrt{64-y^2}}^{\sqrt{64-y^2}}\int_{\sqrt{x^2+y^2}}^{ 8} 1\, dx\, dy \, dz$$
we have 
$$\sqrt{x^2+y^2}\leq z\leq8$$
$$-\sqrt{64-y^2}\leq x\leq \sqrt{64-y^2}$$
$$0\leq y\leq 8$$
clearly, this shape is a cone (incomplete) and with substituting spherical coordinates:
$$0\leq \theta\leq \pi$$
$$0\leq \phi\leq \dfrac{\pi}{4}$$
$$0\leq \rho\leq \dfrac{8}{\cos\phi}$$
so
$${\bf V}=\int_0^{\pi}\int_0^\frac{\pi}{4}\int_0^\frac{8}{\cos\phi}{\rho}^2\sin\phi\,d{\rho}\,d{\phi}\,d{\theta}=\color{blue}{\dfrac{256\pi}{3}}$$
