The solid bounded above by the cylinder z=4-x^2 and below by the paraboloid z=x^2+3y^2.

- Apr 22nd 2010, 01:15 PMasdf122345Need help setting up this triple intergral for the volume of the solid.
The solid bounded above by the cylinder z=4-x^2 and below by the paraboloid z=x^2+3y^2.

- Apr 23rd 2010, 06:05 PMhollywood
For these problems, you need to know what the volume looks like - I find it helps to draw a diagram.

The limits for the first or innermost integral (with respect to z) are, of course, to .

Next, you need to know the intersection of the two surfaces, which is given by , or , an ellipse. This is the area you need to cover with the other two integrals.

Let's integrate with respect to y next. For a given x, what is the range of y? All you have to do is solve the ellipse equation for y as a function of x, which gives you two points - the bottom and the top of the ellipse. Those are your limits of integration with respect to y.

For the outermost integral, you just need the range of x that you're dealing with, which is just where the ellipse crosses the x axis - setting y=0 gives , so , which are your limits of integration for x.

I like to think of these problems as adding up tiny cubes of volume. Each cube is labeled with its coordinates (x,y,z). First you add up the cubes from the bottom to the top to make "sticks" of different lengths, labeled with coordinates (x,y). Then you add up the sticks, and for each x, the sticks have a range of y coordinates. Adding up the sticks makes "plates", which are now labeled with x coordinates only. So the final step is to add up all the plates. The plates have different shapes, but the limits of integration just tell you which plates to choose. I suppose that might not help you, but that's how I think of it.

Post again in this thread if you're still having trouble.

- Hollywood - Apr 23rd 2010, 06:33 PMasdf122345
I get it now thanks.