We have the function $r=\sin(3\theta)$, and I want to calcule the area under its curve.
I think this is easy if we use a double integral. It should be:
$3\displaystyle\int_{0}^{\pi/6}\int_{0}^{\sin(3\theta)}rdrd\theta$
Am I right?
But now, I want to find the solution by using a one-variable integral. Is this possible?
Thanks.
It is indeed possible to find the area enclosed by the curve $r = \sin(3\theta)$ using just one integral. Remember that the formula for the area enclosed by $r=f(\theta)$ between $\theta= \alpha $ and $ \theta = \beta$ in polar coordinates is
A = $\displaystyle\int_{\alpha}^{\beta}\frac{1}{2} r^2 d\theta $
We can use this formula to find the area of our function. To simplify calculations, we can use the fact that the graph of $r=\sin(3\theta)$ is a "rose", and that area each of the "petals" is same and thus the area enclosed by our "3-petaled rose" is three times the area of one petal.
One petal is traced out from $\theta = 0$ to $\theta = \pi/3$ so the area enclosed by the curve is given by: \begin{align} A &= 3\int_{0}^{\pi/3}\frac{1}{2} \sin^2(3\theta) d\theta \\ &= \frac{3}{4}\displaystyle\int_{0}^{\pi/3} (1-\cos(6\theta)) d\theta \\ &= \frac{3}{4}[(\frac{\pi}{3} - \frac{1}{6}\sin(2\pi))-(0-\frac{1}{6}\sin(0))]\\ &= \frac{3}{4} \cdot \frac{\pi}{3} \\ &= \frac{\pi}{4} \end{align}
