How to Find the Area of the Interior Using a Parametrization of a Curve?

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SUMMARY

The discussion focuses on finding the area of the interior of the curve defined by the equation x2/3 + y2/3 = 1 using parametrization techniques. The user initially attempted to compute the area by integrating the function y = (1 - x2/3)3/2 from 0 to 1, resulting in an incorrect area of 3/32 π. After further exploration and learning about Green's Theorem, the correct area was determined to be 3/8 π, highlighting the importance of proper parametrization in such calculations.

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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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Hi mmmboh! :smile:
mmmboh said:
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. :wink:
 

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