Evaluate the triple integral ∭ExdV where E is the solid bounded by the paraboloid x=6y2+6z2

Question

and x=6
.

Answer 1

To evaluate the triple integral, we need to determine the limits of integration for each variable: x, y, and z.

The paraboloid is given by x = 6y^2 + 6z^2. We also have the equation x = 6. Setting these two equations equal to each other, we get:

6y^2 + 6z^2 = 6

Dividing both sides by 6, we have:

y^2 + z^2 = 1

This is the equation of a circle with radius 1 centered at the origin in the yz-plane. Therefore, the limits of integration for y and z are -1 to 1.

Since x = 6, the limits of integration for x are from 6y^2 + 6z^2 to 6. Substituting the values for y and z, we have:

6(1)^2 + 6(1)^2 = 12

So the limits of integration for x are from 0 to 12.

Therefore, the triple integral becomes:

∭Ex dV = ∫[from 0 to 12] ∫[from -1 to 1] ∫[from -1 to 1] dxdydz

Evaluating this triple integral, we can integrate with respect to x first:

∫[from 0 to 12] (2)(1)(2) dydz

Simplifying, we have:

∫[from 0 to 12] 4 dydz

Integrating with respect to y next, we have:

4∫[from 0 to 12] dz

Integrating with respect to z, we get:

4z |[from 0 to 12] = 4(12) - 4(0) = 48

Therefore, the value of the triple integral ∭Ex dV is 48.