Finding the area defined by a curve

**gusztav** · June 28th 2009, 02:09 PM

Hello,

I'm trying to find the area defined by

$4y^2-4x^2+x^4=0$

and would be really grateful for your help!

**Bhajun** · June 28th 2009, 04:20 PM

Life is always made easier if we know how much you know. In any case, this problem can be solved directly. It should be clear by observation that the curve is symmetric over both the x and y axes.

$4y^2 - 4x^2 + x^4 = 0$
$4y^2 = 4x^2 - x^4$
$2y = \sqrt{x^2 \left(4 - x^2\right)}$
$y = \frac{1}{2}x\sqrt{4 - x^2}$

This can be integrated by substitution easily enough. Note that the limit for this integration is +/- 2. By the symmetry of the problem, you can either do the integration from 0 to 2 and multiply the resultant by 4 or from -2 to 2 and multiply by 2 to get the area enclosed in the entire figure.

Hope that helps.

**gusztav** · June 28th 2009, 04:23 PM

Originally Posted by Bhajun

$y = \frac{1}{2}x\sqrt{4 - x^2}$
...

Hope that helps.

Yes, this is very helpful! Just what I needed.

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