1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Arc length help extended to surface area and centroid.

  1. Mar 13, 2008 #1


    User Avatar
    Homework Helper

    1. The problem statement, all variables and given/known data
    A curce,C, has equation [itex]y=x^{\frac{1}{2}}-\frac{1}{3}x^{\frac{3}{2}}+\lambda[/itex] where [itex]\lambda>0[/itex] and [itex] 0\leq x \leq 3[/itex]

    The length of C is denoted by s. Show that [itex]s=2\sqrt{3}[/itex]
    The area of the surface generated when C is rotated through one revolution about the x-axis is denoted by S. Find S in terms of [itex]\lambda[/itex]

    The y-coordinate of the centroid of the region bounded by C,the axes and the line x=3 is denoted by h. Given that [itex]\int _{0} ^{3} y^2 dx=\frac{3}{4}+\frac{8\sqrt{3}}{5}+3\lambda^2[/itex],show that

    [tex]\lim_{\lambda \rightarrow \infty} \frac{S}{hs}=4\pi[/tex]
    2. Relevant equations
    [tex]ds=\sqrt{1+\left (\frac{dy}{dx} \right)^2} dx[/tex]

    Arc Length between [itex]x=x_1 \mbox{and} x_2[/itex] is given by

    [tex]\int _{x_1} ^{x_2} 1 ds[/tex]

    [tex]S=\int _{x_1} ^{x_2} 2\pi y ds[/tex]

    [tex]h= \frac{\int_{x_1} ^{x_2} \frac{y^2}{2}dx}{\int _{x_1} ^{x_2} ydx}[/tex]

    3. The attempt at a solution
    I was able to show [itex]s=2\sqrt{3}[/itex] thanks to some help from the forum yesterday.

    But I believe I am doing something wrong and I don't know where I went wrong so.

    [tex]S=2\pi \int _{0} ^{3} (x^{\frac{1}{2}}-\frac{1}{3}x^{\frac{3}{2}}+\lambda) \times \frac{1}{2}(x^{\frac{-1}{2}}+x^{\frac{1}{2}}) dx[/tex]

    [tex]S=2\pi \left( \int_{0} ^{3} (x^{\frac{1}{2}}-\frac{1}{3}x^{\frac{3}{2}}dx+ \lambda \int_{0} ^{3} \frac{1}{2}(x^{\frac{-1}{2}}+x^{\frac{1}{2}})dx\right) [/tex]

    [tex]S=2\pi \int_{0} ^{3}(1-\frac{x}{3}+x-\frac{x^2}{3})dx +4\pi \lambda \sqrt{3}[/tex]

    [tex]S= 2\pi \left[ \frac{-x^3}{9}+\frac{x^2}{3}+x \right]_{0} ^{3} + 4\pi \lambda \sqrt{3}[/tex]

    Which gives me [itex]6\pi + 4\pi \lambda \sqrt{3}[/itex]
  2. jcsd
  3. Mar 14, 2008 #2

    Gib Z

    User Avatar
    Homework Helper

    From your attempt, how did you get from line 1 to line 2? You didn't expand properly, perhaps you forgot about the brackets in the y term. Try expanding again, I got [tex]3\pi + 4\pi \lambda \sqrt{3}[/tex]
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Arc length help extended to surface area and centroid.
  1. Arc length, area (Replies: 1)