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Polar coordinates finding area between two curves

  1. Apr 20, 2008 #1
    1. The problem statement, all variables and given/known data



    2. Relevant 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
     
  2. jcsd
  3. Apr 20, 2008 #2
    it will be easiest if you first draw out the curves. This will help you figure out what your integral should be.
     
  4. Apr 20, 2008 #3
    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
     
  5. Apr 20, 2008 #4
    first of all over what interval?
     
  6. Apr 20, 2008 #5
    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.
     
  7. Apr 20, 2008 #6
    the interval is 0,2pi
     
  8. Apr 20, 2008 #7
    Multiply both sides by r, then change to rectangular form. Is this Calculus 2 or 3? B/c I did this problem yesterday.
     
  9. Apr 20, 2008 #8
    Calc 2
     
  10. Apr 21, 2008 #9
    how would i change it to rectangular form, using x=rcos(theta) and y=rsin(theta)

    ?
     
  11. Apr 21, 2008 #10

    HallsofIvy

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    Staff Emeritus
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    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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