Integrating (triple) over spherical coordinates

[MorallyObtuse]

Hi,

Set up the triple integral in spherical coordinates to find the volume bounded by \(\displaystyle z = \sqrt{4-x^2-y^2}\), \(\displaystyle z=\sqrt{1-x^2-y^2}\), where \(\displaystyle x \ge 0\) and \(\displaystyle y \ge 0\).

\(\displaystyle \int_0^{2\pi} \int_0^2 \int_{-\sqrt{4-x^2-y^2}}^{\sqrt{4-x^2-y^2}} r\ dz\ dr\ d\theta\)

 

[Ackbach]

I don't think you're using spherical coordinates yet - your integrals look more like cylindrical. This problem could probably be done either way, but the problem did say to use spherical. So, assuming that $\varphi$ is the azimuthal angle, and $\theta$ is the polar angle (some authors switch these two), your volume element is given by
$$dV=r^2 \sin(\theta) \, dr \, d\varphi \, d\theta.$$
Can you continue from here?

 

[Prove It]

Hi,

Set up the triple integral in spherical coordinates to find the volume bounded by \(\displaystyle z = \sqrt{4-x^2-y^2}\), \(\displaystyle z=\sqrt{1-x^2-y^2}\), where \(\displaystyle x \ge 0\) and \(\displaystyle y \ge 0\).

\(\displaystyle \int_0^{2\pi} \int_0^2 \int_{-\sqrt{4-x^2-y^2}}^{\sqrt{4-x^2-y^2}} r\ dz\ dr\ d\theta\)

You should also note that both of the bounding solids are hemispheres centred at the origin, the top being of radius 2 and the bottom of radius 1.

But because we are told x >= 0 and y >= 0, we are only going to be in the first octant.

That means your longitudinal angles do NOT sweep out a full circle, they will only sweep out a quarter circle. Your latitudinal angles only sweep out the top half.