Help with some double integral $$\int_{0}^{\pi/2}\int_{x}^{\pi/2}\frac{\cos y}{y} \, \operatorname{d}\!y\, \operatorname{d}\!x$$
$\iint_R2xy^2 \, \operatorname{d}\!A$ where R is the right half of the unit circle
 A: That first one doesn't converge. Note that if it did, $$\begin{align}\int_0^{\pi/2}\frac{\cos y}y\,dy &= \int_0^{\pi/3}\frac{\cos y}y\,dy+\int_{\pi/3}^{\pi/2}\frac{\cos y}y\,dy\\ &\geq \int_0^{\pi/3}\frac{\cos y}y\,dy\\ &\geq \frac12\int_0^{\pi/3}\frac1y\,dy\\ &=\frac12\lim_{a\to 0^+}\int_a^{\pi/3}\frac1y\,dy.\end{align}$$ Check out that limit.
For the second one, note that in polar coordinates, the right half of the unit circle is $0\leq r\leq 1,-\frac\pi2\leq\theta\leq\frac\pi2.$ Use $x=r\cos\theta,y=r\sin\theta,dA=r\,dr\,d\theta$ to evaluate the second integral.

Since you've fixed your post, note that $$\int_0^{\pi/2}\int_x^{\pi/2}\frac{\cos y}y\,dy\,dx$$ is integrating the function $\frac{\cos y}y$ over the region bounded below by the line $y=x$ and above by the line $y=\frac\pi2,$ with $x$-values ranging from $0$ to $\pi/2$. Equivalently, though, it's bounded on the left by $x=0$ and on the right by the line $x=y$, with $y$-values ranging from $0$ to $\pi/2$. Hence, changing the order of integration gives us $$\begin{align}\int_0^{\pi/2}\int_x^{\pi/2}\frac{\cos y}y\,dy\,dx &= \int_0^{\pi/2}\int_0^y\frac{\cos y}y\,dx\,dy\\ &= \int_0^{\pi/2}\frac{\cos y}y\left(\int_0^y\,dx\right)\,dy.\end{align}$$ Since $\int_0^y\,dx=y,$ then the rest is easy.
A: $$\iint_R2xy^2dA=\iint_{R=\{(x,y)\mid x^2+y^2=1,x\geq0\}}2xy^2dxdy=\int_0^12x\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}y^2dy\\=\int_0^12r\cos(\theta)\int_{-\pi/2}^{\pi/2}r^2\sin^2(\theta)rdrd\theta=\int_0^12r^4dr\int_{-\pi/2}^{\pi/2}\cos(\theta)\sin^2(\theta)d\theta$$
A: Сhange the order of integration in the first integral:
 $$\int_{0}^{\frac{\pi}{2}}\int_{x}^{\frac{\pi}{2}}\frac{\cos y}{y} \, \operatorname{d}\!y\, \operatorname{d}\!x=\int_{0}^{\frac{\pi}{2}}\int_{0}^{y}\frac{\cos y}{y} \, \operatorname{d}\!x\, \operatorname{d}\!y=\int_{0}^{\frac{\pi}{2}}\frac{\cos y}{y}  \int_{0}^{y}\, \operatorname{d}\!x\, \operatorname{d}\!y= \\
= \int_{0}^{\frac{\pi}{2}}\frac{\cos y}{y} \cdot{y}\,\operatorname{d}\!y=\int_{0}^{\frac{\pi}{2}}{\cos y}\,\operatorname{d}\!y=\left.\sin{y} \right|_{0}^{\frac{\pi}{2}}=1$$
A: $$\int_{0}^{\pi/2}\int_{x}^{\pi/2}\frac{\cos y}{y} \, \operatorname{d}\!y\, \operatorname{d}\!x = \int_{0}^{\pi/2}\frac{\cos y}{y}\int_{0}^{y} \, \operatorname{d}\!x\, \operatorname{d}\!y=\ldots.  $$
