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Compute the volume of a solid with given properties

Given a solid with circle base of radius 2 and cross sections which are equilateral triangles, compute the volume of the solid.


We may describe the top half of the circular base of the solid by the equation

    \[ f(x) = \sqrt{4-x^2}. \]

Thus, the length of the base of any equilateral triangular cross section is

    \[ 2 \sqrt{4-x^2} \qquad -2 \leq x \leq 2. \]

Since these are equilateral triangles with side length 2 \sqrt{4-x^2}, the area is given by

    \[ A(x) = \sqrt{3}(4-x^2). \]

Then we compute the volume,

    \begin{align*}  V &= \int_{-2}^2 \sqrt{3}(4-x^2) \, dx \\    &= 4\sqrt{3}(4) - \sqrt{3} \left( \left. \frac{x^3}{3} \right|_{-2}^2 \right) \\    &= 16 \sqrt{3} - \frac{16}{3} \sqrt{3} \\    &= \frac{32 \sqrt{3}}{3}. \end{align*}

(Note: Apostol gives the solution \frac{16 \sqrt{3}}{3} in the back of the book, but I keep getting \frac{32 \sqrt{3}}{3}, as does Edwin in the comments. I’m marking this as an error in the book for now. If you see where my solution is wrong and Apostol is correct please leave a comment and let me know.)

13 comments

  1. Andrei says:

    when you calculated the volume in your solution, you have integered in relation of x. The right thing to do is integer in relation of y.

    (4-x²) = y²

    Then you will find the apostol’s result.

  2. Felipe says:

    Isn’t it equal to \frac{8 \sqrt{3} }{3} i made a calculation of the volume of an cone with a circle base with diameter 4 and “sides” 4, and using the line equation x =2 - \frac{y\sqrt{3}}{3} and therefore calculating the integral

        \[V =\pi \int_{0}^{2\sqrt{3}}2 - \frac{y\sqrt{3}}{3} dy\]

    • Shafil says:

      I don’t think it’s a cone. The figure you get when you cut a cone with a plane perpendicular to its base is a hyperbola (except if you cut it along a diameter).not triangle
      Here we are cutting the figure with any plane that is perpendicular to a fixed diameter.
      It does not have to be along another diameter.

  3. shivam says:

    as it is cone we can rotate a half equilateral travel that help to generate cone and we can compute volume by solid rotation.

  4. cecM says:

    hello, still do not quite understand why 2 root from 4 – x² it is the fixed diameter of our circle, I would greatly appreciate a brief explanation.

    • Anonymous says:

      imagine a circle in a plane ,now image a diameter in that circle (a fixed one), every plane perpendicular to that fixed diameter intersect the solid in an equilateral triangle. By the theory we know that the volume is the integral of the function area(area of each triangle for each position along the fixed diameter ,from -2 to +2 since 2 is the radius) .And its important to notice that the function area depends on the perpendicular segment to the diameter in each triangle.sorry for the bad english.

  5. Edwin says:

    Hey man, me again, I got the same result as you, but Apostol says it’s actually \frac{16 \sqrt{3}}{3}, how come we got double to what’s written on the answers in the book?

    • RoRi says:

      Hey, I’m not sure how he’s getting \frac{16 \sqrt{3}}{3}. I checked back over what I have posted and I can’t find a mistake. (I plugged the integral we have into Wolfram Alpha, and it gives us \frac{32 \sqrt{3}}{3}, so if there is a mistake it’s somewhere before we evaluate the integral.) Maybe it’s a mistake in the solutions (there are a few others). I think I’ll mark this with the “errata” tag unless someone comes along and figures out that Apostol is actually correct (which is definitely possible also).

      • Edwin says:

        Hey, so I thought maybe, because it’s a fixed diameter, the length of the base of the triangle is not actually 2 \sqrt{4-x^2}, rather it’s just \sqrt{4-x^2}, cause it covers the whole diameter, thus, it’s only half of that, is this argument valid? What do you think?

      • RoRi says:

        I’m not sure I understand your argument.

        I looked around more though, and think that our answer is right and Apostol is wrong. One of the examples on this page I think is exactly this problem, and they get the same integral we have \int_{-2}^2 \sqrt{3}(4-x^2) \, dx (it’s most of the way toward the bottom of that pdf). They don’t evaluate the integral, but I’m pretty sure we have that part right (and Wolfram alpha agrees with our evaluation of the integral).

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