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Evaluate the integral after changing to spherical coordinates.∫30∫√9−y2−√9−y2∫√9−x2−y20(x2z+y2z+z3)dzdxdy

1 Answer

4 votes

To change to spherical coordinates, we can use the following formula:

x = ρ sin φ cos θ

y = ρ sin φ sin θ

z = ρ cos φ

We also note that the region of integration is a hemisphere with radius 3, and that the integrand contains x^2z+y^2z+z^3. Since we are integrating over a hemisphere, the bounds of ρ can be from 0 to 3, φ can be from 0 to π/2, and θ can be from 0 to 2π.

Next, we need to express the integrand in terms of ρ, φ, and θ. Substituting x, y, and z, we get:

x^2z + y^2z + z^3 = ρ^4 sin^2 φ cos^2 θ (ρ cos φ) + ρ^4 sin^2 φ sin^2 θ (ρ cos φ) + (ρ cos φ)^3

Simplifying, we get:

x^2z + y^2z + z^3 = ρ^5 cos^2 φ + ρ^3 cos^3 φ

Thus, the new integral is:

∫0^(2π) ∫0^(π/2) ∫0^3 (ρ^5 cos^2 φ + ρ^3 cos^3 φ) ρ^2 sin φ dρ dφ dθ

Integrating with respect to ρ, we get:

∫0^(2π) ∫0^(π/2) [ 1/6 ρ^6 cos^2 φ + 1/4 ρ^4 cos^3 φ ]_|ρ=0^3 sin φ dφ dθ

Simplifying and integrating with respect to φ, we get:

∫0^(2π) [ 9/5 sin^5 φ - 27/14 sin^7 φ ]_|φ=0^(π/2) dθ

Evaluating the limits, we get:

∫0^(2π) [ 9/5 - 27/14 ] dθ

Finally, evaluating the integral, we get:

∫0^(2π) [ 33/35 ] dθ = 66π/35

Therefore, the value of the integral after changing to spherical coordinates is 66π/35.

answered
User MorgoZ
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