Find the area of surface where $x=r\cos(\alpha)$ and $y=r\sin\alpha$ , where $(x^2 + y^2)^3 = x^4 + y^4$
I simplified the equation
$$r^2 = 3/4 + (\cos4\alpha)/4$$
from equation it is clear that
$$2^{-1/2} < r < (3/4 + (\cos4\alpha)/4)^{1/2}$$
and
$$0 < \alpha < \pi/4$$
after combining the integrals
$$\int_0^{\pi/4}d\alpha \int_{2^{-1/2}}^{(3/4 + (\cos4\alpha)/4)^{1/2}}r dr $$ and solving this i get $\pi/32$ which is wrong answer,
Please show me, where is my mistake.
I would say
$A=\frac12\int_{0}^{2\pi} r^2 \,d\alpha=1/8 \int_{0}^{2\pi} (3 + \cos 4\alpha)\,d\alpha=\cdots=\frac{3\pi}4$