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Thread: More Double Integrals Fun - Polar Coordinates

  1. #1
    Member billym's Avatar
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    More Double Integrals Fun - Polar Coordinates

    Use plane polar coordinates to evaluate the double integral:

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

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

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

    My "work":

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

    Correct?

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

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

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

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

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

    Am I finished with this stuff or am I wrong?
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  2. #2
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    Quote Originally Posted by billym View Post
    Use plane polar coordinates to evaluate the double integral:

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

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

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

    My "work":

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

    Correct?

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

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

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

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

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

    Am I finished with this stuff or am I wrong?

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

    Tonio
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  3. #3
    Member billym's Avatar
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    Does that mean I'm going clockwise and I use $\displaystyle y=-r\cdot sin(\theta)$ or do I stick with $\displaystyle y=r\cdot sin(\theta)$ ?
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