Question

Find the area of the region enclosed by one loop of the curve. GIVEN: r=2sin(5theta)

Answer 1

What range of \(\theta\) gives you one complete loop?

Answer 2

do I use 1/2 integral r^2 dtheta?

Answer 3

it doesnt give any.. it just says one loop

Answer 4

Right, that means it's up to you to figure out the interval of integration. The formula is correct, you have
\[A=\frac{1}{2}\int_{\theta_1}^{\theta_2}r^2\,d\theta\]
We know that \(r=2\sin5\theta\), but we don't know (yet) what \(\theta_1\) and \(\theta_2\) will be. Do you know how to plot functions in polar coordinates?

Answer 5

unfortunately no and I don't remember my teacher going over that?

Answer 6

Let's pick \(\theta=0\) as a starting point. Then \(r(0)=2\sin(5\times0)=0\), so we have the point \((r,\theta)=(0,0)\).

For our next point, let's take something that will give us a reasonable sine ratio computation, say \(\theta=\dfrac{\pi}{10}\). Then \(r\left(\dfrac{\pi}{10}\right)=2\sin\dfrac{5\pi}{10}=2\sin\dfrac{\pi}{2}=2\), so we have the point \((r,\theta)=\left(2,\dfrac{\pi}{10}\right)\).

One more: let \(\theta=\dfrac{\pi}{5}\), then \(r\left(\dfrac{\pi}{5}\right)=2\sin\dfrac{5\pi}{5}=2\sin\pi=0\), so we have the point \((r,\theta)=\left(0,\dfrac{\pi}{5}\right)\).

Here's our preliminary sketch given these three points:

It looks like one loop can be covered over \(\left[0,\dfrac{\pi}{5}\right]\), so the area would be given by
\[A=\frac{1}{2}\int_0^{\pi/5}(2\sin5\theta)^2\,d\theta\]