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Integral of Area is Volume?

  1. May 4, 2015 #1
    I may have misinterpreted this but today in calculus (AB) we were forming solids from 2 dimensional equations. One of the methods involved taking an integral of an area equation to solve for a solids volume. I got very excited as I often have difficulty remembering volume equations but am familiar with the basic area ones, so I thought I had found my solution to finding the volume of something with an area equation. However, when I tried to take the integral of the equation for the area of a circle ((pi)r^2) I came up with the equation (((pi)r^3)/3)+C which, as you may know, is not the equation for the volume of a sphere (4(pi)r^3)/3 although it is awfully close. I am very new to this so I'm definitely looking at this the wrong way so if any one could explain this to me or send me any good links that would be awesome!
  2. jcsd
  3. May 4, 2015 #2
    I just looked up the surface area of a sphere and its 4(pi)r^2 so I'm betting that's where the 4 came from but now I may be more confused
  4. May 4, 2015 #3


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    You're trying to make a generalization which is not true, in all cases.

    There are some theorems about how some areas can be turned into volumes by integration:

  5. May 4, 2015 #4
    To get the volume of a sphere by integration, put the center of the sphere at x,y,z=0,0,0.
    For a sphere of radius R, we can integrate along the x-axis from -R to +R.
    We integrate the area (pi)r^2 substituting r^2=R^2-x^2 from the formula for a circle.
    The result is volume=4/3(pi)R^3.
    It's much easier to remember the formula!
  6. May 14, 2015 #5


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    You need to realize that if you're integrating in two variables, that the volume of a solid of revolution is the area underneath the curve being rotated around a specified axis.
  7. May 14, 2015 #6


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    What you did was calculate the volume of a circular cone of height r with base of radius r. Can you see why?

    To get the volume of a sphere, you could consider first a hemisphere, which you can think of as a series of thin disks (circles) of diminishing radius. If these were a finite set, then you'd add up the volume of each thin disk and get (approx) the volume of a sphere. Integration imagines each disk to be infinitesimally thin and you need to think of how big the circle is as you move up the hemisphere. Let's imagine the upper hemisphere (radius R), going from ##z=0## at the equator to ##z=R## at the pole:

    The radius of the circle at each height ##z## is given by:

    ##r^2 = R^2 - z^2##

    (Check this using some basic trigonometry.)

    The area of the circle at each height is ##\pi(R^2 - z^2)##

    This is now what you integrate from ##z = 0## to ##z = R## to get the volume:

    ##V = \int_0^R \pi (R^2-z^2) dz = \pi [R^2z - \frac{z^3}{3}]_0^R = \pi [R^3 - \frac{R^3}{3}] = \frac{2\pi R^3}{3}##

    And the full sphere has twice this volume.
    Last edited: May 14, 2015
  8. May 14, 2015 #7


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    Another way would be to use the shell method.
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