# Find the Area of the Interior

by mmmboh
Tags: area, greens theorem, integral, parametrize
 P: 408 Find a parametrization of the curve x2/3+y2/3=1 and use it to compute the area of the interior. What I did was y=(1-x2/3)3/2 I then integrated this function from 0 to 1 (using maple since it is a crazy integral) and got the answer to be 3/32 $$\pi$$. However this is wrong, I probably wasn't suppose to do it the way I did anyway considering that the integral is so complicated. So what should I do? I am learning Green's theroem right now if that helps, although I may have skipped ahead in it a bit I'm not sure. Thanks. Edit: Hm ok I have figured out what to do, the answer is 3/8 pi. How come my original answer is 1/4 of this though?
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Thanks
P: 26,167
Hi mmmboh!
 Quote by mmmboh Find a parametrization of the curve x2/3+y2/3=1 …
They want you to use a parametrisation.

For example, if it were x2+y2=1, you'd use x = cosθ, y = sinθ.

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