Use polar coordinates to compute the integral I = \int_0^1 dx \int_0^\sqrt{1-x^2} dy\,x^2y . The answer is

Question

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The problem asks us to compute I = ∫_0^1 dx ∫_0^{√(1−x^2)} dy x^2 y using polar coordinates.

First, describe the region and the integrand in polar form. The bounds x from 0 to 1 and y from 0 to √(1−x^2) describe the quarter of the unit disk in the first quadrant: r goes from 0 to 1 and θ goes from 0 to π/2.

Express the integrand and differential in polar coordinates:
- x = r cos θ, y = r sin θ
- x^2 y = (r cos θ)^2 (r sin θ) = r^3 cos^2 θ sin θ
- The area element dx dy becomes the polar Jacobian r dr dθ.
Thus the integrand becomes r^3 cos^2 θ sin θ times the Jacobian r, giving r^4 cos^2 θ sin θ.

Set up the separated integral:
I = ∫_{θ=0}^{π/2} ∫_{r=0}^{1} r^4 cos^2 θ sin θ dr dθ.

Compute the r-integral first:
∫_{0}^{1} r^4 dr = [r^5/5]_{0}^{1} = 1/5.

Next, compute the θ-integral:
∫_{0}^{π/2} cos^2 θ sin θ dθ.
Use a substitution u = cos θ, so du = −sin θ dθ. When θ goes from 0 to π/2, u goes from 1 to 0. The integral becomes ∫_{1}^{0} u^2 (−du) = ∫_{0}^{1} u^2 du = [u^3/3]_{0}^{1} = 1/3.

Combine the results:
I = (1/5) × (1/3) = 1/15.

Note about the provided options: the given data shows an answer of 1/15, but the list of answer choices (answer_options) is missing. The computed value using the polar-coordinate setup confirms the result 1/15.

Use polar coordinates to compute the integral I = \int_0^1 dx \int_0^\sqrt{1-x^2} dy\,x^2y . The answer is单项选择题

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