MHB 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:

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

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

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

Hence:

$$b\theta=k\pi$$ where $$k\in\mathbb{Z}$$

$$\theta=\frac{k}{b}\pi$$

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

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

Let:

$$u=b\theta\,\therefore\,du=b\,d\theta$$

and our integral becomes:

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

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

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

Applying the FTOC, we get:

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

In our given problem, we identify:

$$a=8,\,b=7$$

hence:

$$A=\frac{8^2}{4\cdot7}\pi=\frac{16}{7}\pi$$