Math 241 - Quiz #12

Evaluate the triple integral of x over the domain E, which is between the parabloid x = y^2 + z^2 and the plane x = 4.

Solution

We can describe E two different ways:
First we can let 0 <= x <=4 and then the cross-sections in (y,z) are circles of radius sqrt(x)/2. So we get
0 <= x <= 4, 0 <= r <= sqrt(x)/2, 0 <= theta <= 2 pi,
where y = r cos(theta) and z = r sin(theta)

Integrating, we get:
int(0 to 4) int(0 to 2pi) int(0 to sqrt(x)/2) xr dr dtheta dx

Second way is to let y = r cos(theta) and z = r sin(theta) as in the first, but start with the end circle (at x=4) as the base. So we get:
0 <= r <= 1, 0 <= theta <= 2pi, 4r^2 <=x <= 4

Integrating, we get:
int(0 to 2pi) int(0 to 1) int(4r^2 to 4) xr dx dr dtheta

In either case the final value is 16pi/3

ccollins@math.utk.edu

Last Modified: Apr. 9, 1999