Polar Coordinates Area

  • Thread starter future_phd
  • Start date
1. The problem statement, all variables and given/known data
Find the area inside the lemniscate r = 2sqrt(sin(2theta))



2. Relevant equations
Integral from a to b of (1/2)[f(theta)]^2 d(theta)



3. The attempt at a solution
I tried integrating from 0 to 2pi and got an area of 0. Then I tried integrating from 0 to pi and still got an area of 0. I looked at the answer and they integrated from 0 to pi/2 and then multiplied the integral by 2. I don't understand why they chose to integrate from 0 to pi/2 or why they multiplied the integral by 2? Any help would be greatly appreciated.
 
32,114
3,993
y = sin(2x) has a complete cycle between 0 and pi. If you wanted the area between this curve and the x-axis, it wouldn't do you any good to integrate between 0 and pi -- you would get 0. So instead you would integrate between 0 and pi/2 to get the area under one arch, and then double it, to get the area of both regions.

A similar thing is happening with your polar curve.
 
Ahh that makes sense, thank you!
 

Want to reply to this thread?

"Polar Coordinates Area" You must log in or register to reply here.

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving

Top Threads

Top