Triple integral in cylindrical coordinates?

Let x = rcos(t)

Let y = rsin(t)

Let z = h

The Jacobian for cylindrical coordinates is r

The bounds of integration are:

0<= r <= 4 (representiing distance from the z axis)

pi/4 <= t <= 3pi/4 (representing the angle about the z axis)

-1 <= h <= 1 (representing the height along the z axis)

Now, x^2 + y^2 + z^2 = r^2(cos^2(t) + sin^2(t)) + h^2

= r^2 + h^2

So, after multiplying by the Jacobian r, the integrand becomes

r^3 + r*h^2.

Integrate with respect to r first, obtaining

r^4/4 + r^2*h^2/2 from 0 to 4...so this is

64 + 8*h^2

Integrate this with respect to h, obtaining

64h + 8/3*h^3 from -1 to 1...so this is

128 + 16/3 = 400/3

Integrate this with respect to t, which is simply multiplication by

(3pi/4 - pi/4) = pi/2

So the final answer is 200*pi/3

It's a pain to type this out in plain text, but I hope you can get it:

Integrate (region w) x^2 + y^2 + z^2 dV

Convert to polar coordinates.

Substitute x = r cos θ

Substitute y = r sin θ

Limits:

0 ≤ r ≤ 4

pi/4 ≤ θ ≤ 3pi/4

-1 ≤ z ≤ 1

I'm going to short hand INT for integrate symbol here:

INT 0:4 INT pi/4:3pi/4 INT -1:1 ((rcosθ)^2 + (rsinθ)^2 + z^2) r dz dθ dr

INT 0:4 INT pi/4:3pi/4 INT -1:1 r^3cos^2θ + r^3sin^2θ + rz^2 dz dθ dr

The rest should be fairly straightforward. Don't forget that when you convert to polar coordinates, dx dy dz becomes r dr dθ dz. That extra r is crucial!