How to evaluate this double integral over a semicircle?

Question

$$\iint_{D}\frac{x^4+y^4}{1+e^{3x^2y-y^3}} dxdy$$

$D = \{(x,y):x^2+y^2 \leq 1,x>0\}$. From the shape of region $D$ it seems to me that it's better to convert it to polar coordinate,but with no luck.

Answer 1

The integral may be converted to polar coordinates, $r \in [0,1]$, $\theta \in [-\pi/2,\pi/2]$, as

$$\int_0^1 dr \, r^5 \, \int_{-\pi/2}^{\pi/2} d\theta \frac{\cos^4{\theta}+\sin^4{\theta}}{1+e^{r^3 \sin{3 \theta}}}$$

Let

$$I(a) = \int_{-\pi/2}^{\pi/2} d\theta \frac{\cos^4{\theta}+\sin^4{\theta}}{1+e^{a \sin{3 \theta}}}$$

for some $a$. Consider the quantity

$$\begin{align}I(0)-I(a) &= \frac12 \int_{-\pi/2}^{\pi/2} d\theta \left (\cos^4{\theta}+\sin^4{\theta} \right) \frac{e^{a \sin{3 \theta}}-1}{e^{a \sin{3 \theta}}+1}\\ &= \frac12 \int_{-\pi/2}^{\pi/2} d\theta \left (\cos^4{\theta}+\sin^4{\theta} \right) \tanh{\left(\frac{a}{2} \sin{3 \theta}\right)}\\ &= 0\end{align}$$

because we are integrating an odd integrand over a symmetric interval. Thus, $I(a)$ is independent of $a$ and is equal to

$$I(0) = \frac12 \int_{-\pi/2}^{\pi/2} d\theta \left (\cos^4{\theta}+\sin^4{\theta} \right) = \frac{3 \pi}{8} $$

Therefore, the sought-after integral is then

$$ \frac{3 \pi}{8} \int_0^1 dr \, r^5 = \frac{\pi}{16}$$

Answer 2

Hint 1: $$x^4 + y^4 = r^4 - 2x^2y^2 = r^4-2r^4\cos^2(\theta)\sin^2(\theta)=\frac{1}{2}r^4\sin^2(2\theta)$$ Hint 2: $$3x^2y - y^3 = y(3x^2 - y^2)= r^3\sin(\theta)\left(3\cos^2(\theta) - \sin^2(\theta)\right)$$ $$=2 \cos(2 \theta)+1$$