# Thread: Whats going wrong? Double integrals in polar coordinates

1. ## 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

2. 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

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

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