triple integral intervals for flux calculus

Question

I'd like some help to calculate flux of F={(x, y, z) = x(z^3 − y^2), −yz^3, z(y^2 − 1)} going through surface S = {(x, y, z) : x^2 + 4y^2 + z^2 = 36, x(x + 2) ≤ 8} using gauss. The triple integral to work with should be ∫∫∫-1dV ,but here I get stuck on how to set the integral intervals. Can someone help me? Thank you in advance (Result should be 18π)

Answer 1

I am making the following 3 assumptions regarding the ellipsoid

x² + 4y² + z² = 36

cut off by plains x=-4 and x=2

Assumption 1:

The flux is supposed to include the flat sides where the cuts are made.

This makes the surface closed.

Assumption 2:

The phrase "by Gauss" refers to the divergence theorem, which indeed yields

∫∫∫-1dV

as you said.

Assumption 3:

Your last statement does not imply that the triple integral should evaluate to 18π,

as that integral must be negative.

To evaluate the integral, imagine that the x-axis in the interval (-4, 2)

is divided into slices of thickness dx.

The slice at position x intersects the given surface in the ellipse

4y² + z² = 36 - x² (1)

Using the formula for the area of an ellipse, the area of (1) is

(π/2)(36 - x²)

Therefore at position x

dV = (π/2)(36 - x²)dx

Therefore

∫∫∫-1dV = -∫(π/2)(36 - x²)dx

where the bounds of integration are from -4 to 2

-∫(π/2)(36 - x²)dx = -(π/2)(36x - x³/3) + C

When evaluated between -4 and 2 we get

-(π/2)(36×(2 + 4) - (2³/3 - (-4)³/3) = -96π