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I am trying to derive the sphere/hemisphere volume formula by using the same concept (polar coordinate) when deriving the surface area:

SURFACE

enter image description here

VOLUME

enter image description here

The math looks ok but I cannot explain to my kid what has gone wrong. The usual cartesian coordinate derivation is as below: enter image description here

Can anyone offer an explanation what is wrong with the polar way of deriving the hemisphere volume?

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  • $\begingroup$ The main point is that the height of the blue cylindrical piece is not $ds$ but rather $dx$; see my answer for details. $\endgroup$
    – Mikhail Katz
    Commented Aug 1 at 9:18

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The volume of that little slice you've drawn isn't $\pi y^2 \, ds$; $ds$ is too long and tilted at the wrong angle. This is why the answer you get is an overestimate; we have $\frac{\pi^2}{4} = 2.47 \dots$ while $\frac{2 \pi}{3} = 2.09 \dots $.

The correct volume is just $\pi y^2 \, dx$, the same as in your cartesian coordinate derivation (although since $x$ is decreasing $dx = - r \sin \theta \, d \theta$ is technically negative so we need to multiply by $-1$); this gives a corrected integral

$$V = \int_0^{\frac{\pi}{2}} \pi R^3 \sin^3 \theta \, d \theta$$

which is $\frac{2 \pi}{3} R^3$ as expected.

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  • $\begingroup$ Thank you for going an extra mile to explain on the misuse of $ds$. So $dx=ds\sin\theta=Rd\theta\sin\theta$, and we are back to $dV=\pi y^2dx$. Just wondering we did use $ds$ in $dA=2\pi yds$ and yielded the correct result. If anyone knows of any youtube videos (or websites, books, etc) explaining this issue would be much appreciated. All info I can find so far is to use $dV=\pi y^2dx$ (for volume), $dA=2\pi yds$ (for surface), and just apply integration on them. $\endgroup$
    – Math_Physics
    Commented Jul 22 at 10:23
  • $\begingroup$ If "$ds$ is too long and tilted at the wrong angle", apparently it is not a 1-form but rather something else :-) $\endgroup$
    – Mikhail Katz
    Commented Jul 22 at 14:07
  • $\begingroup$ @Math_Physics: it's a good question. I'm not familiar with this but maybe you could find something on Khan Academy, e.g.: khanacademy.org/math/multivariable-calculus/… I don't know if I have a really satisfying explanation here, other than "well, the surface really is tilted." You can ask the same questions one dimension down, for calculating the circumference and area of a circle in 2d, where the geometry is maybe a little easier to understand. $\endgroup$
    – Qiaochu Yuan
    Commented Jul 22 at 16:03
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    $\begingroup$ There's a subtlety in the volume calculation, which is: why don't we have to take into account the fact that the slice is a little smaller at one end vs. the other end? The answer, and you can check this fairly explicitly, is that if you try to do this you'll find it only changes the volume by an error of order the "square of $dx$" (or $ds$, or any of the other $d$'s), and errors this small can be safely ignored because they disappear in the limit. But the same is not true of the surface area. $\endgroup$
    – Qiaochu Yuan
    Commented Jul 22 at 16:05
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Let's consider the lower-dimensional case of the calculation of the length of the circle and the area of the disk it bounds. The calculation with $ds$ goes through as usual for the length of the circle and gives the expected value. Now consider the calculation of the area of the disk by cutting it up into infinitesimal slices (curvilinear trapezoids) and adding up their areas.

Experienced users know that we can replace each infinitesimal slice by a rectangle of infinitesimal width $dx$ without affecting the final outcome for the disk, but to a newcomer this may seem odd ("why didn't we do that for the circle?"). Therefore it is better to approximate by ordinary trapezoids, without changing the length of the top and bottom sides of the curvilinear trapezoid.

The union of the trapezoids avoids the appearance of the "staircase" (as in the case of the rectangles) and provides a more plausible approximation. The area of each trapezoid is the product of its infinitesimal width (height) $dx$ (not the $ds$) by the average of the top and bottom sides. But now the picture becomes clearer: the top and bottom are of almost the same length, so the average is plausibly approximated by the long side of the infinitesimal rectangle.

This thought experiment should make it intuitively clear that in the end it makes no difference whether one uses rectangles or trapezoids, the main point being that the width is not $ds$ but rather $dx$.

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