Area of a Region, Polar, common to r = 1 + sin(theta), r = 1

flakine

Junior Member
Joined
Aug 24, 2005
Messages
78
Find area that is common to r=1+sin θ and r=1

2∫1/2 (1+sin θ )^2 dθ + 2∫1/2 (1)^2 dθ
(interval 0,-pi/2), and (interval pi/2, 0) repectively

∫ (1+sin θ )^2 dθ + ∫ 1dθ

∫ 1+2sinθ+sinθ^2 dθ + ∫ 1dθ

∫ 1+2sinθ+∫((1-cos2θ)/2) dθ + ∫ 1dθ

θ+2cosθ + ∫1-cos u dθ + ∫ 1dθ (u=2θ)

θ+2cosθ + θ +sin 2θ| (0,-pi/2)+ θ | (pi/2, 0)

=pi

What am I doing wrong!
 
The quadrant of the circle is in the region. Also, the small section of the cardioid below the x-axis.

circle:

\(\displaystyle \L\\2(\frac{1}{2})\int_{0}^{\frac{\pi}{2}}{1^{2}}d{\theta}\)

Just from geometry you know this quarter-circle region is \(\displaystyle \frac{\pi}{4}\)

Cardioid:

\(\displaystyle \L\\2(\frac{1}{2})\int_{\frac{-\pi}{2}}^{0}{(1+sin({\theta}))^{2}d{\theta}\)

Add them.

cardioidcs2.jpg
 
I got that part, the intergration in kind of tricky. Can you help with that?
 
How far have you gotten?

Please reply with specifics. Thank you.

Eliz.
 
Its all listed out in my original post. I don't have to means to put in the correct math symbols, but it all fair clear.
 
The first one is a no-brainer. You probably need help with the last integral. Right?.

Expand it out:

\(\displaystyle \L\\\int{(1+sin({\theta}))^{2}}d{\theta}=\int{1}d{\theta}+\int{2sin({\theta})}d{\theta}+\int{sin^{2}({\theta})}d{\theta}\)

The first 2 are easy.

To integrate \(\displaystyle \L\\sin^{2}({\theta})\), try rewriting it as

\(\displaystyle \L\\\frac{1}{2}(1-cos(2{\theta}))\)

Yes, you do have the means to use the math symbols. I type all the LaTex. You just need to take the time to learn some of the code. Click on 'quote' at the right top corner of the post to see the code I used. There are lots of tutorials on line and you can also use TexAide if you don't want to type it in manually.
 
Top