The curve with equation $\displaystyle a^2 y^2 = x^2 (a^2 - x^2)$ has two loops.

Show, by integration, that the area enclosed by a loop is $\displaystyle \frac {2}{3} a^2$.

Please help me to do this.

Printable View

- Apr 21st 2008, 09:55 PMgetonIntegration problem
The curve with equation $\displaystyle a^2 y^2 = x^2 (a^2 - x^2)$ has two loops.

Show, by integration, that the area enclosed by a loop is $\displaystyle \frac {2}{3} a^2$.

Please help me to do this. - Apr 21st 2008, 10:11 PMTheEmptySet
solving the above equation for y we get

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

$\displaystyle \int_0^{a}\frac{x\sqrt{a^2-x^2}}{a}dx$

is half of the area of one loop. (Draw a graph to see)

$\displaystyle 2\int_0^{a}\frac{x\sqrt{a^2-x^2}}{a}dx=$

is the area of one loop. just let $\displaystyle u=a^2-x^2$

good luck.