Final answer:
When converting the given integral to spherical coordinates, the integrand simplifies to zero due to symmetry properties of sine and cosine over a full period, thus the integral evaluates to 0, which is not an option listed in the question.
Step-by-step explanation:
To evaluate the integral using spherical coordinates, we first need to express the limits of integration for the variables x, y, z into spherical coordinates (ρ, θ, φ). In spherical coordinates, ρ is the radius from the origin, θ is the angle with the positive x-axis (in the xy plane), and φ is the angle from the positive z-axis.
The original integral bounds suggest that we are integrating within a cylinder whose radius squared is 36 (since the y-integral goes from 0 to 36 - x^2) and a half-sphere or dome above it (since the z-integral goes from 0 to 72 - x^2 - y^2).This solid lies above the xy-plane and is symmetric with respect to the z-axis.
Using spherical coordinates, the integrand x^2y^2 becomes ρ^4sin^3φcosθsinθ and the bounds for ρ, θ, and φ are 0 to 6, 0 to 2π, and 0 to π/2, respectively, to cover the cylinder and the half-sphere. During the conversion, we also include the Jacobian determinant, which is ρ^2sinφ for spherical coordinates.
The volume element changes from dxdydz to ρ^2sinφ dρdθdφ. The full integral in spherical coordinates is then:
∫
0
π/2
∫
0
2π
∫
0
6
ρ^6sin^4φcosθsinθ dρdθdφ.
On separating the integral and integrating with respect to ρ, θ, and φ, we find that the θ integrand is zero due to the symmetry property of sine and cosine over a full period which reduces the integral to zero. Therefore, the answer is 0.