i dont understand anything about double intergrals..i'm blur about this..please help me

Question

anonymous · Answer

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?

anonymous · Answer

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

anonymous · Answer

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.

anonymous · Answer

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

anonymous · Answer

ok..thanks a lot guys...