# Math Help - finding the shared area of 2 polar equations

1. ## finding the shared area of 2 polar equations

Question: Given the two polar equations r=5-3cos(θ) and r=5-3sin(θ) find the area of the region common to both curves.
So using the equation A= 1/2∫ r^2 dθ, the limits of integration I get by equating the two equations are θ= π/4, 5π/4. My question is am i setting this up all right and my equation to find the area is A = ∫(5−3cos(θ))^2 dθ + ∫(5−2sin(θ))^2 dθ. Can someone let me know if this is correct or what I'm doing wrong please

2. ## Re: finding the shared area of 2 polar equations

oh i forgot to attach my drawing of the graph, here it is

3. ## Re: finding the shared area of 2 polar equations

$A = \int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} \frac{(5-3\sin{\theta})^2}{2} \, d\theta + \int_{\frac{-3\pi}{4}}^{\frac{\pi}{4}} \frac{(5-3\cos{\theta})^2}{2} \, d\theta$

... which is the hard way. note that you may take advantage of symmetry ...

$A = 2 \int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} \frac{(5-3\sin{\theta})^2}{2} \, d\theta$

$A = \int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} (5-3\sin{\theta})^2 \, d\theta$

4. ## Re: finding the shared area of 2 polar equations

so instead of what i had at the beginning (A = 1-2∫(5−3cos(θ))^2 dθ + ∫(5−2sin(θ))^2 dθ)
i can just take A = ∫(5−2sin(θ))^2)dθ)? and because it was multiplied by two it will give me the area of the yellow and blue area (graph)? would i still keep the same limits of integration?

5. ## Solving definte integral

So i just started learning about how to solve definite integrals and this problem is the last one on my homework. I just need to know how to start it, cause I'm completely lost on how it would start, any hint would be helpful

6. ## Re: Solving definte integral

Didn't you just ask this in another thread? Anyway, expand both sides, then use the fact that $sin^2\theta =\frac{1}{2}-\frac{1}{2}cos{(2\theta)}$

7. ## Re: Solving definte integral

Exapnd it out and have a go at integrating it term by term.

8. ## Re: Solving definte integral

okay so if i expand it out then i would get
(5-3sin(ɵ))(5-3sin(ɵ))
25-15sin(ɵ)-15sin(ɵ)-9sin^2(ɵ)
25+9sin^2(ɵ)
25+9(1/2-1/2cos(2ɵ)
25+9/2-9/2cos(2ɵ)
59/2-9/2cos(2ɵ)

okay so this is what i have so far, is there any mistakes or more simplifying i can do before i go ahead and integrate?

9. ## Re: Solving definte integral

Originally Posted by shortman12012
okay so if i expand it out then i would get
(5-3sin(ɵ))(5-3sin(ɵ))
25-15sin(ɵ)-15sin(ɵ)-9sin^2(ɵ)
25+9sin^2(ɵ)
25+9(1/2-1/2cos(2ɵ)
25+9/2-9/2cos(2ɵ)
59/2-9/2cos(2ɵ)

okay so this is what i have so far, is there any mistakes or more simplifying i can do before i go ahead and integrate?
Since when is \displaystyle \begin{align*} -15\sin{\theta} - 15\sin{\theta} = 0 \end{align*}?

10. ## Re: Solving definte integral

Well, you might want to note that -15-15 is NOT 0.

11. ## Re: Solving definte integral

Originally Posted by Prove It
Since when is \displaystyle \begin{align*} -15\sin{\theta} - 15\sin{\theta} = 0 \end{align*}?
oh right....so then you just put in a -30sin(ɵ)
so it be
59/2-30sin(ɵ)-9/2cos(2ɵ)

12. ## Re: Solving definte integral

So, you're just throwing darts at it? Why not proceed carefully and methodically?