Results 1 to 2 of 2

Math Help - Area under polar graph

  1. #1
    Newbie
    Joined
    Jul 2010
    From
    Palo Alto!
    Posts
    13

    Area under polar graph

    Find the area bounded by:

    r=\frac{1}{\sqrt {1+\theta } }\\

    0\leqslant \theta \leqslant \pi

    I have so far set up the integral like this:

    \int\nolimits_{0}^{\pi } {d\theta }\int\nolimits_{0}^{\frac{1}{\sqrt {1+\theta } }} {rdr}

    Looks alright?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,914
    Thanks
    778
    Hello, thepongofping!

    Why are you using a double integral?



    Find the area bounded by: . r\:=\:\dfrac{1}{\sqrt {1+\theta}},\;\;0 \le \theta \le\pi

    \text{Formula: }\;\displaystyle{A \;=\;\tfrac{1}{2}\!\int^{\beta}_{\alpha}\!\! r^2\,d\theta}


    \displaystyle{\text{We have: }\;A \;=\;\tfrac{1}{2}\!\int^{\pi}_0\! \left(\frac{1}{\sqrt{1+\theta}}\right)^2 d\theta \;=\; \tfrac{1}{2}\!\int^{\pi}_0\frac{d\theta}{1+\theta}  }

    . . A \;=\;\frac{1}{2}\ln|1+\theta|\,\bigg]^{\pi}_0 \;=\;\frac{1}{2}\ln(1+\pi) - \frac{1}{2}\ln(1+0)

    . . A \;=\;\frac{1}{2}\ln(1+\pi)
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: May 25th 2010, 01:01 AM
  2. Polar Graph Area
    Posted in the Calculus Forum
    Replies: 2
    Last Post: April 15th 2010, 10:50 AM
  3. find area of polar graph
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 1st 2010, 05:32 PM
  4. find area of polar graph
    Posted in the Calculus Forum
    Replies: 6
    Last Post: January 30th 2010, 02:13 PM
  5. Area of a Polar Curve Above the Polar Axis
    Posted in the Calculus Forum
    Replies: 6
    Last Post: May 14th 2009, 02:41 PM

Search Tags


/mathhelpforum @mathhelpforum