Green's Theorem and simple closed curve

[Amy Marie]

Homework Statement

Use, using the result that for a simple closed curve C in the plane the area enclosed is:

A = (1/2)∫(x dy - y dx) to find the area inside the curve x^(2/3) + y^(2/3) = 4

Homework Equations

Green's Theorem:
∫P dx + Q dy = ∫∫ dQ/dx - dP/dy

The Attempt at a Solution

I solved the equation of the curve for x:

x = (4 - y^(2/3))^(3/2)

Also, from the original curve equation x^(2/3) + y^(2/3) = 4, when x = 0, y = +/- 8 because 4^(3/2) = 8.

But when I plug x = +/- (4 - y^(2/3))^(3/2) in for the x bounds and y = +/- 8 in for the y bounds in the resulting double integral

(1/2)∫∫ 2 dxdy

I have trouble integrating x = (4 - y^(2/3))^(3/2) it with respect to y.

Does anybody happen to know if there is a more correct way to solve this problem?

Thank you for your help!

 

[LCKurtz]

The question does not want you to work the integral out by doing a double integral. It wants you to find a nice parameterization of the curve and do$$
A = \frac 1 2 \int_C x~dy - y~dx$$ So your first job is to find a nice parameterization. As a hint think about a way to parameterize it so that you can use the identity $(r\cos\theta)^2 + (r\sin\theta)^2 = r^2$.

 

[Amy Marie]

The question does not want you to work the integral out by doing a double integral. It wants you to find a nice parameterization of the curve and do$$
A = \frac 1 2 \int_C x~dy - y~dx$$ So your first job is to find a nice parameterization. As a hint think about a way to parameterize it so that you can use the identity $(r\cos\theta)^2 + (r\sin\theta)^2 = r^2$.

Thank you for your help!