The integral ∫01∫01−𝑥2(𝑥2+𝑦2)𝑑𝑦𝑑𝑥 can be transformed into polar coordinates. The transformation from Cartesian coordinates (x, y) to polar coordinates (r, θ) is given by x = rcosθ and y = rsinθ. The Jacobian determinant of this transformation is r, so the differential area element dxdy becomes rdrdθ in polar coordinates.

The limits of integration also change. The square region 0 ≤ x ≤ 1, 0 ≤ y ≤ sqrt(1 - x^2) in the Cartesian plane corresponds to the region 0 ≤ r ≤ 1, 0 ≤ θ ≤ π/2 in the polar plane.

Substituting x = rcosθ and y = rsinθ into the integrand gives (rcosθ)^2 + (rsinθ)^2 = r^2. Therefore, the given integral becomes ∫ from 0 to π/2 ∫ from 0 to 1 of r^2 * rdrdθ.

So, the correct answer is ∫0𝜋2∫01𝑟2𝑑𝑟𝑑𝜃.

Question

The integral ∫01∫01−𝑥2(𝑥2+𝑦2)𝑑𝑦𝑑𝑥 can be transformed into polar coordinates. The transformation from Cartesian coordinates (x, y) to polar coordinates (r, θ) is given by x = rcosθ and y = rsinθ. The Jacobian determinant of this transformation is r, so the differential area element dxdy becomes rdrdθ in polar coordinates.

The limits of integration also change. The square region 0 ≤ x ≤ 1, 0 ≤ y ≤ sqrt(1 - x^2) in the Cartesian plane corresponds to the region 0 ≤ r ≤ 1, 0 ≤ θ ≤ π/2 in the polar plane.

Substituting x = rcosθ and y = rsinθ into the integrand gives (rcosθ)^2 + (rsinθ)^2 = r^2. Therefore, the given integral becomes ∫ from 0 to π/2 ∫ from 0 to 1 of r^2 * rdrdθ.

So, the correct answer is ∫0𝜋2∫01𝑟2𝑑𝑟𝑑𝜃.

Knowee AI · Accepted Answer

The integral ∫01∫01−𝑥2(𝑥2+𝑦2)𝑑𝑦𝑑𝑥 can be transformed into polar coordinates. The transformation from Cartesian coordinates (x, y) to polar coordinates (r, θ) is given by x = rcosθ and y = rsinθ. The Jacobian determinant of this transformation is r, so the differential area element dxdy becomes rdrdθ in polar coordinates.

The limits of integration also change. The square region 0 ≤ x ≤ 1, 0 ≤ y ≤ sqrt(1 - x^2) in the Cartesian plane corresponds to the region 0 ≤ r ≤ 1, 0 ≤ θ ≤ π/2 in the polar plane.

Substituting x = rcosθ and y = rsinθ into the integrand gives (rcosθ)^2 + (rsinθ)^2 = r^2. Therefore, the given integral becomes ∫ from 0 to π/2 ∫ from 0 to 1 of r^2 * rdrdθ.

So, the correct answer is ∫0𝜋2∫01𝑟2𝑑𝑟𝑑𝜃.

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