Calculate $\int_D\sin(\frac{x\pi}{2y})dxdy$

Question

Calculate $\int_D\sin(\frac{x\pi}{2y})dxdy$ $D=\{(x,y)\in\mathbb{R}^2:y\geq x,y\geq 1/\sqrt{2},y\leq(x)^{1/3}\}$.

I've calculate the limits of the integral $2^{-1/6}\leq x\leq1,1/\sqrt{2}\leq y\leq1$ and after doing integral of x first i got stuck

Answer 1

First, your region of integration is wrong. It looks like this:

This suggests integrating first with respect to $x$: \begin{align*}\int_{1/\sqrt{2}}^1\int_{y^3}^y \sin\left(\frac{x \pi}{2y}\right)\,dx\,dy &= \int_{1/\sqrt{2}}^1\left(-\frac{2 y \cos \left(\frac{\pi x}{2 y}\right)}{\pi }\right)\bigg\lvert_{y^3}^y\,dy\\ &= \int_{1/\sqrt{2}}^1\frac{2 y \cos \left(\frac{\pi y^2}{2}\right)}{\pi }\,dy\\ &= \frac{2 \sin \left(\frac{\pi y^2}{2}\right)}{\pi ^2}\bigg\lvert_{1/\sqrt{2}}^1 \\ &= \frac{2-\sqrt{2}}{\pi ^2}. \end{align*}

Answer 2

Assuming those limits are right (I haven't looked at them but at a glance they look a bit iffy so double check):

$$\int^1_{2^{-\frac{1}{2}}}\int^1_{2^{-\frac{1}{6}}} \sin(\frac{x\pi}{2y})\;dx\;dy\\= \int^1_{2^{-\frac{1}{2}}} \left( -\frac{2y}{\pi}\cos(\frac{x\pi}{2y}) \right)^{x=1}_{x=2^{-\frac{1}{6}}}\;dy\\ = \int^1_{2^{-\frac{1}{2}}} -\frac{2y}{\pi}\left(\cos(\frac{\pi}{2y}) - \cos(\frac{\pi}{2^{\frac{7}{6}}y})\right)\;dy$$

I would do this with integration by parts.