Change of Variables to find the volume of a part of a sphere in CYLINDRICAL coords

[jmm5872]

Make the indicated change of variables (do not evaluate) (Not sure how to write an iterated integral with bounds so I will try and explain by just writing the bounds)
(I also tried using the symbols provided, but everything I tried just put a theta in here so I gave up)

\int\int\intxyz dzdxdy

-1 < z < 1
-sqrt(1-y^2) < x < sqrt(1-y^2) ... circle with radius 1
-sqrt(4-x^2-y^2) < y < sqrt(4-x^2-y^2) ... sphere with radius 4


Relevant equations:
x = rcos\theta
y = rsin\theta
z = z

Attempt:
I understand the bounds, it is a sphere with a radius of 4, however, the bounds constrain this to a cylinder with a radius of 1 along the z-axis. This makes it a cylinder with a rounded top and bottom (the only part of the sphere left).

My initial change of the bounds was this:
-4 < z < 4
0 < r < 1
0 < theta < 2pi

However I know this is wrong because these bounds only make a cylinder with radius 1 and length 8, they do not account for the rounded "caps."

My only idea was that I would have to split into two separate integrals, one for the cylinder and one for the top cap (double that one due to symmetry). But it seems like there should be a better way to account for the endcaps.

Thanks

 

[jmm5872]

Okay I think I figured it out.

The bound of z would simply be this:

-sqrt(4-r^2) < z < sqrt(4-r^2)

and r and theta:

0 < r < 1
0 < theta < 2pi

A lot easier than I was making it, I guess that's why no one answered me.

 

[Mark44]

Okay I think I figured it out.

The bound of z would simply be this:

-sqrt(4-r^2) < z < sqrt(4-r^2)

and r and theta:

0 < r < 1
0 < theta < 2pi

A lot easier than I was making it, I guess that's why no one answered me.

For future reference, here's the LaTeX for your iterated integral, showing the limits of integration. To see what I did, click the integral expression.
\int_{\theta=0}^{2\pi} \int_{r=0}^1 \int_{z=-\sqrt{4 - r^2}}^\sqrt{4 - r^2}}{r^2cos(\theta)sin(\theta) z~ r~dz~dr~d\theta

 

[jmm5872]

Thanks, good to see the LaTex. It will help in the future.