Find area of surface enclosed with given curve Find the area of surface with given equation $$\big(x^2 + y^2\big)^2 = a(x^3+y^3)$$
I tried to use polar coordinates 
$x=r\cos\alpha$ and $y=r\sin\alpha$
and so $$r = a(\cos\alpha + \sin\alpha)\bigg(1-{\sin2\alpha\over{2}}\bigg)$$
Maybe anyone can suggest better way?
 A: Okay i solved, with $\alpha \in [-\pi/4, 3\pi/4]$
and $${a^2\over2}\int_{-\pi\over4}^{3\pi\over4}(\sin\alpha + \cos\alpha)^2 (1-{\sin2\alpha \over 2})^2 d\alpha = \bigg(\dfrac{\cos\left(6x\right)+9\sin\left(4x\right)-9\cos\left(2x\right)-48\sin^4\left(x\right)+48\cos^4\left(x\right)-96\cos^2\left(x\right)+60x}{96}\bigg)_{-\pi\over4}^{3\pi\over4} = {5\pi a^2\over 16}$$
A: Why not directly the double integral (in polar coordinates)?
$$\int_0^{2\pi}\int_0^{a(\cos t + \sin t)\bigg(1-{\sin2t\over{2}}\bigg)}r\,dr\,dt=\frac{a^2}2\int_0^{2\pi}(\cos t+\sin t)^2\left(1-\frac{\sin2t}2\right)^2dt=$$
$$=\frac{a^2}2\int_0^{2\pi}\left(1+\sin2t\right)\left(1-\sin2t+\frac{\sin^22t}4\right)dt=\frac{a^2}2\int_0^{2\pi}\left[1-\frac34\sin^22t+\frac{\sin^32t}4\right]dt=$$
$$=\frac{a^2}4\int_0^{4\pi}\left[1-\frac34\sin^2u+\frac{\sin^3u}4\right]du=$$
$$=\frac{a^2}4\left(4\pi-\frac384\pi+\frac14\overbrace{\int_0^{4\pi}\left(\sin u-\sin u\cos^2u\right)du}^{=0}\right)=\frac{5\pi a^2}8$$
Seeing your solution, at least one of us two is wrong. I wouldn't be surprised at all if it is me...yet in your expression you didn't divide by two $\;\sin2\alpha\;$ ... Check this.
Added. Since it must be $\;t\in\left[-\frac\pi4,\,\frac{3\pi}4\right]\;$ for $\;r\;$ to be positive, the above integral must be changed accordingly to these limits. The outcome indeed is $\;\cfrac{5\pi a^2}{16}\;$
