Evaluate the triple integral ∭ExdV where E is the solid bounded by the paraboloid x=2y2+2z2 and x=2.

Answer 1

Over E the variable x ranges between 0 and 2.

A plane characterized by a fixed value of x in the interval [0, 2] intersects E

in a disk D(x) given by the inequality

2y² + 2z² ≤ x

This disk has a radius √(x/2) and area πx/2.

We can evaluate the triple integral by dividing E into such disk,

and integrating over x in [0,2].

∫∫∫E xdV = ∫0 to 2 (∫∫D(x) x dS) dx

where dS represents integration over the surface of the disk D(x).

Since x is constant over the surface D(x)

(∫∫D(x) x dS) evaluates to the product of x with the area of D(x).

Thus

(∫∫D(x) x dS) = πx²/2

Let F(x) be the indefinite integral

F(x) = ∫ (∫∫D(x) x dS) dx = ∫(πx²/2)dx = πx³/6

Then

∫∫∫E xdV = F(2) - F(0) = 8π/6