# Volume problem (triple integral)

• Aug 3rd 2009, 01:57 PM
alan collins
Volume problem (triple integral)
Use cylindrical coordinates to evaluate the triple integral https://webwork2.uncc.edu/webwork2_f...3d99264841.png, where E is the solid bounded by the circular paraboloid https://webwork2.uncc.edu/webwork2_f...a15e5869f1.png and the https://webwork2.uncc.edu/webwork2_f...4304f3b401.png -plane.

This is giving me trouble and I'm not sure about the limits of integration. The fallowing one is sort of similar and also driving me crazy.

Find the volume of the solid that lies within the sphere x^2 + y^2 + z^2 = 9, above the xy plane, and outside the cone z = 4sqrt{x^2+y^2}.
• Aug 3rd 2009, 04:38 PM
slider142
For the first question, what do you propose for the integrand in cylindrical coordinates? Do you know the geometry of the function in the integrand? If the geometry is elusive, use the cylindrical-Cartesian coordinate relations themselves. What are r, $\theta$ and z in terms of x, y and z? If you are new to changing coordinates, these are the first equations you should write down. It may help to note that cylindrical coordinates are just polar coordinates with the z-axis added.
• Aug 3rd 2009, 05:54 PM
alan collins
For the second problem there, I have calculated the area of the top portion of that sphere to be 486pi/5. Now I need to calculate the area of the cone inside this sphere and subtract that from this value that I already have, is this correct?

Then, should I use spherical coordinates to solve the integral of the cone inside the sphere?
• Aug 4th 2009, 02:56 AM
slider142
Quote:

Originally Posted by alan collins
For the second problem there, I have calculated the area of the top portion of that sphere to be 486pi/5. Now I need to calculate the area of the cone inside this sphere and subtract that from this value that I already have, is this correct?

Then, should I use spherical coordinates to solve the integral of the cone inside the sphere?

Can you show how you got that value? The first equation of the second problem is the equation of a sphere of radius 3 centered at the origin. A short method of getting the volume contained in the hemisphere is to use the formula for the volume of a sphere: $\frac{4}{3}\pi r^3$ and halve it. Then you can take away the volume of the cone, as you have surmised.
Good choice to use spherical coordinates for the conical part; all of your limits will be constant.
• Aug 4th 2009, 12:59 PM
alan collins
Yes, thanks again very much. I have it now and the constant limits make the problem so much easier to work on.