# Thread: Computing the area using parametrization

1. ## Computing the area using parametrization

I got stuck on this question.

Find a parametrization of the curve $\displaystyle x^\frac{2}{3} + y^\frac{2}{3} = 1$ and use it to compute the area of the interior

My solution:
I set
$\displaystyle x=cos^3 {\theta}$
$\displaystyle y=sin^3 {\theta}$

Then I'm stuck on how I should apply Green's Theorem.

2. Originally Posted by qwesl
I got stuck on this question.

Find a parametrization of the curve $\displaystyle x^\frac{2}{3} + y^\frac{2}{3} = 1$ and use it to compute the area of the interior

My solution:
I set
$\displaystyle x=cos^3 {\theta}$
$\displaystyle y=sin^3 {\theta}$

Then I'm stuck on how I should apply Green's Theorem.
Green's Theorem states that

$\displaystyle \iint_D \frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}dA = \oint_{\partial D}Pdx+Qdy$

Now there are a few combinations that work, but for example here is one. Pick $\displaystyle P(x,y)=0,Q(x,y)=x$ then you get

$\displaystyle \iint_DdA=\oint_{\partial D}xdy=\int_{0}^{2\pi}\cos^3(\theta)(3\sin^2(\theta )\cos(\theta))d\theta$