# How do you evaluate the integral sin(x^2+y^2)dr where r is the region 9<= x^2 + y^2 <= 64 in polar form?

Feb 26, 2015

We should indeed use the polar coordinate system:
$x = r \cos \phi$
$y = r \sin \phi$

When you substitute these in ${x}^{2} + {y}^{2}$, you get
${x}^{2} + {y}^{2} = {\left(r \cos \phi\right)}^{2} + {\left(r \sin \phi\right)}^{2}$
$= {r}^{2} {\cos}^{2} \phi + {r}^{2} {\sin}^{2} \phi$
$= {r}^{2} \left({\cos}^{2} \phi + {\sin}^{2} \phi\right)$
$= {r}^{2}$
for ${\cos}^{2} + {\sin}^{2} = 1$

Our area becomes $9 \le {r}^{2} \le 64$ by substituting ${x}^{2} + {y}^{2} = {r}^{2}$.
Therefore, our limits for $r$ are: $3 \le r \le 8$, by taking the square root.

When we use polar coordinates, we should take the Jacobian($= J \left(r , \phi\right)$) into account, which is equal to $r$ for polar coordinates. Our integral becomes:
$\int \sin \left({x}^{2} + {y}^{2}\right) \mathrm{dr} = \int \sin \left({r}^{2}\right) \cdot J \left(r , \phi\right) \mathrm{dr} = {\int}_{3}^{8} r \sin \left({r}^{2}\right) \mathrm{dr}$

To solve this integral, we need to use substitution. We substitute $u = {r}^{2}$, so $\mathrm{du} = 2 r \mathrm{dr}$, in other words $\mathrm{dr} = \frac{1}{2 r} \mathrm{du}$.
${\int}_{3}^{8} r \sin {r}^{2} \mathrm{dr} = {\int}_{9}^{64} r \sin u \frac{1}{2 r} \mathrm{du} = {\int}_{9}^{64} \frac{1}{2} \sin u \mathrm{du} = {\left[- \frac{1}{2} \cos \left(u\right)\right]}_{9}^{64} = \frac{\cos \left(9\right) - \cos \left(64\right)}{2}$