• Support PF! Buy your school textbooks, materials and every day products Here!

Polar Regions: Area, Arc Length, and Surface Area

  • #1

Homework Statement



Consider the graph (see attachment) of r = 1 +2cos[tex]\Theta[/tex] in polar coordinates. SET UP integrals to find
1. the area inside the large loop minus the area of the small loop.
2. the arc length of the small loop
3. the surface area of the surface formed by revolving the large loop about the initial ray.


Homework Equations



area A of the polar region
A = [tex]\int[/tex][tex]\frac{1}{2}[/tex](f([tex]\Theta[/tex]))[tex]^{2}[/tex]d[tex]\Theta[/tex] with upper limit b and lower limit a.

arc length AL of the polar region:
AL = [tex]\int[/tex][tex]\sqrt{r^{2}+(\frac{dr}{d\Theta})^{2}} d\Theta[/tex] with upper limit b and lower limit a.

not sure what equation i need to figure out the surface area one

The Attempt at a Solution



I'm pretty much lost when it comes to the entire problem, and have no ideas where to start. Please help!
 

Attachments

Answers and Replies

  • #2
ideasrule
Homework Helper
2,266
0
You've got the equations right. You just have to figure out what to use for the bounds. For 1, if you want to encompass only the larger arc, what should the upper bound be?
 
  • #3
Well if the 4π/3 ray moved in a counterclockwise motion until the 2π/3 ray, then the ray would sweep the entire area of the large loop. But that would mean doing an integral from a lower bound of 4π/3 to the upper bound of 2π/3. This to me doesn't seem right since usually the lower bound is smaller than the upper bound.
 

Related Threads on Polar Regions: Area, Arc Length, and Surface Area

Replies
2
Views
9K
Replies
3
Views
2K
Replies
1
Views
4K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
9
Views
7K
  • Last Post
Replies
1
Views
745
Replies
4
Views
2K
Replies
1
Views
3K
Replies
6
Views
4K
  • Last Post
Replies
8
Views
1K
Top