Results 1 to 6 of 6
Like Tree3Thanks
  • 1 Post By Reckoner
  • 1 Post By Reckoner
  • 1 Post By Reckoner

Math Help - Application of Integration Problem

  1. #1
    Newbie andvaka's Avatar
    Joined
    Jun 2012
    From
    Owings Mills, MD
    Posts
    18

    Question [Solved] Application of Integration Problem

    A ceramic bowl is turned on a potter's wheel to approximate the surface obtained by revolving the curve y= (2/3)((x^2) - 1 ) from x=1 to x=2 about the y axis. What is the approximate outside area of the bowl? [assume that each unit on the x and y axes represents 10 cm]

    Any thoughts?

    I'd really appreciate any help, especially if it would lead to the solution.

    Thanks.
    Last edited by andvaka; June 24th 2012 at 08:49 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Reckoner's Avatar
    Joined
    May 2008
    From
    Baltimore, MD (USA)
    Posts
    1,024
    Thanks
    75
    Awards
    1

    Re: Application of Integration Problem

    Quote Originally Posted by andvaka View Post
    A ceramic bowl is turned on a potter's wheel to approximate the surface obtained by revolving the curve y= (2/3)((x^2) - 1 ) from x=1 to x=2 about the y axis. What is the approximate outside area of the bowl? [assume that each unit on the x and y axes represents 10 cm]

    Any thoughts?
    We are finding the area of a surface of revolution:

    S = 2\pi\int_a^br(x)\sqrt{1+\left(\frac{dy}{dx}\right)  ^2}\,dx

    = 2\pi\int_1^2x\sqrt{1+\left(\frac43x\right)^2}\,dx
    Thanks from andvaka
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie andvaka's Avatar
    Joined
    Jun 2012
    From
    Owings Mills, MD
    Posts
    18

    Re: Application of Integration Problem

    Quote Originally Posted by Reckoner View Post
    We are finding the area of a surface of revolution:

    S = 2\pi\int_a^br(x)\sqrt{1+\left(\frac{dy}{dx}\right)  ^2}\,dx

    = 2\pi\int_1^2x\sqrt{1+\left(\frac43x\right)^2}\,dx

    But shouldn't r(x) be y instead of x? Correct me if I am wrong. I don't understand how you got x from r(x).

    =2\pi\int_{1}^{2}(\frac{2}{3}(x^2 -1))\sqrt{1+(\frac{4x}{3})^2} dx
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Reckoner's Avatar
    Joined
    May 2008
    From
    Baltimore, MD (USA)
    Posts
    1,024
    Thanks
    75
    Awards
    1

    Re: Application of Integration Problem

    Quote Originally Posted by andvaka View Post
    But shouldn't r(x) be y instead of x? Correct me if I am wrong. I don't understand how you got x from r(x).

    =2\pi\int_{1}^{2}(\frac{2}{3}(x^2 -1))\sqrt{1+(\frac{4x}{3})^2} dx
    r(x) represents the distance between the curve and the axis of revolution at a given x-value. Since we are revolving the curve about the y-axis, that distance is just x. If we were revolving about the x-axis, then you would be correct: the distance from the curve to the axis would be y = \frac23\left(x^2-1\right).

    So if we were revolving about the x-axis, the area of the surface would be

    S = 2\pi\int_1^2\frac23\left(x^2-1\right)\sqrt{1+\left(\frac{4x}3\right)^2}\,dx.

    Consider another example, let's say we wanted to revolve the curve about the line x = 3. Then the distance from the curve to the axis of revolution would be r(x) = 3 - x and the area would be

    S = 2\pi\int_1^2(3 - x)\sqrt{1+\left(\frac{4x}3\right)^2}\,dx.

    Give me a second and I'll make a picture to help you visualize it.
    Thanks from andvaka
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie andvaka's Avatar
    Joined
    Jun 2012
    From
    Owings Mills, MD
    Posts
    18

    Re: Application of Integration Problem

    Ah. Understood. I got confused because of the notation used in my textbook vs yours.
    Thanks!
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor Reckoner's Avatar
    Joined
    May 2008
    From
    Baltimore, MD (USA)
    Posts
    1,024
    Thanks
    75
    Awards
    1

    Re: Application of Integration Problem

    Please click the thumbnail below for a larger version.

    The graph shows the curve along with the three different axes of revolution that I mentioned in my last post. The distance between the point (x, y) and each axis of revolution is also depicted. The problem that you posted involved revolution about the line x = 0 (that is, the y-axis), which is shown in red.

    Sorry for the notational confusion. I should have explained the formula I was using more clearly.
    Attached Thumbnails Attached Thumbnails Application of Integration Problem-mhf_20120624b.png  
    Thanks from andvaka
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Application of Integration
    Posted in the Calculus Forum
    Replies: 4
    Last Post: July 27th 2011, 07:55 AM
  2. Help in Application of integration.
    Posted in the Calculus Forum
    Replies: 0
    Last Post: May 17th 2009, 03:55 PM
  3. Application of integration : Work problem
    Posted in the Calculus Forum
    Replies: 3
    Last Post: February 1st 2009, 11:27 AM
  4. integration of application
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 25th 2008, 12:50 PM
  5. application of integration problem
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 19th 2007, 11:04 PM

Search Tags


/mathhelpforum @mathhelpforum