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

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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:
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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