Results 1 to 3 of 3

Math Help - Surface area by revolution: Why is this comming out negative?

  1. #1
    Member sinewave85's Avatar
    Joined
    Oct 2006
    From
    Lost in a series of tubes.
    Posts
    218

    Surface area by revolution: Why is this comming out negative?

    Find the area of the surface generated when the arc of the curve x=\frac{3}{5}y^{3/5}-\frac{3}{4}y^{1/3} between y = 0 and y = 1 is revolved around (a) the y-axis, (b) the x-axis, and (c) the line y = -1.

    I am stuck on the first one and thought that I should figure out what I am not getting before I press on. My answer keeps comming up negative, and I can't figure out why. Here is what I have:

    S=2\pi\int_{c}^{d}g(y)\sqrt{1+[g'(y)]^{2}}dy
    S=2\pi\int_{0}^{1}\left(\frac{3}{5}y^{5/3}-\frac{3}{4}y^{1/3}\right)\sqrt{1+\left(y^{2/3}-\frac{1}{4}y^{-2/3}\right)^{2}}dy
    S=2\pi\int_{0}^{1}\left(\frac{3}{5}y^{5/3}-\frac{3}{4}y^{1/3}\right)\left(y^{2/3}+\frac{1}{4}y^{-2/3}\right)dy
    S=2\pi\int_{0}^{1}\left(\frac{3}{5}y^{7/3}-\frac{3}{5}y-\frac{3}{16}y^{-1/3}\right)dy
    S=2\pi\left(\frac{9}{50}y^{10/3}-\frac{3}{10}y-\frac{9}{32}y^{2/3}\right)\mid_{0}^{1}
    S=\frac{-321\pi}{400}

    Inverting the limits of integration (integrating from 1 to 0 instead of from 0 to 1) would make the answer positive, but is there a reason to do that? I so, could you explain why? Or is there some other mistake I am making? Like I said, I would like to get this sorted before I go on to the other problems. (I am not looking for help on the other parts -- b and c -- of the problem at this point, except for some general guidance, especially relative to what ever mistake I am making in this problem.)
    Last edited by sinewave85; December 2nd 2009 at 06:57 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Calculus26's Avatar
    Joined
    Mar 2009
    From
    Florida
    Posts
    1,271
    See the graph below --note x = g(y) is negative so use -g(y) and go from there
    Attached Thumbnails Attached Thumbnails Surface area by revolution: Why is this comming out negative?-surfacearea.jpg  
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member sinewave85's Avatar
    Joined
    Oct 2006
    From
    Lost in a series of tubes.
    Posts
    218
    Quote Originally Posted by Calculus26 View Post
    See the graph below --note x = g(y) is negative so use -g(y) and go from there
    Ok, I see. Thanks so much for the help!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Surface Area of Revolution
    Posted in the Calculus Forum
    Replies: 2
    Last Post: December 10th 2011, 03:57 AM
  2. Surface area of a revolution
    Posted in the Calculus Forum
    Replies: 7
    Last Post: March 27th 2011, 02:47 PM
  3. Surface Area of a Revolution
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 11th 2010, 11:22 AM
  4. surface area of revolution
    Posted in the Calculus Forum
    Replies: 1
    Last Post: May 6th 2008, 10:31 PM
  5. Area of a surface of revolution
    Posted in the Calculus Forum
    Replies: 5
    Last Post: March 20th 2008, 02:55 PM

Search Tags


/mathhelpforum @mathhelpforum