Answer to Question #139340 in Calculus for Promise Omiponle

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

$\iint_R \sin(x^2+y^2)dA, R=\{16\leq x^2+y^2\leq 25\}$

$x = r\cos \phi \\y = r \sin\phi\\ I = r\\$

Find the integration limits:

$\phi \in [0,2\pi]$

$16 \leq x^2 + y^2 \leq 25\\ 16 \leq r^2\cos^2\phi + r^2\sin^2\phi \leq 25\\ 16\leq r^2\leq 25\\ 4\leq r \leq 5$

$r \in[4,5]$

So:

$\iint_R sin(x^2 + y^2)dA = \int_4^5dr\int_0^{2\pi} rsin(r^2)d\phi = \\ =\int_4^5rsin(r^2)dr \phi|_0^{2\pi} = 2\pi\int_4^5 rsin(r^2)dr = \\ = \left[ \begin{array}{ccc} r^2 = u & 5 \to 25\\ 2rdr = du& 4 \to 16\\ \end{array} \right] = \pi \int_{16}^{25}sin(u)du =\\ =-\pi cos(u)|_{16}^{25} =\pi(cos16 - cos25) = \\=2\pi sin(\frac{41}{2})sin(\frac{9}{2})$