below and above by sphere x^2+y^2+z^2=9 and inside cylinder x^2 + y^2=4
using cylindrical, are these limits correct?
r: 0 to 2
theta: 0 to 2pi
z: - sqrt(9-r^2) to sqrt(9-r^2)
below and above by sphere x^2+y^2+z^2=9 and inside cylinder x^2 + y^2=4
using cylindrical, are these limits correct?
r: 0 to 2
theta: 0 to 2pi
z: - sqrt(9-r^2) to sqrt(9-r^2)
no, again this has to be done in sections. You're going to have a cylindrical portion and then 2 spherical end cap portions and you're going to have to write it as one integral plus twice another (the end caps are identical)
Looking at the picture you see you need to just integrate a cylinder of radius 2 from $z=-\sqrt{5}$ to $\sqrt{5}$
Then from $z=\sqrt{5}$ to $z=3$ you have a spherical cap that's the top portion of a sphere of radius 3.
The cylinder is easily integrated in cylindrical coordinates.
The spherical caps (there's an identical one on the bottom) are easily integrated in spherical coordinates.
So just break the integral up into those 3 pieces which are really just 2 pieces since two are identical.
The integral for the cylinder is trivial.
Finding the upper limit for $\theta$ for the caps is similar to the other problem you had.
I am sorry, but I am still unsure. So the first integral is:: r: 0 to 2 theta: 0 to 2pi z: - sqrt(5) to sqrt(5). And the second is:: r: 0 to 2 theta: 0 to 2pi z...
Or the second integral is r: 0 to 2 theta: 0 to 2pi z: - sqrt(9-r^2) to sqrt(9-r^2), and that is where hollywood gets $\displaystyle 2\pi \int_0^2 2r\sqrt{9-r^2}\, dr = 4\pi\left(9-\frac{5\sqrt{5}}{3}\right)$, the $\displaystyle 2\sqrt{9-r^2}$ is from the z.
Are you saying this can be done in one integral hollywood?