Question

Sin(x2 y2 da r , where r is the region in the first quadrant between the circles with center the origin and radii 1 and 5

Answer 1

I assume there's a plus sign missing above...

Convert to polar coordinates, using

$\begin{cases}x(r,\theta)=r\cos\theta\\y(r,\theta)=r\sin\theta\end{cases}$

Then the Jacobian is

$\dfrac{\partial(x,y)}{\partial(r,\theta)}=\begin{vmatrix}x_r&y_r\\x_\theta&y_\theta\end{vmatrix}=\begin{vmatrix}\cos\theta&\sin\theta\\r\sin\theta&-r\cos\theta\end{vmatrix}=-r$

Then

$\mathrm dA=\mathrm dx\,\mathrm dy=|-r|\,\mathrm dr\,\mathrm d\theta=r\,\mathrm dr\,\mathrm d\theta$

so the integral can be written as

$\displaystyle\iint_R\sin(x^2+y^2)\,\mathrm dA=\int_0^{\pi/2}\int_1^5r\sin(r^2)\,\mathrm dr\,\mathrm d\theta$

Let $s=r^2$, so that $\dfrac{\mathrm ds}2=r\,\mathrm dr$.

$\displaystyle\frac12\int_0^{\pi/2}\int_1^{25}\sin s\,\mathrm ds\,\mathrm d\theta=-\frac12(\cos25-\cos1)\int_0^{\pi/2}\mathrm d\theta=\frac\pi4(\cos1-\cos25)$