# Thread: Volume Integral between two parabloids

1. ## Volume Integral between two parabloids

I am asked to find the volume between the two paraboloids z=9-x^2-y^2 and z=x^2 + y^2. Converting to cylindrical polar coordinates, these become z=9-r^2 and z=r^2. Clearly z ranges between 0 and 9 and I think r ranges from -sqrt(z) to + sqrt(z)? This given the integral which evaluates to 0! Have I made a mistake in my working?

2. There seems to be quite a lot of guessing going on in there.

Solve r^2 = 9 - r^2 to find the limits of 'r'.

You may wish to do 'z' first and have limits [ r^2 , 9 - r^2 ]

3. Originally Posted by TKHunny
There seems to be quite a lot of guessing going on in there.

Solve r^2 = 9 - r^2 to find the limits of 'r'.

You may wish to do 'z' first and have limits [ r^2 , 9 - r^2 ]
Ok, I see what you are saying. Solving that equation, one obtains the limits [-3/sqrt(2),3/sqrt(2)] for r. Then z has limits [r^2,9-r^2] and theeta limits [0,2pi]. I then get the volume integral which still equals 0!

4. 'r' limits are no good. You are thinking rectangular, not polar. Try [0, 3/sqrt(2)]

5. Originally Posted by StaryNight
Ok, I see what you are saying. Solving that equation, one obtains the limits [-3/sqrt(2),3/sqrt(2)] for r. Then z has limits [r^2,9-r^2] and theeta limits [0,2pi]. I then get the volume integral which still equals 0!
No, r does NOT go from $-3/\sqrt{2}$ to $3/\sqrt{2}$. In cylindrical or polar coordinates r is the distance from 0 to the point and is never negative.

$\int_0^{3/\sqrt{2}}\int_{r^2}^{9- r^2} r dzdr= \int_0^{3/\sqrt{2}} (9- 2r^2)rdr$