# Integrating 2

1. Jul 8, 2009

### jeff1evesque

1. The problem statement, all variables and given/known data
Using Spherical coordinates, find the volume of the solid enclosed by the sphere $$x^2 + y^2 + z^2 = 4a^2$$ and the planes z = 0 and z = a.

2. Relevant equations
I have the solutions to this problem, and it is done by integrating two parts:
$$V = V_{R=const.} + V_{z = const.}$$

The limits for $$V = V_{R=const.}$$ are
$$[0 \leq \phi \leq 2\pi], [\frac{\pi}{2} - sin^{-1}(\frac{1}{2}) \leq \theta \leq \frac{\pi}{2}], [0 \leq R \leq 2 \pi]$$

The limits for $$V_{z = const.}$$ are
$$[0 \leq \theta \leq \frac{\pi}{2} - sin^{-1}(\frac{1}{2})], [0 \leq R \leq \frac{a}{cos(\theta)}], 0 \leq \phi \leq 2\pi]$$

3. The attempt at a solution

Could someone explain to things to me:
1. Why there are two things we are integrating: $$V = V_{R=const.} + V_{z = const.}$$ I would think there should be only one integral, one that is bounded between z = 0, and z = a within the given sphere.
2. Why the limits are defined as it is- more specifically, the limits for $$\theta$$ for $$V = V_{R=const.}$$, and $$\theta, R$$ for $$V_{z = const.}$$

Thanks so much,

JL

2. Jul 9, 2009

### benorin

In above plot, green surface is V_R=const. and, imagining the blue cone surface filled to the brim would be the Vz=const. Two integrals are needed because why? What is the equation of the blue surface in spherical coords?

3. Jul 9, 2009

### jeff1evesque

Oh, ok thanks- I think I understand this problem now. By the way, that's a really cool picture.

JL