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Im trying to find the area of $y^2=x^2(a^2-x^2)$ where $a>0$.

From my calculations it seems to be $$ 2\int_{0}^{a} x\sqrt{a^2 - x^2}dx=\frac{2}{3}a^3, $$ but I am not sure if it is right.

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  • $\begingroup$ You are right wolframalpha.com/input/…. $\endgroup$ – eranreches Jun 13 at 12:43
  • $\begingroup$ You know you can put entire equations in dollar signs, right? Not only is it easier to write $y^2=x^2(a^2-x^2)$ than it is to write $y^2$=$x^2$($a^2$-$x^2$), the end result also looks a whole lot better. $\endgroup$ – Arthur Jun 13 at 12:45

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