Evaluate the triple integral over the region E, where E lies above the cone $z=\frac{r}{3}$ and below the sphere $z^2+r^2=1$, of the function $y^2{z}^2$ $dV$. Which of the following is the correct value of the integral?

• A. $0$
• B. $\frac{\pi }{12}$
• C. $\frac{\pi }{18}$
• D. $\frac{\pi }{36}$

Answer: (D)

Verified step by step guidance

1. Step 1: Recognize that the region E is bounded by two surfaces: the cone z = r/3 and the sphere z^2 + r^2 = 1. The integral is given in cylindrical coordinates, where r is the radial distance, θ is the angular coordinate, and z is the vertical coordinate.
2. Step 2: Convert the function y^2 z^2 into cylindrical coordinates. Recall that y = r sin(θ) in cylindrical coordinates. Thus, y^2 becomes r^2 sin^2(θ), and the function becomes r^2 sin^2(θ) z^2.
3. Step 3: Set up the limits of integration for the region E. The cone z = r/3 gives the lower bound for z, and the sphere z^2 + r^2 = 1 gives the upper bound for z. For r, the sphere's equation implies r ranges from 0 to 1. For θ, the full rotation around the z-axis means θ ranges from 0 to 2π.
4. Step 4: Write the triple integral in cylindrical coordinates. The volume element dV in cylindrical coordinates is r dr dθ dz. The integral becomes: ∫(θ=0 to 2π) ∫(r=0 to 1) ∫(z=r/3 to sqrt(1 - r^2)) r^3 sin^2(θ) z^2 dz dr dθ.
5. Step 5: Evaluate the integral step by step. First, integrate with respect to z, keeping r and θ constant. Then, integrate with respect to r, and finally integrate with respect to θ. Use trigonometric identities and symmetry properties of sin^2(θ) to simplify the angular integral.