Use greens theorem to evaluate the line integral

use green's theorem to evaluate the following line integral: 2ydx + x^2 ydy where C is the circle of radius 2 centered at the origin oriented clockwise

By Green's Theorem, if D is the curve enclosed by C, then:
∫C (P dx + Q dy) = ∫∫D (∂Q/∂x - ∂P/∂y) dA,

where P and Q are some functions of x and/or y.

So, in this case:
∫C (2y dx + x^2*y dy) = ∫∫D [(∂/∂x)(x^2*y) - (∂/∂y)(2y)] dA
= ∫∫D (2xy - 2) dA,

where D is the region enclosed by C, the circle with radius 2 centered at the origin.

It would be awkward to evaluate ∫∫D (2xy - 2) dA using Cartesian coordinates; it is significantly easier to use polar coordinates to evaluate this integral since we can easily represent the circle in polar coordinates (r = 2).

In polar coordinates, we can describe all points in D with the following inequalities:
D = {(r, θ) | 0 ≤ r ≤ 2, 0 ≤ θ ≤ 2π}.

Therefore, the required line integral equals:
∫∫D (2xy - 2) dA = ∫∫ r[2(r*cosθ)(r*sinθ) - 2] dr dθ (from r=0 to 2) (from θ=0 to 2π)
= ∫∫ r(2r^2*sinθcosθ - 2) dr dθ (from r=0 to 2) (from θ=0 to 2π)
= ∫∫ r[r^2*sin(2θ) - 2] dr dθ (from r=0 to 2) (from θ=0 to 2π), since sin(2θ) = 2sinθcosθ
= ∫∫ [r^3*sin(2θ) - 2r] dr dθ (from r=0 to 2) (from θ=0 to 2π)
= ∫ {[(1/4)r^4*sin(2θ) - r^2] (evaluated from r=0 to 2)} dθ (from θ=0 to 2π)
= ∫ [4sin(2θ) - 4] (from θ=0 to 2π)
= [-2cos(2θ) - 4θ] (evaluated from θ=0 to 2π)
= -2 - 8π - (-2 + 0)
= -8π.

I hope this helps!

By Green's Theorem, the line integral can be expressed as the following
two variable integral:

∬(2xy-2)dxdy As 2xy is the partial derivative of x^2y with respect to x
And 2 is the partial derivative of 2y with respect to y.

∬(2xy)dxdy-∬(2)dxdy

Due to the symmetry of the area, the first term equals zero, and the
second term comes out to be -2(area of the circle).

Thus, the line integral equals -8π.

I solved your problem a couple hours ago and you also gave me best answer. Was this problem wrong that you're reposting it? Just wondering.

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