Question

Using polar coordinates, evaluate the integral ∫∫rsin(x2+y2)da where r is the region 16≤x2+y2≤49

Answer 1

Let :
`I=\iint_R\sin(x^2+y^2)da` where :
`R=\{(x,y)\in{\mathbb R}^2~,~16\le x^2+y^2\le 49\}` .By using the polar coordinates :
`\begin{cases}x=\rho \cos\theta\\y=\rho\sin\theta \end{cases}` we get :
`x^2+y^2=\rho^2` and :

`(x,y)\in\matbb R\iff \begin{cases}16\le\rho^2\le49\\0\le\theta\le 2\pi\end{cases}`
Therefore :

`I=\int_(16)^(49)\int_0^(2\pi)\sin(\rho^2)\rho d\rho d\theta=2\pi\int_(16)^(49)\rho\sin(\rho^2) d\rho`
Now , let
`u=\rho^2` , then :
`\rhod\rho=\frac12 du` so we get :


`I = 2\pi\int_4^7\frac12\sin u du=\pi* \left[\cos u\right]_4^7=\pi(\cos7-\cos4)`