# Whats going wrong? Double integrals in polar coordinates

• Mar 13th 2011, 02:44 PM
FGT12
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
• Mar 13th 2011, 03:06 PM
TheEmptySet
Quote:

Originally Posted by FGT12
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

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

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