Question

Evaluate The Triple Integral ∭ExdV Where E Is The Solid Bounded By The Paraboloid X=3y2+3z2 And X=3.

Answer 1

**Final Answer:**

The triple integral
`\(\iiint_E E_x \, dV\)`, where E is the solid bounded by the paraboloid
`\(x = 3y^2 + 3z^2\)` and x = 3, is evaluated to be zero.

**Explanation:**

To evaluate the given triple integral, we need to set up the integral using the provided limits based on the region defined by the paraboloid
`\(x = 3y^2 + 3z^2\)` and the plane x = 3. Since the integrand is
`\(E_x\)`, the component of the electric field in the x-direction, we integrate over the volume E.

The paraboloid
`\(x = 3y^2 + 3z^2\)` intersects the plane x = 3 in a circular region. In cylindrical coordinates, the limits for y and z would be determined by the intersection of the paraboloid and the plane. However, the integrand
`\(E_x\)` does not depend on y or z, and the shape of E is symmetric with respect to the x-axis. Therefore, the integral of
`\(E_x\)` over E is zero.

Understanding the symmetry and characteristics of the region helps simplify the integral, resulting in a final value of zero. This type of analysis is common in evaluating integrals over symmetric regions in multivariable calculus and mathematical physics.