Thread: Where have I gone wrong here? (Area enclosed by polar coords)

1. Where have I gone wrong here? (Area enclosed by polar coords)

Hi guys,

I'm asked to show that the area enclosed by the curve

$\displaystyle r=2(1-\sin\theta)\sqrt{\cos\theta}$

where

$\displaystyle -\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}$ is 16/3

I proceeded as follows

$\displaystyle A = \frac{1}{2}\int^{\frac{\pi}{2}}_{\frac{-\pi}{2}}\left(2(1-\sin\theta)\sqrt{\cos\theta}\right)^2\, d\theta$

after some working I arrived at

$\displaystyle A = 2\left[\sin\theta+\frac{1}{2}\cos(2\theta)+\frac{1}{3}\si n^3\theta\right]^{\frac{\pi}{2}}_{-\frac{\pi}{2}}$

$\displaystyle =2\left[\frac{1}{3} - (-2\frac{1}{3})\right] = 2\left[\frac{8}{3}\right] = \frac{16}{3}$

as required.

I was then asked to show that the initial line divides the area enclosed by the curve in the ratio 1:7. I thought this would simply be a case of putting in pi/2 and zero into the above integral and obtaining 2/3 (thus the other part would be 14/3 - a ratio of 1:7) but instead I got -1/3!

So I tried putting in 0 and -pi/2 into the formula and I got 17/3.

These two sum to give the required whole area but do not split the area into 1:7. I have absolutely no idea where to go from here.

Any help would be much appreciated

2. MUst be simple arithmetic since we get

2[ sin(t)+1/2cos(2t)+1/3sin^3(t)] (0 to pi/2)=

2[(1 -1/2 +1/3) -1/2]= 2/3

3. Originally Posted by Calculus26
MUst be simple arithmetic since we get

2[ sin(t)+1/2cos(2t)+1/3sin^3(t)] (0 to pi/2)=

2[(1 -1/2 +1/3) -1/2]= 2/3
...

*facepalm*

thanks