Question

Evaluate integration integration y^2+3x dA where R is the region in the third quadrant between x^2+y^2=1 and x^2+y^2=3

Answer 1

$\iint (y^2+3x) dR,$ where R is the region in the third quadrant between $x^2+y^2=1$ and $x^2+y^2=3$

We can use the polar coordinates

$\begin{cases} x = r\cos \phi\\ y = r\sin \phi \end{cases}$

the finction will be $f(r, \phi) = r(r^2 \sin^2 \phi+3r \cos \phi)$

Then

$\iint (y^2+3x) dR = \int_{1}^{\sqrt 3} dr \int_{\pi}^{\frac{3\pi}{2}} (r^3 \sin^2 \phi+3r^2 \cos \phi)d\phi=\\ =\int_{1}^{\sqrt 3} dr \int_{\pi}^{\frac{3\pi}{2}}(\cfrac{r^3}{2}-\cfrac{r^3\cos 2\phi}{2}+3r^2\cos\phi)d\phi =\\ =\int_{1}^{\sqrt 3} (\cfrac{r^3\phi}{2}-\cfrac{r^3\sin2\phi}{4}+3r^2sin\phi|_{\pi}^{\frac{3\pi}{2}})dr = \\ =\int_{1}^{\sqrt 3}(\cfrac{r^3}{2}\cfrac{3\pi}{2}-3r^2-\cfrac{r^3}{2}\pi)dr =\\ =\int_{1}^{\sqrt 3}(\cfrac{\pi r^3}{4}-3r^2)dr= \\ =(\cfrac{\pi r^4}{16}-r^3)|_{1}^{\sqrt 3} =\\ = \cfrac{9 \pi}{16}-3\sqrt 3 -(\cfrac{\pi}{16}-1) = \cfrac{\pi}{2}+1-3\sqrt 3$