finding the shared area of 2 polar equations

[shortman12012]

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

Also here is my graph

 

[renegade05]

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

Also here is my graph
View attachment 1511

Well, I think you may have it right. The only thing i can see wrong is the limits of integration of the blue region. It will be from \(\displaystyle \frac{5\pi}{4}\) to \(\displaystyle \frac{9\pi}{4}\). You must add a \(\displaystyle \pi\) if you want to do this with only two integrals.

So the area will be:

\(\displaystyle \int_\frac{\pi}{4}^\frac{5\pi}{4} (5-3sin(\theta))^2 \,d\theta\) + \(\displaystyle \int_\frac{5\pi}{4}^\frac{9\pi}{4} (5-3cos(\theta))^2 \,d\theta\)

So using the equation A= 1/2∫ r^2 dθ

This actually comes from a double integral as you will see in CALC III.

 

[shortman12012]

a quick question, are the 2 areas symmetric?
if they are couldnt i just find the area of one and multiply it by two to find the total shared region?

 

[renegade05]

a quick question, are the 2 areas symmetric?
if they are couldnt i just find the area of one and multiply it by two to find the total shared region?

In this case, yes they are symmetric. So yes, you could do that.

 

[shortman12012]

okay thank you so much for your help!