Question

Use Green’s Theorem to evaluate the integral I = Z C 2y 3 dx − 2x 3 dy where C is the circle of radius 2 centered at the origin, oriented counterclockwise

Answer 1

Answer:

-48π

Step-by-step explanation:

Recall that Green's theorem states the following:

Let P, Q be $\bf C^1$-functions on a region A, which is the interior of a closed piecewise $\bf C^1$-path C, parametrized counterclockwise. Then  

$\bf \displaystyle\oint_CPdx+Qdy=\displaystyle\iint_A(\displaystyle\frac{\partial Q}{\partial x}-\displaystyle\frac{\partial P}{\partial y})dydx$

We have

$\bf P(x, y) = 2y^3\\\\Q(x,y)=-2x^3$

$\bf \displaystyle\frac{\partial Q}{\partial x}=-6x^2\\\\\displaystyle\frac{\partial P}{\partial y}=6y^2$

Therefore

$\bf \oint_C2y^3dx-2x^3Qdy=\displaystyle\iint_A(-6x^2-6y^2)dydx$

where A is the interior of the circle of radius 2 centered at the origin.

If we change to polar coordinates

x = rcos(t)

y = rsin(t)

Then A can be described with the inequalities

0 < r < 2

0 < t < 2π  

and

$\bf \displaystyle\iint_A(-6x^2-6y^2)dydx=-6\displaystyle\int_{0}^{2\pi}\displaystyle\int_{0}^{2} (r^2cos^2t+r^2sin^2t)rdrdt=\\\\-6\displaystyle\int_{0}^{2\pi}\displaystyle\int_{0}^{2} (r^2)rdrdt=-6\left(\displaystyle\int_{0}^{2\pi}dt\displaystyle\int_{0}^{2}r^3dr \right)=\\\\=\boxed{-48\pi}$