Question

The integral $\frac{1}{2\pi }\int {\int }_D(x+y+10)dxdy,$ where $D$ denotes the disc: $x^2+y^2\le 4,$ evaluates to

• A. 20

Answer 1

The correct option is A 20
Let $x=rcos\theta$
$y=rsin\theta$
$because{x}^2+y^2\le 4$
$r^{2}co{s}^2\theta +r^{2}si{n}^2\theta \le 4$


$\Rightarrow r^2\le 4$
$\therefore r\le 2$
Now, $I=\frac{1}{2\pi }\int \int (x+y+10)dxdy$
$\Rightarrow I=\frac{1}{2\pi }\int \int (rcos\theta +rsin\theta +10)rdrd\theta$
$\Rightarrow I=\frac{1}{2\pi }{\int }_{r=0}^2{\int }_{\theta =0}^{2\pi }{r}^{2}cos\theta drd\theta$
$+\frac{1}{2\pi }{\int }_{r=0}^2{\int }_{\theta =0}^{2\pi }{r}^{2}sin\theta drd\theta$
$+\frac{1}{2\pi }{\int }_{r=0}^{r=2}{\int }_{\theta =0}^{\theta =2\pi }10rdrd\theta$
$\Rightarrow I=\frac{1}{2\pi }[{\begin{array}{c}\frac{r^3}{3}\end{array}|}_0^2.sin\theta {|}_0^{2\pi }+{\begin{array}{c}\frac{r^3}{3}\end{array}|}_0^2.-cos\theta {|}_0^{2\pi }+10{\begin{array}{c}\frac{r^2}{2}\end{array}|}_0^2\theta {|}_0^{2\pi }]$
$\Rightarrow I=20$