# Area in Polar Coordinates 2

• Dec 9th 2007, 08:06 PM
shogunhd
Area in Polar Coordinates 2
Ok so the equation is

r= 6 +4 sin theta

Its is a limacon so the area will be symmetrical from 0 to pi/2

Plugging this into the area formula I get

Integral (6+4 sin theta)^2
Integral (36 + 48 sin theta + 16 sin^2 theta)
Integral (36 + 48 sin theta + 16 [(1-cos 2 theta)/2]
Integral (36 + 48 sin theta + 8 (1 - cos 2 theta)
Integral (36 + 48 sin theta + 8 - 8 cos 2 theta)
Integral (44 + 48 sin theta - 8 cos 2 theta)

which gives me

44 theta - 48 cos theta - 8 sin 2 theta + C

I worked it out and I am still unable to arrive at the answer. Can anyone tell me what I'm doing wrong?
• Dec 9th 2007, 10:00 PM
Soroban
Hello, shogunhd!

Quote:

Find the area enclosed by: .$\displaystyle r \:= \:6 +4\sin\theta$

It is a limacon, symmetric to the "y-axis".
We can integrate from $\displaystyle -\frac{\pi}{2}$ to $\displaystyle \frac{\pi}{2}$ . . . and double.

We have: .$\displaystyle A \;=\;2 \times \frac{1}{2}\int^{\frac{\pi}{2}}_{\text{-}\frac{\pi}{2}} (6 + 4\sin\theta)^2\,d\theta$

The function is: .$\displaystyle 36 + 48\sin\theta + 16\sin^2\!\theta \;=\;36 + 48\sin\theta + 16\left(\frac{1-\cos2\theta}{2}\right)$

. . $\displaystyle = \;36 + 48\sin\theta + 8 - 8\cos2\theta \;=\;44 + 48\sin\theta - 8\cos2\theta$

The integral is: .$\displaystyle A \;\;=\;\;\int^{\frac{\pi}{2}}_{\text{-}\frac{\pi}{2}}\left(44 + 48\sin\theta - 8\cos2\theta\right)\,d\theta \;\;=\;\;44\theta - 48\cos\theta - 4\sin2\theta\,\bigg|^{\frac{\pi}{2}}_{\text{-}\frac{\pi}{2}}$

. . $\displaystyle = \;\;\bigg[44\left(\frac{\pi}{2}\right) - 48\cos\left(\frac{\pi}{2}\right) - 4\sin(\pi)\bigg] - \bigg[44\left(\text{-}\frac{\pi}{2}\right) - 48\cos\left(\text{-}\frac{\pi}{2}\right) - 4\sin(-\pi)\bigg]$

. . $\displaystyle = \;\;\bigg[22\pi - 0 - 0\bigg] - \bigg[-22\pi - 0 - 0\bigg] \;\;=\;\;\boxed{44\pi}$

• May 5th 2010, 09:31 PM
Ife
Why is this doubled though? If the integral is between $\displaystyle \pi/2 and -\pi/2$ isn't that for the entire region?