Triple integral in cylindrical/spherical

[frozenguy]

Homework Statement

Use cylindrical coordinates to find the volume of the solid.
The solid is enclosed by the paraboloid z=x²+y² and the plane z=9

Homework Equations

z=r²

The Attempt at a Solution

So I'm getting close to the answer but not quite, and I keep getting a negative which doesn't make sense. And I think my limit on the second integration needs to be a function of theta.

I chose z=9 as my z upper limit and z=r2 as my lower and just used rdzdrd(theta) as my integrand. Used 0 for lower limit for both dr and dtheta and started to use sqrt(9/2) as my upper limits for both dr and d(theta) but then changed d(theta's) upper limit to 2pi.

I need help lol.
BTW.. My notation on this is sloppy. I originally started with the z limits reversed but changed it once I integrated.. sorry.

 

[elect_eng]

Homework Statement

Use cylindrical coordinates to find the volume of the solid.
The solid is enclosed by the paraboloid z=x²+y² and the plane z=9

I believe you have the limits for z reversed, and the limits for r should be 0 and 3.

 

[elect_eng]

I thought I would mention that there is a simple way to solve this problem by considering the total volume of the cylinder with radius R=3 and height H=9 and then subtracting off the excess volume. The excess volume can be expressed as an area spun around the z axis, and the result is very simple as follows.

$$V=2\pi R^2 H - 2 \pi \int_0^R f(x) x dx =81\pi-2\pi \int_0^3 x^3 dx ={{81 \pi}\over{2}$$

Little tricks like this can save time, or provide a way to check your answer when you are required to demonstrate your ability to do the triple integral.