# Math Help - Need help setting up this triple intergral for the volume of the solid.

1. ## Need 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.

2. 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, $x^2+3y^2$ to $4-x^2$.

Next, you need to know the intersection of the two surfaces, which is given by $4-x^2=x^2+3y^2$, or $2x^2+3y^2=4$, 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 $2x^2=4$, so $x=\pm\sqrt{2}$, 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 $x^2+3y^2$ to the top $4-x^2$ 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

3. I get it now thanks.