1. anonymous

Are you confused about choosing your limits of integration or evaluating them? (or both)? Do you have an example of a problem to work out?

2. anonymous

both of it..ok.. eg : z=16-x^2 + 2Y^2

3. anonymous

Ouch that's a tough one. Are you trying to find the volume underneath the surface? Is that all the information given in the problem? I found the region in the xy-plane to be a hyperbola, but the parabola in the yz-plane makes the surface increase without bound. Unless there are specific y-value boundaries, I can't evaluate the volume. Someone else should chime in if I am forgetting about something. Anyway, you should let z = 0 to find the region in the xy-plane we want. Set the equation as the equation of a hyperbola to get x^2/16-y^2/8=1. If you sketch this hyperbola, we want to look at the region between the two parts of the hyperbola. We need y-value boundaries, but the x-value boundaries are x = g1(y) and x = g2(y). To find these, solve for x.$x = \pm \sqrt{2y^2+16}$ The plus version is the g2(y) and the minus version is the g1(y). So, our double integral should be... $\int\limits_{a}^{b}\int\limits_{-\sqrt{2y^2+16}}^{\sqrt{2y^2+16}}(16-x^2+2y^2)dxdy$ I'll admit this was difficult, though so someone else should check over my work and continue. I'll continue working this one though.

4. anonymous

do it this way, integrate wrt x first an dthen substitute the limits...then you are left with y term, do it the same way again and you will get the solution

5. anonymous

ok..thanks a lot guys...