# Thread: use triple integral to find volume of solid enclosed between the surfaces

1. ## use triple integral to find volume of solid enclosed between the surfaces

surfaces x=y^2+z^2 and x = 1-y^2. I drew them below:

I know I need to find the region projected on the xy plane and integrate over that area, as well as integrate the height on the z-axis; however I am not sure how to proceed. I think I need some equations, possibly setting the two equal to each other to find where they intersect.

2. ## Re: use triple integral to find volume of solid enclosed between the surfaces

set two equal: y^2 + z^2 = 1 - y^2. Then 2y^2 + z^2 = 1, an ellipse on the yz plane

also, on yz plane, z^2+y^2=0 & 1 = y^2 or y = +/- 1
on xy plane, x = y^2 & x = 1- y^2
and on xz plane, x=z^2 & x=1

3. ## Re: use triple integral to find volume of solid enclosed between the surfaces

The surface $x=y^2+z^2$ is a paraboloid opening in the x-direction, and the surface $x=1-y^2$ is a parabolic cylinder extending in the z direction. Your "drawings" look correct. You can see that the paraboloid "pokes through" the parabolic cylinder, creating an enclosed volume. That's your integration volume.

The top and bottom of the volume is the paraboloid - solving for z gives $z = \sqrt{x-y^2}$ and $z=-\sqrt{x-y^2}$. Now projecting onto the xy plane gives the parabola $x=y^2$ and another parabola $x=1-y^2$.

The left boundary is $x=y^2$ and the right boundary is $x=1-y^2$. Projecting onto the y-axis gives the interval $\left[ \frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2} \right]$.

So the integral is
$\int_{-\frac{\sqrt{2}}{2}}^{\frac{\sqrt{2}}{2}} \int_{y^2}^{1-y^2} \int_{-\sqrt{x-y^2}}^{\sqrt{x-y^2}} \,dz \,dx \,dy$

The general procedure is to find the boundaries in the x, y, or z direction (i.e. top and bottom, front and back, or left and right), then project in that same direction and do the same with one less dimension.

- Hollywood