Results 1 to 2 of 2

Thread: Help setting up integral to find surface area

  1. #1
    Newbie
    Joined
    Aug 2008
    Posts
    20

    Help setting up integral to find surface area

    I need help with the following problem:

    Let C be the curve defined by y=2sin(3x) on the interval [0,Pi/2]. I need help setting up the integral required to find the surface area of the surface that results when C is revolved about the specified axis:

    a.) The x-axis
    b.) The y-axis

    I think for a you would use S=2Pi∫(from 0 to Pi/2) 2sin(3x)*sqrt(1+(2cos(3x) *3)^2)

    I think for b you would use S=2Pi∫(from 0 to Pi/2) x*sqrt(1+2cos(3x)*3)^2)

    Have I set these integrals up correctly or have I done something wrong? Thanks in advance to anyone who can help me.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Aug 2008
    Posts
    903
    For the equation $\displaystyle x=t,;\ y(t)=2\sin(3t)$, the surface areas of the figures around the x and y axes respectively are:

    $\displaystyle A_y=2\pi\int x ds$

    $\displaystyle A_y=2\pi\int y ds$

    with $\displaystyle ds=\sqrt{(x')^2+(y')^2}$

    Then for $\displaystyle f(x)=2\sin(3x);\; 0\leq x\leq \pi/2$, in parametric form that's $\displaystyle x(t)=t,\;y(t)=2\sin(3t);\;0\leq t\leq \pi/2$ and thus:

    $\displaystyle A_y=2\pi\int_0^{\pi/2} t\sqrt{1+(6\cos(3t))^2}dt$

    $\displaystyle A_x=2\pi\int_0^{\pi/2} 2\sin(3t)\sqrt{1+(6\cos(3t))^2}dt$
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: Jul 6th 2010, 01:00 PM
  2. Replies: 1
    Last Post: Apr 18th 2010, 01:28 PM
  3. Find the area of the surface ...
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Nov 11th 2008, 06:19 PM
  4. Find the area of the surface ...
    Posted in the Calculus Forum
    Replies: 0
    Last Post: Nov 11th 2008, 05:17 PM
  5. Find the area of a surface (double integral)
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Feb 13th 2007, 01:44 PM

Search Tags


/mathhelpforum @mathhelpforum