Transforming a Triple Integral: From Ellipsoid to Sphere

[Amy Marie]

Homework Statement

Evaluate ∫∫∫[W] xz dV, where W is the domain bounded by the elliptic cylinder (x^2)/4 + (y^2)/9 = 1 and the sphere x^2 + y^2 + z^2 = 16 in the first octant x> or = 0, y> or = 0, z> or = 0.

Homework Equations

First, I tried to find the bounds for z:
z = 0 (because z is greater than or equal to zero) to z = sqrt(16 - x^2 - y^2).

Then setting z = 0, I tried to find the x bounds:
x = sqrt(4 - (4y^2)/9) to x = sqrt(16 - y^2).

Finally with both x and z set to 0, I tried to find the y bounds:
y = 3 to y = 4.

The Attempt at a Solution

∫[/B]3 to 4 ∫sqrt(4 - (4y^2)/9) to sqrt(16 - y^2) ∫0 to sqrt(16 - x^2 - y^2) xz dzdxdy

When I tried to solve this, it didn't work. I'm wondering if I have the bounds wrong.

 

[STEMucator]

EDIT:

You should find the intersection of the sphere and cylinder by setting them both to be zero and solving. This will give you an ellipsoid.

Using the transformation $x = \sqrt{20} u$, $y = \sqrt{\frac{135}{8}} v$, and $z = \sqrt{15}w$ will transform the ellipsoid into a sphere of radius 1. Computing the Jacobian of this transformation will allow you to transform the original integral:

$$\iiint_V xz \space dV = \iiint_{V'} (\sqrt{20} u)(\sqrt{15}w) \space |J| \space dV'$$

I believe a change to spherical co-ordinates from here will clean up the limits and the integral.