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


`(\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
`\frac{\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)`