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Integrals of surface area/volume

  1. Apr 22, 2006 #1
    Argh I just typed up an entire question and accidentally closed the window. Let's try this again.

    My friend and I have been pondering this for a while, and hopefully someone will be able to help us out.

    When you find the volume of a solid of revolution, you can use the disk method:

    [tex]V = 2\pi \int_a^b [R(x)]^2\,dx[/tex]

    The volume element in the integral is the volume of a cylinder with height dx.

    When you find the area of a surface of revolution, you use the following formula:

    [tex]A = 2\pi \int_a^b R(x)\sqrt{1+(\frac{dy}{dx})^2}\,dx[/tex]

    The area element in the integral is the lateral surface area of a frustum of a cone. Why doesn't this integral, too, use a cylinder, as the volume integral did? Why does one method use cylinders, and one method use frustums?

    Thank you in advance :)
  2. jcsd
  3. Apr 22, 2006 #2


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    Your formula for V shouldn't have a 2 in it. You have a sideways stack of cylinders of height dx, as you say. The volume of each cylinder is the surface area of the cross-section, times the height. The surface area of the cross-section is, of course, just [itex]\pi [R(x)]^2[/itex], so the total volume is:

    [tex]V = \pi \int _a ^b [R(x)]^2\, dx[/tex]

    Anyways, the reason is that looking at the lateral surface area of a cylinder will give an area element of [itex]2\pi R(x)\, dx[/itex], and comparing it to the given area element, we see that the ratio of the frustrum-area-element to the cylinder-area-element is:

    [tex]\sqrt{1 + \left (\frac{dy}{dx}\right )^2}[/tex]

    which can be quite large if |dy/dx| is large. That's why it matters whether we look at it as a frustrum rather than a cylinder. It is closer to being a frustrum than it is to being a cylinder, and when we look at surface area, this geometric dissimilarity makes a big numerical difference.

    The volume element if regarded as a frustrum of a cone is:

    [tex]\pi R(x)^2dx \pm \pi R(x) dydx[/tex]

    Dividing the frustrum-volume-element by the cylinder-volume-element gives:

    [tex]1 \pm \frac{dy}{R(x)}[/tex]

    which is practically 1 because dy is quite small. So although it really is a frustrum, it is much simpler to conceptualize and do computations when treating it as a cylinder, and we can do this without problem when we're looking at volume because the volume element is practically the same no matter how we regard it, i.e. this geometric dissimilarity makes an infinitessimal numerical difference.
  4. Apr 22, 2006 #3
    Awesome, thank you!

    I am curious how you got the volume element of the frustum, though. I tried to derive it using [itex]\frac{\pi h}{3}\left(R^2 + Rr + r^2\right)[/itex], where [itex]h = dx[/itex], [itex] R = R\left(x\right)[/itex], and [itex]r = R\left(x\right) + dy[/itex], and I ended up with something like

    [tex]\pi \left[R\left(x\right)\right]^2 + \pi R\left(x\right)dydx + \frac{\pi dx dy^2}{3}[/tex]

    Did you just leave out the last term [because it, too, has no impact]? Or am I way off?

    Thank you again.
  5. Apr 22, 2006 #4


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    Your formula is correct, mine was an approximation, but the difference is negligible. We see that treating at as a cylinder instead of a frustrum makes an infinitessimal difference. The extra term in your calculation makes an even smaller difference, as it makes an "infinitessimal²" difference.

    Basically, I took the volume of the frustrum to be the volume of the cylinder, plus or minus the volume of a triangular ring. The volume of this ring is the area of the triangle, multiplied by the distance travelled by the centroid. The distance travelled by the centroid is [itex]2\pi[/itex] times the distance from the centroid to the x-axis. I approximated this distance to be R(x), which is really the distance from the base of the triangle to the x-axis. Technically, the distance from the x-axis to the centroid is [itex]R(x) + \frac{dy}{3}[/itex] (to get the "dy/3" part, you have to remember something about the geometry of triangles - it has to do with centroids, medians, and proportions). The area is [itex]\frac{1}{2}dxdy[/itex], so the contribution of this triangular ring to the volume element is

    [tex]\frac{1}{2}dxdy\left [2\pi \left (R(x) + \frac{dy}{3}\right )\right ] = \pi R(x)dydx\ +\ \frac{\pi dxdy^2}{3}[/tex]
    Last edited: Apr 22, 2006
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