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Using Calculus to find Volume of 3-D Solids

  1. Oct 17, 2005 #1
    Hey folks... I've been thinking about using integration to deal with the volumes of various objects such as:

    Cylinders: For cylinders, I used integration to find the volume. I imagined the area of a circle, which is [tex]\pi*r^2[/tex] and then decided that if I stacked a series of solid disks (which have the area of a circle) to a height of h, then I would get the volume of a cylinder of height h:

    [tex]\int_{0}^{h} \pi*r^2*dh[/tex]

    And that yields the correct volume for a cylinder, [tex]V=\pi*r^2*h[/tex].

    My question is, can I apply this method of thinking to find the area of a sphere and a cone?

    Every time I try, I get the wrong answer.

    I assume that I need to take a series of disks and add them up, as I did for the cylinder, but the cross sectional area will change as they get slightly smaller. So I tried this:

    [tex]2 * \int_{0}^{r} \pi*r^2*dr[/tex]

    My logic was that the radius will shrink to zero as you get to the outer most disk of the sphere, and then you have to double the answer to get the second half of the sphere. The problem is that this gives me [tex](2/3)\pi*r^2[/tex] and the real volume is 4/3, not 2/3.

    And then comes the cone, and I'm not even sure where to start since I would think it would be similar to the sphere, but obviously it's not since they look so different.

    So yes, I'm a bit confused, but if anyone can decipher what I've written and point me in the right direction, that would be greatly appreaciated. Thanks!
    Last edited: Oct 17, 2005
  2. jcsd
  3. Oct 17, 2005 #2


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    But the radius of each disk doesn't decrease linearly.
    Since the equation of a sphere is x2+ y2+ z2= R2, if we let z range from -R to R, we can measure the radius of each disk along either x or y axis- let's take the x-axis so y= 0 and x2+ z2= 0 or [itex]x= \sqrt{R^2- z^2}[/itex].
    The area of the disk is [itex]\pi \left(\sqrt{R^2-z^2}\right)^2[/itex]
    The volume of the sphere is given by
    [tex]\pi \int_{-R}^R R^2- z^2 dz[/tex].

    Similarly, if a cone has height h and base radius R, we can look at the xz-plane. One side of the cone is the line running from (0,0,h) to (R, 0, 0). That has equation x= R- (R/h)z with y= 0. for any z, x is the radius of a disc with area [itex]\pi x^2= \pi(R- (R/h)z)^2= \piR^2(1- \frac{z}{h})^2[/tex]. The volume is given by
    [tex]\pi R^2\int_0^h(1- \frac{z}{h})^2 dz[/tex]
  4. Oct 17, 2005 #3
    edit: nevermind, I found my mistake. i was confusing the decreasing radius of the disc with the radius of the sphere.
    Last edited: Oct 17, 2005
  5. Oct 17, 2005 #4
    that makes sense, thank you!
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