Area Enclosed by One Petal of a Rose Curve: r = 8sin7θ

[MarkFL]

Here is the question:

Find the area of the region enclosed by one loop of the curve.?

Find the area of the region enclosed by one loop of the curve:

r = 8 sin 7θ

I have posted a link there to this thread so the OP can view my work.

 

[MarkFL]

Hello Katie,

Let's generalize a bit and use the function:

\(\displaystyle r=a\sin(b\theta)\) where neither $a$ nor $b$ is zero.

We should set the function to zero to find our limits of integration:

\(\displaystyle a\sin(b\theta)=0\)

Hence:

\(\displaystyle b\theta=k\pi\) where \(\displaystyle k\in\mathbb{Z}\)

\(\displaystyle \theta=\frac{k}{b}\pi\)

Taking two consecutive values of $k$ (0 and 1 will do) we may now state the area $A$ enclosed by one petal is:

\(\displaystyle A=\frac{1}{2}\int_0^{\frac{\pi}{b}} \left(a\sin(b\theta) \right)^2\,d\theta=\frac{a^2}{2}\int_0^{\frac{\pi}{b}} \sin^2(b\theta)\,d\theta\)

Let:

\(\displaystyle u=b\theta\,\therefore\,du=b\,d\theta\)

and our integral becomes:

\(\displaystyle A=\frac{a^2}{2b}\int_0^{\pi} \sin^2(u)\,du\)

Applying a double-angle identity for cosine on the integrand, we obtain:

\(\displaystyle A=\frac{a^2}{4b}\int_0^{\pi}1-\cos(2u)\,du\)

Applying the FTOC, we get:

\(\displaystyle A=\frac{a^2}{4b}\left[u-\frac{1}{2}\sin(2u) \right]_0^{\pi}=\frac{a^2}{4b}\pi\)

In our given problem, we identify:

\(\displaystyle a=8,\,b=7\)

hence:

\(\displaystyle A=\frac{8^2}{4\cdot7}\pi=\frac{16}{7}\pi\)