Finding area between two curves Polar Coordinates

  • #1
784
11

Homework Statement


Find the area inside the circle r = 3sinθ and outside the carotid r = 1 + sinθ




The Attempt at a Solution


Alright so I graphed it and found that they intersect at ∏/6 and 5∏/6.
I can't think of a good way to approach the problem. The carotid has some of it's area beneath the x-axis otherwise I would take the area of the 3sinθ - the area of the other one. Will it work if set the limits of integration to pi/6 and 5pi/6 and take the integral of (3sinθ - (1+sinθ)? I'm a bit lost
 

Answers and Replies

  • #2
STEMucator
Homework Helper
2,075
140

Homework Statement


Find the area inside the circle r = 3sinθ and outside the carotid r = 1 + sinθ




The Attempt at a Solution


Alright so I graphed it and found that they intersect at ∏/6 and 5∏/6.
I can't think of a good way to approach the problem. The carotid has some of it's area beneath the x-axis otherwise I would take the area of the 3sinθ - the area of the other one. Will it work if set the limits of integration to pi/6 and 5pi/6 and take the integral of (3sinθ - (1+sinθ)? I'm a bit lost

Yes, that all seems reasonable. Just compute your integral and you're done.
 
  • #3
784
11
I got 1.369705. Is there any online integral doer I can use to check my work?
 
  • #4
STEMucator
Homework Helper
2,075
140
I got 1.369705. Is there any online integral doer I can use to check my work?

Yes, that's the correct answer. Wolfram alpha is good for checking your work afterwards.
 
  • #5
LCKurtz
Science Advisor
Homework Helper
Insights Author
Gold Member
9,557
767

Homework Statement


Find the area inside the circle r = 3sinθ and outside the carotid r = 1 + sinθ




The Attempt at a Solution


Alright so I graphed it and found that they intersect at ∏/6 and 5∏/6.
I can't think of a good way to approach the problem. The carotid has some of it's area beneath the x-axis otherwise I would take the area of the 3sinθ - the area of the other one. Will it work if set the limits of integration to pi/6 and 5pi/6 and take the integral of (3sinθ - (1+sinθ)? I'm a bit lost

NO. That's the wrong integrand. Look up the formula for polar area.
 

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