Yes, exactly! You're welcome.

[Huski]

Hi, everyone. I had an example from my book, but I wasn't sure how they got $\dfrac{1}{2}cos\theta$ on step 7? It seems like once they combined the constants, they ended up with just $cos2\theta$. Although, they have a $\dfrac{1}{2}$ in front. Can someone help me understand where that constant came from? Thank you.

1. Homework Statement
Find the area of the region enclosed by the polar curve $r=3-3cos\theta$

Homework Equations

Area in Polar Coordinates
$dA=\dfrac{1}{2}r^2d\theta$

Double Angle Identity:
$cos2\theta=2cos^2\theta-1$

The Attempt at a Solution

1. $dA=\dfrac{1}{2}(3-3cos\theta)^2d\theta$

factor

2. $dA=\dfrac{1}{2}(9-18cos\theta+9cos^2\theta)d\theta$

write the integral and factor $\dfrac{1}{2}$ through

3. $\displaystyle A=\dfrac{1}{2}\int_{0}^{2\pi}(9-18cos^2\theta+9cos^2\theta)d\theta$

factor out 9

4. $\displaystyle A=9\cdot\dfrac{1}{2}\int_{0}^{2\pi}(1-2cos\theta+cos^2\theta)d\theta$

rewrite double-angle identity

5. $cos^2\theta=\dfrac{cos^2\theta+1}{2}$

replace $cos^2\theta$ with the double-angle identity

6. $\displaystyle A=\dfrac{9}{2}\int_{0}^{2\pi}\left(1-2cos\theta+\dfrac{cos^2\theta+1}{2} \right)d\theta$

combine the constants to get $\dfrac{3}{2}$

7. $\displaystyle A=\dfrac{9}{2}\int_{0}^{2\pi}\left(\dfrac{3}{2}-2cos\theta+\dfrac{1}{2}cos2\theta \right)d\theta$

integrate

8. $\dfrac{9}{2}\left[\dfrac{3}{2}\theta-2sin\theta+\dfrac{1}{4}sin\theta \right]_{0}^{2\pi}$

answer

9. $\dfrac{27\pi}{2}$

 

[blue_leaf77]

Check again the double angle identity you used in step 5.

 

[Huski]

I meant to write $\dfrac{cos2\theta+1}{2}$ for step 5 and 6. Although I do have it written for step 7.

 

[blue_leaf77]

So are the steps above the one given in your book or your own? I don't see any flaw in it except for the typo in writing the double angle identity.

 

[Huski]

Not my own. From the book. I just don't get where the$\dfrac{1}{2}$ comes from in step7?

 

[blue_leaf77]

$$
\cos^2\theta=\dfrac{\cos 2\theta+1}{2} = \frac{1}{2}\cos 2\theta + \frac{1}{2}$$

 

[Ray Vickson]

I meant to write $\dfrac{cos2\theta+1}{2}$ for step 5 and 6. Although I do have it written for step 7.

Just for the record: don't write $cos 2 \theta$ in LaTeX; it looks ugly and is hard to read. Write $\cos 2 \theta$ instead. You do that by typing "\cos" instead of "cos", and that goes for most other elementary functions as well.

 

[Huski]

Just for the record: don't write $cos 2 \theta$ in LaTeX; it looks ugly and is hard to read. Write $\cos 2 \theta$ instead. You do that by typing "\cos" instead of "cos", and that goes for most other elementary functions as well.

Alright, noted.

$$
\cos^2\theta=\dfrac{\cos 2\theta+1}{2} = \frac{1}{2}\cos 2\theta + \frac{1}{2}$$

Oh, you split up the $\dfrac{\cos2\theta+1}{2}$ into two denominators like $\dfrac{\cos2\theta}{2}+\dfrac{1}{2}$ That makes sense, thanks.