# Polar graph area

1. Feb 1, 2013

### phospho

Question is attached:

working:

$$r^2 = a^2 + 6acos(\theta) + 9cos^2(\theta)$$

using $$\frac{1}{2}\displaystyle\int^{2\pi}_0 r^2d\theta$$

using this I get a = 7

are my limits right, as it says theta can't be 2pi?

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2. Feb 1, 2013

### Dick

a=7 looks ok. It doesn't really matter if they write the limit as <2pi or <=2pi. Including or excluding a single point doesn't change the integral.

Last edited: Feb 1, 2013
3. Feb 1, 2013

### phospho

thanks

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