1. ## Polar coordinates

Hey, I need help!!! Could anyone help me to solve this? I need to change this functions to polar ant than I have to calculate area which limited by two curves....

Thank you!!!

2. ## Re: Polar coordinates

I don't see a function, I see an equation. Do you want to solve this equation for $r=\sqrt{x^2+y^2}$ ?

3. ## Re: Polar coordinates

Hello, raplys!

I need to change this function to polar: .$\displaystyle (x^2+y^2)^2 \:=\:a^2(x^2+2y^2)$

and then I have to calculate area which limited by two curves. . What two curves?

Conversions: .$\displaystyle \begin{Bmatrix} x \:=\:r\cos\theta \\ y \:=\:r\sin\theta \\ x^2+y^2 \:=\:r^2 \end{Bmatrix}$

We have: .$\displaystyle (x^2+y^2)^2 \:=\:a^2(x^2+2y^2)$

. . . . . . . . $\displaystyle (x^2+y^2)^2 \:=\:a^2(x^2+y^2 + y^2)$

. . . . . . . . . . . . $\displaystyle (r^2)^2 \;=\;a^2 ( r^2 + r^2\sin^2\!\theta)$

. . . . . . . . . . . . . . $\displaystyle r^4 \;=\;a^2r^2(1 + \sin^2\!\theta)$

. . . . . . . . . . . . . . $\displaystyle r^2 \;=\;a^2(1+\sin^2\!\theta)$

. . . . . . . . . . . . . . .$\displaystyle r \;=\;\pm\:\!a\sqrt{1 + \sin^2\!\theta}$