Let : [tex]I=\iint_R\sin(x^2+y^2)da[/tex] where :[tex]R=\{(x,y)\in{\mathbb R}^2~,~16\le x^2+y^2\le 49\}[/tex] .By using the polar coordinates : [tex]\begin{cases}x=\rho \cos\theta\\y=\rho\sin\theta \end{cases} [/tex] we get : [tex]x^2+y^2=\rho^2[/tex] and :
[tex](x,y)\in\matbb R\iff \begin{cases}16\le\rho^2\le49\\0\le\theta\le 2\pi\end{cases}[/tex]
Therefore :
[tex]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[/tex]
Now , let [tex]u=\rho^2[/tex] , then : [tex]\rhod\rho=\frac12 du[/tex] so we get :
[tex]I = 2\pi\int_4^7\frac12\sin u du=\pi\times \left[\cos u\right]_4^7=\pi(\cos7-\cos4)[/tex]
