If a Rose leaf is described by the equation $r = \sin 3\theta$, find the area of one petal.
[Math] Find the area of the Rose’s petal.
polar coordinates
Related Solutions
This is the polar graph of your function $r=4\cos(3\theta)$:
As mentioned in the comment from @coffeemath; by domain I will assume that you mean the values for $\theta$ to make each petal. So the interval in this case is simply $\cfrac{2\pi}{3}$ for each petal as the full domain ($2\pi$) must be divided equally across each petal. So this means that $-\cfrac{\pi}{6} \le \theta \le \cfrac{\pi}{6}$ is the domain for the first petal (centred on the $x$-axis).
I will work this out assuming that the curve was supposed to lie in the plane $x+y+z = 3$, because in $\mathbb{R}^3$, the orthogonal complement of the associated plane through the origin has dimension $1$.
Notice that if the curve were supposed to lie in the $xy$-plane instead, we could set $$ (x(t), y(t)) = 4 \cos(5t) (\cos t, \sin t) = 4 \cos(5t)[\cos t \, \textbf{e}_1 + \sin t \textbf{e}_2 ]. $$ This function gives a rose with $5$ petals because it is $\pi$-periodic, and there are precisely $5$ values of $t$ in $[0, \pi)$ for which $|\cos(5t)| = 1$. (Note that it is $\pi$-periodic, because $(\cos(t + \pi), \sin (t + \pi)) = -(\cos t, \sin t)$ and $\cos(5(t + \pi)) = \cos(5t + 5 \pi) = \cos(5t+ \pi) = - \cos(5t)$.)
This easier problem suggests a way forward in the case of the plane $x+y+z=3$: if we find an orthonormal basis $\{ \textbf{w}_1, \textbf{w}_2 \}$for the orthogonal complement of $(1,1,1)$ so that $\textbf{w}_1$ points in the direction of $(0,0,3) - (1,1,1)$, then we can replace $\textbf{e}_1$ and $\textbf{e}_2$ by the new basis vectors and translate by $(1,1,1)$ to give $$ (x(t), y(t), z(t) ) = (1,1,1) + 4 \cos(5t) [\cos t \ \textbf{w}_1 + \sin \, \textbf{w}_2]. $$
Note that if we plug in $t = 0$, then we get $$ (x(0), y(0), z(0)) = (1,1,1) + 4 \textbf{w}_1, $$ so there is a petal in the direction of $\textbf{w}_1$, which is the direction of the line from the center $(1,1,1)$ to the point $(0,0,3)$.
Best Answer
A sketch is useful here, but the only important observation is that $r=0$ when $\theta=0$, and again at $\frac{\pi}{3}$. These are your limits for one petal.
Since the area of a polar curve between the rays $\theta=a$ and $\theta=b$ is given by $\int_{a}^{b}\frac{1}{2}r^{2}d\theta$, we have
$$A=\int_{0}^{\pi/3}\frac{1}{2}\sin^{2}(3\theta)d\theta=\frac{1}{2}\int_{0}^{\pi/3}\frac{1-\cos(6\theta)}{2}d\theta$$ $$=\frac{1}{4}\left[\theta-\frac{\sin(6\theta)}{2}\right]^{\pi/3}_{0}=\frac{1}{4}\left(\frac{\pi}{3}-\frac{1}{2}\sin\left(\frac{6\pi}{3}\right)\right)=\frac{\pi}{12}$$