Calculating Polar Graph Area with a=7 and Limited Theta

[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?

 

[Dick]

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?

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.

 

[phospho]

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.

thanks