Christopher F. answered • 04/06/20

M.S. in Mathematics with 500+ hours in Calculus tutoring

Green's Theorem turns a line integral into a double integral. It helps simplify hard line integrals into easier double integral problems.

First, with our given field F, we have:

∫F•dr = ∫(-3y)dx + (3x)dy

Green's Theorem says: ∫Mdx + Ndy (over curve C) = ∫∫∂N/∂x - ∂M/∂y dA (over region R)

Our M is -3y and our N is 3x

The partial derivative of M with respect to y (aka ∂M/∂y) is -3

The partial derivative of N with respect to x (aka ∂N/∂x) is 3

Applying Green's to our line integral above, we get: ∫(-3y)dx + (3x)dy = ∫∫3-(-3) dA = ∫∫6dA

∫∫6dA is over region R: x^{2 }+ y^{2 }≤ a^{2 }which is the disk at the origin with radius a

∫∫6dA = 6 ∫∫dA where ∫∫dA is the area of region A, i.e., the disk at the origin with radius a

Area of the disk at the origin with radius a is π a^{2}

So we have ∫(-3y)dx + (3x)dy = ∫∫3-(-3) dA = ∫∫6dA = 6 ∫∫dA = 6*π a^{2}

Had we done the line integral directly:

x= a*cost

y=a*sint

dx= a*(-sint)dt

dy= a*(cost)dt

Then by substitution, ∫(-3y)dx + (3x)dy = ∫ (-3a*sint)*a*(-sint)dt + (3*a*cost)*a*(cost)dt =∫

∫(3a^{2}sin^{2}t)dt + (3a^{2}cos^{2}t)dt = ∫3a^{2}(sin^{2}t+cos^{2}t)dt

But sin^{2}t+cos^{2}t = 1, so we get ∫3a^{2}(sin^{2}t+cos^{2}t)dt = ∫3a^{2}dt (over the curve C: 0≤ t ≤ 2π)

= 3a^{2} ∫dt (0 to 2π) = 3a^{2}(2π - 0) = 6*π a^{2}

Our Line integral and Double Integral match; they both equal 6*π a^{2}. This verify's green's theorem by evaluating both sides of the equation for the given field F.

Hope this helps! If you have any questions about the steps, please comment. :)