Question on Area between Polar Curves

[Edder]

Hey everyone,
I have two questions regarding the area of polar curves.

1. Find the area of the region lying the polar curve r=1 + cos(theta), and outside the polar curve r= 2cos(theta)

2. Find the area of the shaded region inside the graph of r= 1+2cos(theta); (the graph shows the top half of the cardioid shaded)

Basically, I know how to solve these problem and how to draw the graphs. What I need help on is finding the limits of integration. For the 1st problem, I picked my limits of integration to be 0 to pi, and then I multiplied the integral by 2 to find the total area. (Is this method correct)
For the 2nd, I solved for r, and got cos(theta)= -1/2. So would my limits of integration be 2pi/3 to 4pi/3?

Any help and feedback appreciated.

Thanks

 

[Deleted member 4993]

Hey everyone,
I have two questions regarding the area of polar curves.

1. Find the area of the region lying the polar curve r=1 + cos(theta), and outside the polar curve r= 2cos(theta)

2. Find the area of the shaded region inside the graph of r= 1+2cos(theta); (the graph shows the top half of the cardioid shaded)

Basically, I know how to solve these problem and how to draw the graphs. What I need help on is finding the limits of integration. For the 1st problem, I picked my limits of integration to be 0 to pi, and then I multiplied the integral by 2 to find the total area. (Is this method correct)
For the 2nd, I solved for r, and got cos(theta)= -1/2. So would my limits of integration be 2pi/3 to 4pi/3?

Any help and feedback appreciated.

Thanks

Did you plot the functions?

Where did those intersect?

 

[Edder]

Did you plot the functions?

Where did those intersect?

For the 1st one, they intersected at 0 and with both graphs starting at distance 2. On the 2nd, it was just one function of r that was a cardioid where I had to find the area of the top half.

 

[HallsofIvy]

For the 1st one, they intersected at 0 and with both graphs starting at distance 2

It doesn't make sense to say "they intersected at 0". Graphs intersect at a point, not a a number. In Cartesian Coordinates the origin is (0, 0): x= 0, y= 0. But in polar coordinates, r= 0 while \(\displaystyle \theta\) can be any angle. Yes, when \(\displaystyle \theta= 0\), r= 2 for both. BUT for \(\displaystyle r= 1+ cos(\theta)\), as \(\displaystyle \theta\) goes from 0 to \(\displaystyle \pi\), r goes form 2 to 0, half of the loop, while for \(\displaystyle r= 2cos(\theta)\), r goes from 2 to 0 as \(\displaystyle \theta\) goes from 0 to just \(\displaystyle \pi/2\). I would suggest that you integrate \(\displaystyle r= 1+ cos(\theta)\) from 0 to \(\displaystyle \pi\) to find the area between the graph and the x axis in quadrant I, integrate \(\displaystyle 2cos(\theta)\) from 0 to \(\displaystyle \pi/2\) to find the area between that graph and the positive x-axis, then subtract to find the area between them. Of course, that is half of the total area you want.

On the 2nd, it was just one function of r that was a cardioid where I had to find the area of the top half.

This cardiod has one loop inside another. when you say the "top half" do you mean above the inner loop of the entire area above the positive x-axis, ignoring the upper loop? r goes from 3 to 0 as \(\displaystyle \theta\) goes from 0 to \(\displaystyle (2/3)\pi\) so the area above the x-axis, ignoring the inner loop, would be \(\displaystyle \int_0^{2\pi/3} (1+ 2cos(\theta))r drd\theta\). To find the are above that inner loop, I would find the area of the inner loop, above the x-axis, separately, then subtract.

 

[Edder]

It doesn't make sense to say "they intersected at 0". Graphs intersect at a point, not a a number. In Cartesian Coordinates the origin is (0, 0): x= 0, y= 0. But in polar coordinates, r= 0 while \(\displaystyle \theta\) can be any angle. Yes, when \(\displaystyle \theta= 0\), r= 2 for both. BUT for \(\displaystyle r= 1+ cos(\theta)\), as \(\displaystyle \theta\) goes from 0 to \(\displaystyle \pi\), r goes form 2 to 0, half of the loop, while for \(\displaystyle r= 2cos(\theta)\), r goes from 2 to 0 as \(\displaystyle \theta\) goes from 0 to just \(\displaystyle \pi/2\). I would suggest that you integrate \(\displaystyle r= 1+ cos(\theta)\) from 0 to \(\displaystyle \pi\) to find the area between the graph and the x axis in quadrant I, integrate \(\displaystyle 2cos(\theta)\) from 0 to \(\displaystyle \pi/2\) to find the area between that graph and the positive x-axis, then subtract to find the area between them. Of course, that is half of the total area you want.


This cardiod has one loop inside another. when you say the "top half" do you mean above the inner loop of the entire area above the positive x-axis, ignoring the upper loop? r goes from 3 to 0 as \(\displaystyle \theta\) goes from 0 to \(\displaystyle (2/3)\pi\) so the area above the x-axis, ignoring the inner loop, would be \(\displaystyle \int_0^{2\pi/3} (1+ 2cos(\theta))r drd\theta\). To find the are above that inner loop, I would find the area of the inner loop, above the x-axis, separately, then subtract.

Regarding the 2nd problem, I know about the inner loop, I just forgot to mention it. The problem wants me to find the area of the function above the entire x-axis, so the upper loop included. However, your explanation makes sense and cleared up some things for me.

Thanks man, appreciate it.