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(Volume)' = Area

  1. Apr 25, 2007 #1


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    h = any dimension of the figure, A(h) = area of the cross-sections perpendicular to h described as a function of the position along h
    this will work for any figure (no matter if the prism is slanted or the cross-sections change shape).

    Can anyone maybe explain this to me? I want to find out how to integrate any forumula for Area of an object into a formula for Volume of an object.

    The formula is this: [tex]\int A(h) dh[/tex]

    This sentence that I don't understand is this: area of the cross-sections perpendicular to h described as a function of the position along h

    Last edited: Apr 25, 2007
  2. jcsd
  3. Apr 25, 2007 #2
    The statement is clearly written, I will help you decode its meaning.

    Imagine the shape existing in an (x,y,z) coordinate system, and let us choose to integrate along the z-axis (we could integrate along any path, so we choose a simple one) so imagine your shape as a potato being skewered by the z-axis.

    With this setup, the cross-sections will be parallel to the x-y-plane. Just like slices of a potato, we can add these up, each multiplied by its thickness, to determine the volume of the shape.
  4. Apr 26, 2007 #3


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    I understand the concept, but not how to do it... Let's say you want to integrate the area of a cylinder to the volume of it:

    [tex]V = \int (2 \pi r^2 + 2 \pi r h) dh[/tex]

    How would you do this?

    The volume is: [tex] \pi r^2 h[/tex]

    EDIT: Or maybe link me to a page with an explanation of integration/derivation of two variables...
    Last edited: Apr 26, 2007
  5. Apr 26, 2007 #4
    What are you doing with that integral? In a cylinder, h is constant everywhere , so to calculate the volume of that, you just multiply the area of the bottom by h.
    In fact the cylinder volume is equal to integration of (hdA), not A(h)dh
    Last edited: Apr 26, 2007
  6. Apr 26, 2007 #5


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    A(h) = The area with all variables, applies for all objects.

    The general rule is that the integral of the area equals the volume. It says so in wikipedia. So.. I guess it applies for objects with more than one variables too.

    My question is still, how do you integrate a function with more than one variable?
    Last edited: Apr 26, 2007
  7. Apr 26, 2007 #6


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    A cylinder has a constant radius, r isn't a variable.

    Imagine you look at a deck of cards. If you know the volume of a face of a card, adding that up over all the cards gives you the volume. Now, as your volume becomes very narrow, your volume of each piece becomes area (similiar to how in normal integration, one side of the rectangle becomes the value at that point), times the incredibly small height to give you your miniscule volume. So back to the cylinder:

    The cross section area is Pi*r2, as h ranges from 0 to H let's say. Pi*r2 is a constant w.r.t. h, so the integral is Pi*r2*h, and your final volume is Pi*r2H.

    Now suppose it was a cone, with its point at the origin. The radius at height h would be (say) h, so you're actually integrating from 0 to H: Pi*h2

    This means your integral is Pi/3*h3 from 0 to H, and your final volume is Pi/3*H3
  8. Apr 26, 2007 #7


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    Well, I'm sure we could find some "objects" (with a very general concept of "object") that do not have area! But that really has nothing to do with you question which was about "area of the cross-sections perpendicular to h described as a function of the position along h"

    The crucial words there are "cross-sections perpendicular to h". What you gave before, "[itex]2\pi r^2+ 2\pi rh"[/itex], is the total surface area of a cylinder, not the area of some cross section. If you take h to be along the central axis of the cylinder, then the cross section is a circle. Its area is a constant: [itex]\pi R^2[/itex] where R is the radius of the base of the cylinder. You are really integrating
    [tex]\_0^H \pi R^2 dh= 2\pi R^2H[/tex]
    the usual formula for area of a cylinder.

    To make it a little more interesting, take cross sections perpendicular to a diameter of the cylinder's base. (Imagine the water surface as you are draining a horizontal cylindrical tank.) Now each cross section is a rectangle with one side of length H and the other side running from across a chord of the circle. If we take h now to be the distance up from the bottom of the circle, then (Taking a coordinate system in which the circle is [itex]x^2+ y^2= R^2[/itex]) one end of that side is is at [itex]\sqrt{R^2- x^2}[/itex] and the other is at [itex]-\sqrt{R^2- x^2}[/itex] so that side has length [itex]2\sqrt{R^2- x^2}[/itex] or using our "h", [itex]2\sqrt{R^2- h^2}[/itex] so the rectangle has area [itex]2H\sqrt{R^2- x^2}[/itex]. Now you are saying that the volume is the integral of that with respect to h (which goes from -R to R):
    [tex]\int_{-R}^R 2H\sqrt{R^2- h^2}dh[/itex]
    To do that use a "trig substitution": let h= Rsin t so that dh= Rcos t dt and [itex]\sqrt{R^2- x^2}= \sqrt{R^2- R^2 sin^2 t}= R cos t[/itex]. When h= R= R cos t, cos t= 1 so [itex]t= \pi/2[/itex] and when h= -R= R cos t, cos t= -1 so [itex] t= -\pi/2[/itex]. With that substitution the integral is
    [tex]2HR^2\int_{-\pi/2}^{\pi/2} cos^2 t dt[/itex]
    Now use the trig identity [itex]cos^2 t= (1/2)(1+ cos 2t)[/itex]. The integral is
    [tex]2HR^2\int_{-\pi/2}^{\pi/2}(1/2)(1+ cos 2t)dt= HR^2 (t+ (1/2)sin(2t)[/tex]
    evaluated between [itex]-\pi/2[/itex] and [itex]\pi/2[/itex] Of course [itex]sin(2(1/2)\pi])= sin(\pi)= 0[/itex] and [itex]sin(2(-1/2)\pi)= sin(-\pi)= 0[/itex] so that part gives 0. The "t" evaluated between [itex]-\pi/2[/itex] and [itex]\pi/2[/itex] is simply [itex]\pi/2-(-\pi/2)= \pi[/itex]. In other words that whole integral evaluates to [itex]\pi HR^2[/itex] just as before.
  9. Apr 27, 2007 #8
    In this example, I think you're using the word "variable" just to mean a symbol (a letter). Other people above are trying to point out that for a cylinder, r is a constant, not a variable, with respect to h.

    A symbol is a number. There's nothing special about it. You don't treat letters any differently than other numbers. Just do the integral. (For example, how would you integrate 2 pi h? You wouldn't say "oh, pi is a Greek letter, I don't know how to integrate functions with Greek letters." Pi is constant. So it's just like any other number.)

    The integral of 2 pi r^2 + 2 pi r h with respect to h is just the sum of those two integrals. 2 pi r^2 is just a constant (it doesn't contain anything that depends on h, they're all just constant numbers) so its integral is 2 pi r^2 h. 2 pi r h is a constant (2 pi r) times the variable of integration (h), so its integral is just 1/2 x 2 pi r h^2 = pi r h^2. What's the problem?

    If you want to integrate a function that contains a symbol which really is variable in h, then you need to rewrite that variable, everywhere it appears, in terms of h. Well you don't need to I guess... but it might help. :-)
    Last edited: Apr 27, 2007
  10. Apr 29, 2007 #9


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    All right, thanks.
    Halls of Ivy, sorry, I didn't understand all that so it gave little help...
    But Xezlec, Pi, can never be anything else than approx 3.14. The radius of a cylinder is not defined as one number. Both the volume and area of a figure depends on both h and r. That's why I thought both were variables. But if you tell me to treat the radius as a constant, I must believe you. If I want to integrate the function of area to a function of volume, I must find the function for area of the "cross-sections perpendicular to h". Does that mean the area of the "vertical" side of a cylinder standing up? And then just exclude the area of top and bottom.
  11. Apr 29, 2007 #10

    Gib Z

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    When Integrating, you are integrating ONE cylinder. Sure r varies and isn't constant with different cylinders, but when integrating ONE cylinder, it is constant in that cylinder.
  12. Apr 29, 2007 #11


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    Think about it then, if you are integrating ONE cylinder, h is no variable no more is it?
  13. Apr 29, 2007 #12
    If you have a cylinder with radius r and height h, then both h and r are constant for that cylinder. What makes you think otherwise? Perhaps Halls of Ivy's use of h as the variable of integration throws you off, but if you look closely, he uses H for the fixed height of the cylinder, not h.

    For a cylinder of radius r and height h,

    [tex]V=\int_0^h{\pi r^2 dx}=\int_0^h{\pi r^2 dy}=\int_0^h{\pi r^2 dt}=...[/tex]

    The variable of integration doesn't matter.
  14. Apr 29, 2007 #13
    Not really. You shouldn't have to do anything you don't understand. Only treat something as a constant if it is a constant.

    I'm not sure I know what you're talking about. Integrate anything you want to get anything you want, as long as it makes sense. I think integrating slices perpendicular to the height (if that means what I think it means) is the easiest way to solve the cylinder problem, but mathematics doesn't care what you do. It will give you whatever you ask for (even if that isn't what you really wanted, of course).

    I think the best assumption I can make at this point is that you don't really get what an integral is, and therefore don't understand how to apply it. Forgive me if I'm mistaken, but I'm going to speak from that perspective in the hopes of saying something helpful.

    An integral is just the sum of the "values" of an infinite number of infinitely small "pieces" of "something". We take the "something" in question, and define some kind of coordinates for referring to every "piece" of that "something".

    Above and/or below the integral sign, we write the range of coordinates (in the coordinate system we chose) of the "pieces" whose "values" we want to sum up. We can't name the coordinate of every single "piece" we want, because there are infinity of them. Instead, we just name the boundaries. In the 1-D case, this means the start and end coordinates.

    After the integral sign, we write a formula for the "value" of any "piece" in terms of its coordinates in our chosen coordinate system. In order to make the integral easy to solve, this formula should take the form of something times "d"something. The "d"something should be the size of the "piece", in terms of the coordinate system we're using. In other words, the amount of coordinate space occupied by the "piece". This sounds strange, since I already said the pieces are infinitely small, but actually the "d"something does represent an "infinitely small number". But not small enough to be zero. Don't think too hard about that detail; there's a more accurate description of it, but it wouldn't make things any clearer here.

    So, if we decide to make the "something" a 3-D shape, and to make the "pieces" cross-sectional slices of that shape, and to make the "values" the volumes of the individual slices, then we can just sum up the volumes of an infinite number of those infinitely small slices to get the total volume of the shape, no matter how curvy it is. That's one integral that will give you a volume.

    For our coordinate system, we use the position of each slice in the direction (call it "x") perpendicular to the slices we are taking. Then, the formula for the volume of each slice is just the formula for the area of that slice times dx (which is the infinitely small thickness of the slice in the x direction). In other words, volume = integral from min x to max x of Area(x) times dx. Now it should be clear that "integral" is little more than a fancy word for "sum", plus the idea that the things you're summing are itty-bitty.

    Of course, Area(x) must be written in terms of x (that's why it's written with the (x) at the end of it). Any "variable" in Area(x) that does not vary with x is not a variable. Anything that does, is, and needs to be rewritten in terms of x. To me, that's part of the concept of a "function of x".

    It seems odd to use an integral of circular cross-sections to find the volume of a cylinder, because all you're doing is adding an infinity of infinitely small cylinders to get one big cylinder. So you're using the formula in order to get the formula. Which is kinda silly. But does this help at all?
  15. May 3, 2007 #14


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    Well, thanks for your post, the reason for me to find out this was of mere interest, no particular use, since it's easier to find the formula for the volume, than integrating the area..
  16. May 4, 2007 #15

    Gib Z

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    Not always. How about two spheres of radius 4cm that are 1cm "into" each other...There is no general(mathematical) method of finding the volume of any given shape. Immersing them in water does good though.

    EDIT: No general EXACT method I should say. Numerical Integration can get you arbitrarily close to the volume. Any putting them in water works to the precision of the measurement tools.
    Last edited: May 4, 2007
  17. May 4, 2007 #16


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    Jarle's problem appears to be confusing "area of a cross section" with "surface area".
  18. May 4, 2007 #17


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    Area of the cross section is the area that is defined by h?

    It gets me confused when you say only to integrate the variable h. But when i look at the formula for volume of a cylinder i see that h is not integrated, but the radius is.
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