CSCE 315 Lecture 21

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Integrals in Polar Coordinates

• Simplify the region
• Simplify the integral

$\iint 4y\,\mathrm {d} A$ around a quarter-circle of radius 2 in the first quadrant.

There are three pieces to fix:

1. the range: $0\leq r\leq 2$, and $0\leq \theta \leq {\tfrac {\pi }{2}}$
2. The Integrand: $x=r\cos {\theta }$, and $y=r\sin {\theta }$
3. and the differentials: $\mathrm {d} A=\mathrm {d} x\mathrm {d} y=r\mathrm {d} r\mathrm {d} \theta$. This is because $x$ "translates" into $\mathrm {d} r$, and $y$ "translates" into $r\mathrm {d} \theta$

Therefore the new integral is "simplified" to

$\int _{0}^{\frac {\pi }{2}}\int _{0}^{2}4r^{2}\sin {\theta }\,\mathrm {d} r\,\mathrm {d} \theta$

Another Example

$\iint _{R}\left(x^{2}-y^{2}\right)\,\mathrm {d} x\,\mathrm {d} y$

1. $x^{2}-y^{2}=r^{2}\cos {(2\theta )}$
2. R: $0\leq \theta \leq \arctan(2)$, $0\leq r\leq {\tfrac {2}{\cos {\theta }}}$