Double Integral Volume Problem

1. Sep 8, 2010

jumbogala

1. The problem statement, all variables and given/known data
Use double integrals to find the volume of the region in the first octant (x, y, z all more than or equal to zero) bounded by the vertical plane 2x + y = 2 and the surface z = x2

2. Relevant equations

3. The attempt at a solution
I'm having major problems visualizing this, which is stopping me from even getting started.

z = x2 I think I can visualize by itself.

But the plane is confusing me. My prof taught us that to sketch a plane, you find the zeros of the equation. So setting y and z to zero, we find the plane crosses the x axis at 1 and similarly the y axis at 2. But then the plane would be horizontal, not vertical...

Help?

2. Sep 8, 2010

Dick

Yes, the plane crosses the x axis at 1 and the y axis at 2. It doesn't cross the z axis at all. Doesn't that make it parallel to the z axis? I.e. vertical?

3. Sep 8, 2010

jumbogala

Oh, for some reason I was visualizing it passing through the z axis at x = 0. But okay, now I see it.

I think the setup is going to be ∫∫ x 2 dA.

I'm not really sure how the bounds on the integral here work though. Usually I draw a diagram and look at what shape the "base" should have in the xy plane. It kind of looks like it should be a triangle of some kind, but I'm not sure.

Maybe a triangle with vertices (1, 0), (0, 0) and (0, 2)?

Last edited: Sep 8, 2010
4. Sep 8, 2010

Dick

You are definitely right about that triangle. And sure, integrate z (i.e. x^2) over it.

5. Sep 8, 2010

jumbogala

I figured out the equations for the sides of the triangle and then did the double integral like normal. It worked. Thanks!!