Integrating 2

[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

[benorin]

http://i306.photobucket.com/albums/nn249/arinkovsky/sphere_cone.jpg

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?

[jeff1evesque]

http://i306.photobucket.com/albums/nn249/arinkovsky/sphere_cone.jpg

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?

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

JL