Hello,

I'm trying to find the area defined by

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

and would be really grateful for your help!

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- Jun 28th 2009, 02:09 PMgusztavFinding the area defined by a curve
Hello,

I'm trying to find the area defined by

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

and would be really grateful for your help! - Jun 28th 2009, 04:20 PMBhajun
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.

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

$\displaystyle 4y^2 = 4x^2 - x^4$

$\displaystyle 2y = \sqrt{x^2 \left(4 - x^2\right)}$

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

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. - Jun 28th 2009, 04:23 PMgusztav