# Thread: calculus, double integral in polar coordinates

1. ## calculus, double integral in polar coordinates

The problem is:
Use polar coordinates to find the volume of the given solid bounded by the paraboloid z=10-3x^2-3y^2 and the plane z=4.

Normally I can do these integrals, but what's throwing me off is that the region isn't bounded by the xy-plane, and the whole region lies above it. I'm not sure how to set it up. I know that it will be the double integral of 10-3r^2 and with theta going from 0 to 2pi and r going from 0 to ... I'm not sure. I'm really just confused. Thank you!

2. Originally Posted by sfitz
The problem is:
Use polar coordinates to find the volume of the given solid bounded by the paraboloid z=10-3x^2-3y^2 and the plane z=4.

Normally I can do these integrals, but what's throwing me off is that the region isn't bounded by the xy-plane, and the whole region lies above it. I'm not sure how to set it up. I know that it will be the double integral of 10-3r^2 and with theta going from 0 to 2pi and r going from 0 to ... I'm not sure. I'm really just confused. Thank you!
Just subtract 4 from both equations and do the area under z=6-3r^2 and above z=0 (the xy plane). if you plug in z=0 to the first equation you get 0=6-3r^2. so you have 3r^2=6 which leads to r=±sqrt(2). So your r goes from -sqrt(2) to sqrt(2).