# Thread: evaluate the double integral

1. ## evaluate the double integral

Using polar coordinates, $\int\int_{R} \sin(x^2+y^2)dA$ evaluate the integral where R is the region $1 \leq x^2+y^2 \leq 64.$

Heres what I got sofar, I need help integrating the inner integral. I'm lost from here.
$1 \leq x^2+y^2 \leq 64$
$1 \leq r^2 \leq 8$
$1 \leq r \leq 8$

$\int_{0}^{2pi}\int_{1}^{8}\sin(r^2) rdr d\theta$

2. Try a u-substitution $u = r^2$.

3. Because ...

$\frac {d}{dr} \cos (r^{2}) = - 2\cdot r\cdot \sin (r^{2})$

... is...

$\int r\cdot \sin (r^{2}) \cdot dr = - \frac{1}{2}\cdot \cos (r^{2}) + c$

... so that...

$\int_{0}^{2\cdot \pi} \int_{1}^{8} \sin (r^{2})\cdot r\cdot dr\cdot d\theta = \pi \cdot (\cos 1 - \cos 64) \simeq .46635958\dots$

Kind regards

$\chi$ $\sigma$