Question

Using polar coordinates, evaluate the integral ∫∫Rsin(x^2+y^2)dA where R is the region 16≤x^2+y^2≤25.

Answer 1

$\iint_R \sin(x^2+y^2)dA, R: 16 \leq x^2+y^2 \leq25$

$x=r\cdot \cos \phi, y=r \cdot \sin \phi$

$x^2+y^2=r^2\cos^2 \phi+r^2\sin^2 \phi=r^2 \cdot (cos^2\phi+sin^2\phi)=r^2\implies$

$\implies 16 \leq r^2\leq 25 \implies 4 \leq r \leq 5$

$R=\{(r,\phi): 4\leq r \leq 5, 0 \leq\phi\leq2\pi\}$

$\iint_R \sin(x^2+y^2)dA=\int_4^5\int_0^{2\pi}\sin r^2 \cdot rd\phi dr=$

$=\int_4^5 r\sin r^2dr \int_0 ^{2\pi}d\phi=2\pi\int_4^5 \frac{1}{2}\sin r^2 d(r^2)=$

$=-\pi \cos r^2|_4^5 =-\pi(\cos 25-\cos 16)=\pi (\cos 16-\cos 25)$

$\iint_R \sin(x^2+y^2)dA=\pi (\cos 16-\cos 25)$