Evaluate the triple integral ∭ExdV where E is the solid bounded by the paraboloid x=6y^2+6z^2

Question

and x=6
.

Answer 1

To evaluate the triple integral ∭ExdV, we first need to determine the limits of integration for each variable.

The paraboloid equation can be written as x = 6y^2 + 6z^2.

To determine the limits for y and z, we solve for y and z in terms of x:

x = 6y^2 + 6z^2

6y^2 = x - 6z^2

y^2 = (x - 6z^2)/6

y = ±√((x - 6z^2)/6)

Since the paraboloid is symmetric about the yz-plane, we only need to consider the positive square root of y.

Now, we find the limits of integration for y:

When x = 6 (the upper bound), we have:

6 = 6y^2 + 6z^2

y^2 + z^2 = 1

This is the equation of a unit circle in the yz-plane. Therefore, we can let y and z vary over the entire circle, which gives us the limits 0 ≤ y ≤ √1 = 1 and -1 ≤ z ≤ 1.

When x = 6y^2 + 6z^2 = 0 (the lower bound), we have:

0 = 6y^2 + 6z^2

y^2 + z^2 = 0

This equation only holds when y = z = 0. Therefore, the lower bound is a single point (0, 0).

Finally, the limits of integration for x are given by the equation of the paraboloid:

x = 6y^2 + 6z^2

Substituting y = 1 and z = 1, we have:

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

Therefore, the limits of integration for x are 0 ≤ x ≤ 12.

Now we can set up the triple integral:

∭ExdV = ∫(0 to 12)∫(0 to √1)∫(-1 to 1) Exdydzdx

Note that the integrand does not depend on y or z, so we can bring it outside the integral:

∭ExdV = Ex ∫(0 to 12)∫(0 to √1)∫(-1 to 1) dydzdx

Since the integrand does not depend on y or z, the innermost integral becomes:

∫(-1 to 1) dy = 2

The limits of integration for y and z remain the same, so we have:

∭ExdV = 2Ex ∫(0 to 12)∫(0 to √1) dzdx

Now we can evaluate the remaining integrals:

∫(0 to √1) dz = z ∣(0 to √1) = √1 - 0 = √1 = 1

∫(0 to 12) 1 dx = x ∣(0 to 12) = 12 - 0 = 12

Substituting these values back into the expression:

∭ExdV = 2Ex * 1 * 12 = 24Ex