# Polar Coordinates - Areas

1. May 7, 2012

### captainquarks

I am asked to consider the following graph:

r2=a+sin(θ), where a=2

I have a picture of this plot, which I have attached,

We are asked to find the area of the upper 'cresent' of the curve, contained at the top

How would I go about calculating that?

I've found that if I plot r=√(2+sinθ) and r=-√(2+sinθ) that this gives me the separate graphs individually, and that if I integrate from 0-π on the first one, then integrate from π-2π on the second one, if i subtract, i get the correct answer geometrically, using my graphing programme, but i dont know how to do this analytically? (i have attached another image to show the to sections - its the blue section i need)

Any help would be vastly appreciated, thank you

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2. May 7, 2012

### captainquarks

From my graphing programme, ive found the area should be 4.06-2.139 = 1.921 approximately, if this helps anyone?

3. May 7, 2012

### hivesaeed4

I think your going to have to use cylindrical coordinates. From the graph it's evident that the limits of theta would be 0 to pi. The limits of r would be sqrt(2) to sqrt(3) (if your confused about how did I get these limits try finding the max and min values of r by playing around with the theta value).

4. May 7, 2012

### captainquarks

I get where you have your limits by maximising sin in the range of 0-pi, giving r^2= 2 or 3... Ive never done cylindrical coordinates before? We've never encountered them in lessons yet. Worrying

5. May 7, 2012

### captainquarks

Never mind then lol. I know its definitely not cylindrical coordinates. Though.

6. May 10, 2012

### jssamp

Is there any reason you are plotting $r = -\sqrt{2+sinθ}$? It would be easier if you use $r = \sqrt{2-sinθ}$. It plots the same circle but the intersecting points are coincident. The you could simply integrate

$∫\frac{1}{2}((2+sinθ)-(2-sinθ))$dθ

Since these two curves have the same period and starting point you can integrate from 0 to ∏