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Math Help - Whats going wrong? Double integrals in polar coordinates

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    Whats going wrong? Double integrals in polar coordinates

    Find the area enclosed by one leaf of the rose r=12cos(3\theta)

    \displaystyle\int^\frac{\pi}{6}_0 r\ dr \ d\theta
    \displaystyle\int^\frac{\pi}{6}_0 \left[ \frac{r^2}{2}\right]\limits_{0}^{12cos(3\theta)} \ d\theta
    \displaystyle\int^\frac{\pi}{6}_0 144 cos^2(3\theta) \ d\theta
    36 \displaystyle\int^\frac{\pi}{6}_0 3cos(\theta) + cos(3\theta) \ d\theta
    \left[ 3sin(\theta) +\displaystyle\frac{sin(3\theta)}{3}\right]\limits_{0}^{\pi/6}

    the problem is that this equals 33 and not the answer of 12pi as in my book
    Last edited by mr fantastic; March 13th 2011 at 04:17 PM. Reason: Title.
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    Quote Originally Posted by FGT12 View Post
    Find the area enclosed by one leaf of the rose r=12cos(3\theta)

    \displaystyle\int^\frac{\pi}{6}_0 r\ dr \ d\theta
    \displaystyle\int^\frac{\pi}{6}_0 \left[ \frac{r^2}{2}\right]\limits_{0}^{12cos(3\theta)} \ d\theta
    \displaystyle\int^\frac{\pi}{6}_0 144 cos^2(3\theta) \ d\theta
    36 \displaystyle\int^\frac{\pi}{6}_0 3cos(\theta) + cos(3\theta) \ d\theta
    \left[ 3sin(\theta) +\displaystyle\frac{sin(3\theta)}{3}\right]\limits_{0}^{\pi/6}=

    the problem is that this equals 33 and not the answer of 12pi as in my book
    Hmmm your title says double integrals but I only see one but here it goes.

    Setting 0=12\cos(3\theta) gives \displastyle \frac{\pi}{2}+\pi n=3\theta \iff \frac{2n+1}{6}\pi=\theta

    Now since we are using a double integral to find area we want to integrate

    \displaystyle \iint 1 dA=\int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \int_{0}^{12\cos(3\theta)}rdrd\theta

    Now taking the integrals gives

    \displaystyle \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \frac{r^2}{2}\bigg|_{0}^{12\cos(3\theta)}d\theta =72 \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \cos^2(3\theta)d\theta =36 \int_{\frac{\pi}{6}}^{\frac{\pi}{2}}1+ \cos(6\theta)d\theta

    \displaystyle 36\left[ \theta+\frac{1}{12}\sin(6\theta)\right]\bigg|_{\frac{\pi}{2}}^{\frac{\pi}{6}}=36\left( \frac{\pi}{2}-\frac{\pi}{6}\right)=12\pi
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