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Integral (area) of a revolved parabola

  1. Nov 25, 2015 #1
    Whats's the volume under the revolving parabola y=ax2. R is xmax.
    Snap1.jpg
    $$V=\int_0^R\pi xdx\cdot y=\pi\int_o^R x\cdot Ax^2dx=\pi A\frac{R^4}{4}$$
    Volume should be relative to R3. and if i had, for example, y=Ax3 then, according to my calculation i would get relative to R7 and so on.
     
  2. jcsd
  3. Nov 26, 2015 #2

    Samy_A

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    Indeed (proportional would probably be a better term).
    So clearly something is wrong here.

    What formula are you using to find the volume of a solid of revolution?
     
  4. Nov 26, 2015 #3

    Ssnow

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    Why there is ##x\cdot y## in your formula, this must be ##V=\int_{0}^{R}\pi f^{2}(x)\,d\,x##...
     
  5. Nov 26, 2015 #4
    The volume is the sum of the vertical tubes of changing height y and infinitesimally small thickness dx. the perimeter of the tubes is πr.
    $$V=\int dv=\int (\pi x)y\cdot dx$$
    y, in turn, equals Ax2. but according to this approach i would get a bigger error if i had, for example, y=Bx3
     
  6. Nov 26, 2015 #5
    There is no reason for the volume to be proportional to ##R^3##. In fact you will discover that it isn't!
    I'm not sure why you are considering the perimeter of the disks centred about the x-axis. Would you not want to be using the area of the disks and then integrating over x? If you revolve your parabola about the x-axis, you get lots of solid disks centred about the x-axis which grow in radius as you move further along in x, right? What is the radius of a given disk as a function of x? Once you have this, you then know the area of each disk as a function of x. Once you have an expression for the disk area as a function of x, you can integrate over x, effectively summing up the infinitesimal cylinders centred about the x-axis. Does this make sense?
     
  7. Nov 29, 2015 #6
    $$V=\int_0^{y=AR^2}\pi(R^2-x^2)dy=\pi\int_0^{AR^2}R^2-\frac{y}{A}dy=\pi\left[ R^2y-\frac{y^2}{2A} \right]_0^{AR^2}=\frac{\pi A}{2}R^4$$
    Also with my previous method of tubes:
    $$V=\int_0^R2\pi xdx\cdot y=2\pi\int_o^R x\cdot Ax^2dx=\frac{\pi A}{2}R^4$$
    I found in a site the volume under a paraboloid, an inverse of mine, is ##V=\frac{\pi}{2}hr^2##, and according to that my volume is:
    $$V=\pi R^2h-\frac{\pi}{2}hR^2=...=\frac{\pi A}{2}R^4$$
    In all 3 methods i get the same.
    But the units are ##R^4[m^4]## while volume is [m3]
     
  8. Nov 30, 2015 #7

    Samy_A

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    I made the same mistake as you. In what units is A?
     
  9. Nov 30, 2015 #8

    Svein

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    Yes. Your trouble lies with the interpretation of the constant A. Since you are trying to find a volume, you are assigning a "length dimension" to x. Assume that x has the dimension m (meter). Then y also needs to have the dimension m. Using the formula y = Ax2 means that A must have the dimension m-1. With that in mind the dimension of your volume becomes m-1⋅m4 = m3.
     
  10. Nov 30, 2015 #9
    I understood, thanks
     
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