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Solid of revolution problem

  1. Sep 15, 2006 #1
    Hello I have a question here which I've been thinking for quite sometime. And I am still very puzzled. It goes:

    Find the volume generate by the curve y=lnx rotating about the line y=x. Within the interval of x [4,10].

    I would like to know the approach to this problem as I have been asking some of my lecturers how to do this. Some said for this question, it involves interpolation while some said double integration etc.
    Hope you can help me with this.:smile: Thankx!
  2. jcsd
  3. Sep 15, 2006 #2


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    My first thought was to rotate 45 degrees so y= x became the y-axis. The problem would be determining what y= ln x becomes.

    The other thing that occured to me was to write the equation of the line through (x0,x0) perpendicular to y= x (that's easy, its y= -x+ 2x0) and finding where that intersects y= ln(x) so we can use the distance between the two points.

    Unfortunately, both of those methods involve esentially solving an equation of the form ln x= -x+ 2x0 and there is no simple way of doing that. Lambert W function perhaps?
  4. Sep 19, 2006 #3
    Aligning the shape so that it is axially symmetrical with respect to the [itex]z[/itex]-axis, the volume of your shape will then by given by

    V=\int_{0}^{2\pi}d\phi\int_{z_{l}}^{z_{u}}dz\int_{0}^{r\left( z\right)
    }rdr=\pi\int_{z_{l}}^{z_{u}}dzr^{2}\left( z\right) .

    Next we have to determine [itex]r\left( z\right)[/itex]. You've been given a radial profile [in a coordinate sytem [itex]\left( x,y\right) ][/itex], of a body that is axially aligned with the line [itex]y=x[/itex].

    y=\ln\left( x\right) \text{, \ \ \ \ \ }4\leq x\leq10.

    Next we want to rotate our [itex]\left(x,y\right)[/itex] coordinate frame so that the profile (the body rather) is axially alined with [itex]z[/itex]. We do this by applying the transformation

    z=\frac{x+y}{\sqrt{2}}\text{, \ \ \ \ \ \ \ \ }r=\frac{y-x}{\sqrt{2}}.


    If you want to know where this transformation comes from then we can show it in the following way. If we map the curve into the complex plane by saying that


    (so that [itex]\operatorname{Re}\left(\zeta\right) =x[/itex] and [itex]\operatorname{Im}\left(\zeta\right)=y[/itex]) then we will plot out exactly the same curve as in the [itex]\left(x,y\right)[/itex] plane, except now we're looking at the complex plane. This has the advantage that working with polar coordinates and rotations is simple. The function curve described by [itex]\zeta[/itex] can be written as:

    \zeta=r\left( x,y\right) e^{i\phi\left( x,y\right) }\text{.}

    Now if we want to rotate this curve by [itex]-\pi/4[/itex] [which turns the line [itex]y=x[/itex] in the [itex]\left( x,y\right)[/itex] coordinate system, onto the line [itex]\ y^{\prime}=0[/itex] in a new coordinate system, [itex]\left( x^{\prime},y^{\prime}\right)[/itex]], then we simply multiply by [itex]e^{-i\pi/4}[/itex]. This gives

    \zeta^{\prime}=\zeta e^{-i\pi/4}=\zeta\frac{\left[ 1-i\right] }{\sqrt{2}
    }=\frac{\left[ x+iy\right] \left[ 1-i\right] }{\sqrt{2}}=\frac{x+y}
    {\sqrt{2}}+i\left( \frac{y-x}{\sqrt{2}}\right) .

    If we now take real and imaginary parts of [itex]\zeta^{\prime}[/itex] our new basis becomes:

    x^{\prime}=\frac{x+y}{\sqrt{2}}\text{, \ \ \ \ \ \ \ }y^{\prime}=\frac

    ****End Aside****

    Parametrically the curve [itex]y=\ln\left(x\right)[/itex] (in the [itex]\left(x,y\right) [/itex] frame) can be written as

    \nu\left( t\right) =\left( t,\ln\left( t\right) \right)


    x\left( t\right) =t\text{, \ \ \ \ \ \ \ \ \ \ \ \ \ \ }y\left( t\right)
    =\ln\left( t\right) \text{, }

    and the upper and lower limits for [itex]t[/itex] are

    t_{l}=4\text{, \ \ \ \ \ \ \ \ }t_{u}=10\text{.}

    Hence, in the rotated frame (the [itex]\left( r,z\right)[/itex] frame), the parameterised curve is

    z\left( t\right) =\frac{t+\ln\left( t\right) }{\sqrt{2}}\text{,
    \ \ \ \ \ \ \ \ }r\left( t\right) =\frac{\ln\left( t\right) -t}{\sqrt{2}}.

    Looking at the expression for the volume,

    V=\pi\int_{z_{l}}^{z_{u}}dzr^{2}\left( z\right) ,

    since [itex]z[/itex] is a function of [itex]t[/itex] we can write

    V=\pi\int_{t_{l}}^{t_{u}}dt\frac{dz}{dt}r^{2}\left( t\right) =\frac{\pi
    }{2\sqrt{2}}\int_{4}^{10}dt\left( 1+\frac{1}{t}\right) \left( \ln\left(
    t\right) -t\right) ^{2}=\frac{\pi}{2\sqrt{2}}\int_{\ln\left( 4\right)
    }^{\ln\left( 10\right) }\left( e^{s}+1\right) \left( s-e^{s}\right)
    =\frac{\pi\left(190.\,51\right) }{2\sqrt{2}}=211.6\left( \text{units of volume}\right) .
    Last edited: Sep 19, 2006
  5. Sep 19, 2006 #4
    Sorry for the _w i d e s c r e e n_ effect. I'll edit the post if it's hard to read without a 25 inch monitor. I think the solution should look like this, but I may be wrong. If you find any errors let me know.

    -- It's not looking so widescreen anymore.

    -- I re-read the initial post and I'm not quite sure what is meant by "Within the interval of x [4,10]" means. If this means that the original profile you are given is drawn from x=4 to x=10, with lines connecting each end point perpendicularly to the line y=x, (which is what I assumed) and then rotated about y=x, then I think the above solution should be correct.

    Alternatively (and most probably) if it means that the profile you're given has a line from (4,ln(4)) to (4,4) and another line from (10,ln(10)) to (10,10), then the answer should be larger by the volume of a cone (at the end closer to the origin) and smaller by the volume of a second cone at the end furthest from the origin. These corrections to the area shouldn't be difficult to calculate though.

    -- To find the line perpendicular to y=x that intersects the endpoints of the cone, just say that y=-x+c. We require that at x=4, y=ln(4), so y=-x + 4 + ln(4). This intercepts y=x at x = (4+ln(4))/2 = 2+ln(2). So the radius of the bottom cone-too-many is r^2 = [ln(2) - 2]^2 + [2+ln(1/2)]^2 , and its height is given by h = sqrt(2)(ln(2) - 2). From this you can calculate the volume of the erroneous cone. The same method can be applied for the missing cone at the top.
    Last edited: Sep 19, 2006
  6. Sep 23, 2006 #5
    Oops...Well, what I mean was your second alternative. Which was (4,ln4) to (4,4) and (10, ln10) to (10,10) => i.e. 2 straight lines perpendicular to the x-axis.

    But anyway, thanks for your help :cool:
  7. Sep 24, 2006 #6
    It's okay, you can just calculate the additional volume by working out the volume of the cones at the end of the body.
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