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Polar coordinates finding area between two curves

  • Thread starter th3plan
  • Start date
92
0
1. Homework Statement



2. Homework Equations

r=sinx
r= cosx

Ok , i need help how to properly select the integral to evaluate the area they make. Can someone please show me how , i know how to evaluate it just having hard times with integrals

3. The Attempt at a Solution
 

Answers and Replies

208
0
it will be easiest if you first draw out the curves. This will help you figure out what your integral should be.
 
92
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i graphed it on calculator, and i did set them equal to each other to get tanx=1 but from here on idk what to do
 
1,703
5
first of all over what interval?
 
208
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if you're given those in parametric form in polar, you are going to get two circles in the x,y (or r,theta) plane, i believe. But as ice109 said, its also important that you know how x varies for this one.
 
92
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the interval is 0,2pi
 
1,750
1
Multiply both sides by r, then change to rectangular form. Is this Calculus 2 or 3? B/c I did this problem yesterday.
 
92
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Calc 2
 
92
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how would i change it to rectangular form, using x=rcos(theta) and y=rsin(theta)

?
 
HallsofIvy
Science Advisor
Homework Helper
41,738
898
the interval is 0,2pi
No, it isn't. Since sine and cosine are negative for half that interval using 0 to 2[itex]\pi[/itex] gives you each circle twice. And, in fact, the area you want only requires [itex]\theta[/itex] going from 0 to [itex]\pi/2[/itex].

However, you are correct that the circles intersect when tan[itex]\theta[/itex]= 1- that is, at [itex]\theta= \pi/4[/itex] as well as at 0. For [itex]0\le \theta\le \pi/4[/itex], a radius goes from 0 to cos([itex]\theta[/itex]) while from [itex]\pi/4\le \theta\le \pi/2[/itex] it goes from 0 to sin([itex]\theta[/itex]). From symmetry, you should be able to integrate cos([itex]\theta[/itex]) from 0 to [itex]\pi/4[/itex] and double.
 

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