determine $\iiint zdv$ using spherical coordinates Let $D$ be the region enclosed by the surface $z = \sqrt{4-x^2-y^2}$ and $z = -\sqrt{x^2+y^2}$
Determine:
$$\iiint zdv$$
using Spherical coordinates.
I'm trying to solve the above but I'm unsure how to go about it.
My attempt has been:
$$\int_0^{2\pi}\int_0^{2}\int_{-\sqrt{x^2+y^2}}^{\sqrt{4-x^2-y^2}} zdv$$
which I then integrate the first layer leaving me with:
$$\int_0^{2\pi}\int_0^{2} 2-x^2-y^2dv$$
Should I have subbed in the spherical coordinates right at the start and integrated it as:
$$\int_0^{2\pi}\int_0^{2}\int_{-\sqrt{x^2+y^2}}^{\sqrt{4-x^2-y^2}}p\cos\phi p^2\sin\phi dp d\theta d\phi$$
Based on a users input, would I have to first rewrite $z = \sqrt{4-x^2-y^2}$ and $z = -\sqrt{x^2+y^2}$
In terms of spherical coordinates?

Combine $z = \sqrt{4-x^2-y^2}$ and $z = -\sqrt{x^2+y^2}$ to get $z=-\sqrt2$, which corresponds to the boundary $\cos\theta = \frac zr =-\frac{\sqrt2}2$, or $\theta =\frac{3\pi}4$, in spherical coordinates.
Then, the integral is
$$\int_V zdv = 2\pi \int_{\frac{3\pi}4}^\pi\int_0^{2}(r\cos\theta )\>r^2\sin\theta drd\theta=- 2\pi
$$
