- #1
DWill
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Homework Statement
Set up triple integrals for the volume of the sphere rho = 2 in (a) spherical, (b) cylindrical, and (c) rectangular coordinates.
Homework Equations
Volume in cylindrical coordinates: Triple integral of dz r dr d(theta) over region D.
Volume in spherical coordinates: Triple integral of [p^2 sin(phi) dp d(phi) d(theta)] over region D.
p = rho
The Attempt at a Solution
Trying to find limits of integration:
For (a), I looked at the limits starting with theta, then phi and then p. Because the region is a sphere, starting from the positive x-axis I thought theta would range from 0 to 2pi (make one complete revolution).
phi I thought should range from the positive z-axis to the negative z-axis, so from 0 to pi.
And finally p should range from 0 to 2.
However, the answers are the following:
8 ||| p^2 sin(phi) dp d(phi) d(theta) (note: using "|||" for triple integral)
with these limits of integration:
0 <= theta <= pi/2
0 <= phi <= pi/2
0 <= p <= 2
Of these, I only got the limit for p correct. Can someone explain with theta and phi are only from 0 to pi/2? Also, where did the 8 come from before the triple integral?
(b) Using a similar approach, I thought these should be limits of integration: 0 <= theta <= 2pi, 0 <= r <= 2, 0 <= z <= sqrt(4-r^2)
The correct answers are:
8 ||| r dz dr d(theta)
with these limits of integration:
0 <= theta <= pi/2
0 <= r <= 2
0 <= z <= sqrt(4-r^2)
Once again I have no idea how to get those correct limits for theta and z. Since the graph is a sphere, I thought theta has to range from 0 to 2pi. And again I don't know where the 8 is coming from.
I figure if I can understand these then rectangular coordinates shouldn't be much harder. Can anyone explain these to me? Thanks