More Double Integrals Fun - Polar Coordinates

**billym** · November 24th 2009, 09:45 AM

Use plane polar coordinates to evaluate the double integral:

$I=\int\int_Ax-y-1 \, dxdy$

where A is the region defined by $x \ge 0$ and $1\le x^2+y^2 \le 4$

*************

My "work":

$\int_{\pi/2}^{3 \pi/2}\int_{1}^{2}r^2cos(\theta)-r^2sin(\theta)-r \,drd\theta$

Correct?

$=\int_{\pi/2}^{3 \pi/2}\left[ \frac{r^3(cos(\theta)-sin(\theta))}{3}-\frac{r^2}{2}\right]_{1}^{2} \, d(\theta)$

$=\int_{\pi/2}^{3 \pi/2}\frac{7}{3}cos(\theta)-\frac{7}{3}sin(\theta)-3/2$

$=\left[\frac{7}{3}sin(\theta)+\frac{7}{3}cos(\theta)-\frac{3}{2}(\theta)\right]_{\pi/2}^{3\pi/2}$

$=\frac{7}{3} sin \left( \frac{3 \pi}{2} \right)+\frac{7}{3} cos \left( \frac{3 \pi}{2}\right)-\frac{3}{2}\left(\frac{3 \pi}{2}\right)- \left[ \frac{7}{3} sin \left(\frac{\pi}{2}\right)+\frac{7}{3} cos\left( \frac{\pi}{2}\right)-\frac{3}{2}\left(\frac{\pi}{2}\right)\right]$

$=-\frac{7}{3}-\frac{9 \pi}{4}-\frac{7}{3}+ \frac{3 \pi}{4}=-\frac{14}{3}-\frac{3 \pi}{2}$

Am I finished with this stuff or am I wrong?

**tonio** · November 24th 2009, 10:38 AM

Originally Posted by billym

Use plane polar coordinates to evaluate the double integral:

$I=\int\int_Ax-y-1 \, dxdy$

where A is the region defined by $x \ge 0$ and $1\le x^2+y^2 \le 4$

*************

My "work":

$\int_{\pi/2}^{3 \pi/2}\int_{1}^{2}r^2cos(\theta)-r^2sin(\theta)-r \,drd\theta$

Correct?

$=\int_{\pi/2}^{3 \pi/2}\left[ \frac{r^3(cos(\theta)-sin(\theta))}{3}-\frac{r^2}{2}\right]_{1}^{2} \, d(\theta)$

$=\int_{\pi/2}^{3 \pi/2}\frac{7}{3}cos(\theta)-\frac{7}{3}sin(\theta)-3/2$

$=\left[\frac{7}{3}sin(\theta)+\frac{7}{3}cos(\theta)-\frac{3}{2}(\theta)\right]_{\pi/2}^{3\pi/2}$

$=\frac{7}{3} sin \left( \frac{3 \pi}{2} \right)+\frac{7}{3} cos \left( \frac{3 \pi}{2}\right)-\frac{3}{2}\left(\frac{3 \pi}{2}\right)- \left[ \frac{7}{3} sin \left(\frac{\pi}{2}\right)+\frac{7}{3} cos\left( \frac{\pi}{2}\right)-\frac{3}{2}\left(\frac{\pi}{2}\right)\right]$

$=-\frac{7}{3}-\frac{9 \pi}{4}-\frac{7}{3}+ \frac{3 \pi}{4}=-\frac{14}{3}-\frac{3 \pi}{2}$

Am I finished with this stuff or am I wrong?

If you integrate between $\frac{\pi}{2}\mbox{ and }\frac{3\pi}{2}$ you\re integrating on the left semi-circle ( $x<0$ ) , NOT on on the right one $x>0$ , as you have to.
You have to integrate between $-\frac{\pi}{2}\,\,\,and\,\,\,\frac{\pi}{2}$

Tonio

**billym** · November 25th 2009, 09:59 AM

Does that mean I'm going clockwise and I use $y=-r\cdot sin(\theta)$ or do I stick with $y=r\cdot sin(\theta)$ ?

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