Question

Let \( V \) be the solid region in \( \mathbb{R}^3 \) bounded by the paraboloid \( y = x^2 + z^2 \) and the plane \( y = 4 \). Then the value of \[ \iiint_V 15 \sqrt{x^2 + z^2} \, dV \] is

• A. \( 128 \pi \)
• B. \( 64 \pi \)
• C. \( 28 \pi \)
• D. \( 256 \pi \)

Hint:

In cylindrical coordinates, the integral becomes simpler when dealing with circular symmetry. Use the appropriate bounds for \( r \) and the corresponding expression for \( y \) in the given region.

Answer 1

The Correct Option is A

We are given a solid region bounded by the paraboloid \( y = x^2 + z^2 \) and the plane \( y = 4 \). To solve for the volume integral, we first set up the integral in cylindrical coordinates. Let \( x = r \cos \theta \) and \( z = r \sin \theta \), so \( x^2 + z^2 = r^2 \). The equation of the paraboloid becomes \( y = r^2 \), and the plane is given by \( y = 4 \). The bounds for \( r \) are from 0 to \( 2 \) (since \( r^2 = 4 \) at the plane). The volume integral is: \[ \iiint_V 15 \sqrt{x^2 + z^2} \, dV = 15 \int_0^{2\pi} \int_0^2 \int_{r^2}^4 r^2 \, dy \, dr \, d\theta \] After performing the integration, we get the result \( 128 \pi \). Thus, the correct answer is (A).