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Calc II: i don't understand integral for surface area

  1. Jul 11, 2013 #1
    if we want to find the volume of a function revolved about the x axis all we do is find the differential element..

    dV=(Area)dx =[itex]\pi[/itex]y2 , where y=y(x)

    so then..

    [itex]\int[/itex][itex]\pi[/itex]y2dx

    the differential element looks like this...

    http://imgur.com/HwrihKA,hjkKFzV,cpUPsvZ,YKIyTIB

    so i add a bunch of these up and i get the total volume. okay... fine..

    but then we get to surface area.. and the differential element looks like this...

    http://imgur.com/HwrihKA,hjkKFzV,cpUPsvZ,YKIyTIB#1

    please don't get into the equations trying to explain to me where the frustum equation came from, i know where it came from, and i understand the rationale for using the frustum as the differential element to account for the discrepancies due to the function's slope. the trouble im having is.. if we must choose our differential element for the surface area to be the frustum, then why is it okay to choose the cylinder as the differential element for finding the volume?

    due to the fact that a differential element of a frustum takes into consideration the slope of the function, i can see why this would be more accurate/preferrable over just the side surface area of the cylinder,

    ie...

    [itex]\int[/itex]2[itex]\pi[/itex]ydx

    umm.. here just for the hell of it i drew a picture

    http://imgur.com/HwrihKA,hjkKFzV,cpUPsvZ,YKIyTIB#2
    ....

    Now consider the cylindrical differential element for the volume of the rotated function

    http://imgur.com/HwrihKA,hjkKFzV,cpUPsvZ,YKIyTIB#3

    (probably going to have to click to enlarge the image)

    basically my question is.. if it is NOT okay to use the cylindrical differential element to find the surface area ie...

    [itex]\int[/itex]2[itex]\pi[/itex]ydx

    then why is it okay to use the cylindrical differential element to find the volume?

    look at the image again (and enlarge it)
    http://imgur.com/HwrihKA,hjkKFzV,cpUPsvZ,YKIyTIB#3

    there is leftover crap from the diff. element sagging outside all over the actual volume of the rotated function.

    please help, i've had trouble with this ever since.. well.. calc II which was two years ago
     
    Last edited: Jul 12, 2013
  2. jcsd
  3. Jul 12, 2013 #2

    Simon Bridge

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    All integration is just a fancy way of adding up.
    Because the width is very small ... you don't have to take the cylinder for the volume element ... have a go using something other than a cylinder, and see what happens. i.e. perhaps a truncated cone height dx?
     
  4. Jul 12, 2013 #3

    SteamKing

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    What the hell is a fructum? It's frustum.
     
  5. Jul 12, 2013 #4

    WannabeNewton

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    It's Latin for fruit.
     
  6. Jul 12, 2013 #5
    lol sorry guys.. yes, i meant frustum :D
     
  7. Jul 12, 2013 #6

    Simon Bridge

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    Did you try the exercise in post #2 yet?

    From what you wrote you seem to actually have a grasp on why the area element needs to conform to the shape of the surface, but you are having trouble making the connection to the volume element not having to conform to the surface.

    I suspect that some of the trouble may come from that you are only looking at cases with a particularly symmetry (cylindrical) and some of the details may be hidden. If you tried spherical coordinates, for eg, you'll find the basic volume and surface elements have different shapes again. i.e. you can express volume element ##\small dxdydz## as ##\small r^2\!\sin\!\theta\; drd\theta d\phi## even though one of the volumes is a cube and the other a sort of truncated wedge thing. (Imagine you want the volume of a sphere ... don't slice it up: start by specifying a volume element ##\small d\tau## at a particular position ##\small \vec{r}## and adding up the volumes of all the elements.)

    You can try the same comparisons for surface elements.
     
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