Results 1 to 12 of 12

Math Help - finding the shared area of 2 polar equations

  1. #1
    Newbie
    Joined
    Dec 2011
    Posts
    6

    finding the shared area of 2 polar equations

    Question: Given the two polar equations r=5-3cos(θ) and r=5-3sin(θ) find the area of the region common to both curves.
    So using the equation A= 1/2∫ r^2 dθ, the limits of integration I get by equating the two equations are θ= π/4, 5π/4. My question is am i setting this up all right and my equation to find the area is A = ∫(5−3cos(θ))^2 dθ + ∫(5−2sin(θ))^2 dθ. Can someone let me know if this is correct or what I'm doing wrong please
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Newbie
    Joined
    Dec 2011
    Posts
    6

    Re: finding the shared area of 2 polar equations

    oh i forgot to attach my drawing of the graph, here it is
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,621
    Thanks
    426

    Re: finding the shared area of 2 polar equations

    A = \int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} \frac{(5-3\sin{\theta})^2}{2} \, d\theta + \int_{\frac{-3\pi}{4}}^{\frac{\pi}{4}} \frac{(5-3\cos{\theta})^2}{2} \, d\theta

    ... which is the hard way. note that you may take advantage of symmetry ...

    A = 2 \int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} \frac{(5-3\sin{\theta})^2}{2} \, d\theta

    A = \int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} (5-3\sin{\theta})^2 \, d\theta
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Dec 2011
    Posts
    6

    Re: finding the shared area of 2 polar equations

    so instead of what i had at the beginning (A = 1-2∫(5−3cos(θ))^2 dθ + ∫(5−2sin(θ))^2 dθ)
    i can just take A = ∫(5−2sin(θ))^2)dθ)? and because it was multiplied by two it will give me the area of the yellow and blue area (graph)? would i still keep the same limits of integration?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Dec 2011
    Posts
    6

    Solving definte integral

    So i just started learning about how to solve definite integrals and this problem is the last one on my homework. I just need to know how to start it, cause I'm completely lost on how it would start, any hint would be helpful

    finding the shared area of 2 polar equations-png.latex.pn.png
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member Quacky's Avatar
    Joined
    Nov 2009
    From
    Windsor, South-East England
    Posts
    901

    Re: Solving definte integral

    Didn't you just ask this in another thread? Anyway, expand both sides, then use the fact that sin^2\theta =\frac{1}{2}-\frac{1}{2}cos{(2\theta)}
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,236
    Thanks
    28

    Re: Solving definte integral

    Exapnd it out and have a go at integrating it term by term.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    Dec 2011
    Posts
    6

    Re: Solving definte integral

    okay so if i expand it out then i would get
    (5-3sin(ɵ))(5-3sin(ɵ))
    25-15sin(ɵ)-15sin(ɵ)-9sin^2(ɵ)
    25+9sin^2(ɵ)
    25+9(1/2-1/2cos(2ɵ)
    25+9/2-9/2cos(2ɵ)
    59/2-9/2cos(2ɵ)

    okay so this is what i have so far, is there any mistakes or more simplifying i can do before i go ahead and integrate?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,404
    Thanks
    1293

    Re: Solving definte integral

    Quote Originally Posted by shortman12012 View Post
    okay so if i expand it out then i would get
    (5-3sin(ɵ))(5-3sin(ɵ))
    25-15sin(ɵ)-15sin(ɵ)-9sin^2(ɵ)
    25+9sin^2(ɵ)
    25+9(1/2-1/2cos(2ɵ)
    25+9/2-9/2cos(2ɵ)
    59/2-9/2cos(2ɵ)

    okay so this is what i have so far, is there any mistakes or more simplifying i can do before i go ahead and integrate?
    Since when is \displaystyle \begin{align*} -15\sin{\theta} - 15\sin{\theta} = 0 \end{align*} ?
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,367
    Thanks
    1313

    Re: Solving definte integral

    Well, you might want to note that -15-15 is NOT 0.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Newbie
    Joined
    Dec 2011
    Posts
    6

    Re: Solving definte integral

    Quote Originally Posted by Prove It View Post
    Since when is \displaystyle \begin{align*} -15\sin{\theta} - 15\sin{\theta} = 0 \end{align*} ?
    oh right....so then you just put in a -30sin(ɵ)
    so it be
    59/2-30sin(ɵ)-9/2cos(2ɵ)
    Follow Math Help Forum on Facebook and Google+

  12. #12
    MHF Contributor
    Joined
    Aug 2007
    From
    USA
    Posts
    3,110
    Thanks
    2

    Re: Solving definte integral

    So, you're just throwing darts at it? Why not proceed carefully and methodically?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: May 3rd 2011, 10:38 PM
  2. Replies: 2
    Last Post: March 21st 2010, 09:04 AM
  3. Finding area of a polar curve help please.
    Posted in the Calculus Forum
    Replies: 1
    Last Post: December 7th 2009, 03:11 PM
  4. Finding polar equations
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 3rd 2008, 06:58 PM
  5. Area Shared by Polar Curves
    Posted in the Calculus Forum
    Replies: 7
    Last Post: April 21st 2008, 08:06 PM

Search Tags


/mathhelpforum @mathhelpforum