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: $0\le r\le 2$, and $0\le \theta \le \frac{\pi }{2}$
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

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

Another Example

$\underset{R}{\iint }(x^2-y^2)\mathrm{d}x\mathrm{d}y$

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