polar loop: area of one loop of r = 4cos4theta

Re: polar loop

Hello, soccerball3211!

I'll set it up for you . . .

Find the area of one loop of the polar equation \(\displaystyle r\:=\:4\cos4\theta\)
Click to expand...
Polar area formula: \(\displaystyle \L\,A\;=\;\frac{1}{2}\int^{\;\;\;\beta}_{\alpha}r^2\,d\theta\)

We are given \(\displaystyle r\) . . . we can plug that in.
\(\displaystyle \;\;\)We need the limits of integration.

What the limits of one loop of that curve?
When does the curve go into the Pole (origin)?
\(\displaystyle \;\;\)Answer: when \(\displaystyle r\,=\,0\)

Solve: \(\displaystyle \,4\cos4\theta\,=\,0\)

We have: \(\displaystyle \,\cos4\theta\,=\,0\;\;\Rightarrow\;\;4\theta\,=\,\pm\frac{\pi}{2}\;\;\theta\,=\,\pm\frac{\pi}{8}\)

The limits are \(\displaystyle \,-\frac{\pi}{8}\) and \(\displaystyle \frac{\pi}{8}\)

The petal of this rose curve is symmetric to the "x-axis"
\(\displaystyle \;\;\)so we can integrate from \(\displaystyle 0\) to \(\displaystyle \frac{\pi}{8}\) and multiply by 2.

Set-up: \(\displaystyle \L\;A \;= \;2\,\times\,\frac{1}{2}\int^{\;\;\;\frac{\pi}{8}}_0\left(4\cos4\theta)^2\,d\theta\)

Can you finish it now?
 
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