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I had this questions about volume of a hemisphere that I figured out. But now I'm being confused by why the same method doesn't apply when integrating to find the area of a hemisphere? If we're stacking disks of different radii to find the volume of a solid hemisphere, by analogy, I feel like, we should be able to stack rings of different radii to find the surface area of a hemispherical shell. So if earlier we had $\pi r^2 = \pi R^2 \sin (\theta)^2 $ and a height $ R \sin (\theta) d\theta$, the only thing that should change for a ring is that instead of an area of a circle, we're using a circumference $2\pi r = 2 \pi R \sin (\theta) $.

This made sense to me because the height of a disk or a ring/cylinder should be perpendicular to the other dimensions. However, it appears that in the case of a ring the height element is suddenly just $R d \theta$, which is what I mistakenly used when trying to find the volume of a solid hemisphere, where it was wrong, but now for some reason that blows my mind completely, it's appropriate when finding an area...

Could someone please explain this?

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To remove all the distractions of three dimensions and curved surfaces, let's just consider calculating the area of a right isosceles triangle and the length of its diagonal. Take the triangle formed by the points $(0,0)$, $(0,1)$ and $(1,0)$ and cut it into $n$ horizontal slices of height $1/n$.

The area of the $k$-th trapezoidal slice from the top is

$$\frac1n\frac{\frac kn+\frac{k-1}n}2\;,$$

and summing this for $k$ from $1$ to $n$ yields $\frac12$, as it should. The length of the $k$-th slice of the diagonal is

$$ \frac{\sqrt2}n\;, $$

and summing this for $k$ from $1$ to $n$ yields $\sqrt2$, as it should. No approximations so far. Now let's imagine $n\gg1$ and approximate. In the area calculation, let's leave off the little triangles near the diagonal and sum only the rectangular parts of the trapezoidal slices, whose area is

$$ \frac1n\frac{k-1}n\;. $$

Summing this for $k$ from $1$ to $n$ yields $\frac12(n-1)/n$, so it's off by $\frac1{2n}$, the total area of the $n$ triangles with area $\frac1{2n^2}$ each that we neglected. Taking $n\to\infty$, we recover the correct area $\frac12$.

By contrast, if we were to approximate the diagonal the way you proposed to, we'd estimate the contribution from each slice to be

$$ \frac1n\;, $$

and then summing for $k$ from $1$ to $n$ yields $1$, an estimate that's off not by a small additive term that goes to zero with $n\to\infty$, but by a constant factor $\sqrt2$.

The reason is that the error we make in omitting the triangles is quadratic in $n^{-1}$, so $n$ times such an error is still linear in $n^{-1}$, whereas the error we make in approximating the diagonal is linear in $n^{-1}$, so $n$ times that error is a constant. As other answers have pointed out, most of the area is in the rectangles, the triangles are just a small fraction, not just of the whole ($1/n$) but of even of each slice ($1/n^2)$, whereas the error in estimating the diagonal is of course small per slice since the slices are small ($1/n$), but it's not small compared to the contribution of each slice; it's a constant part of that contribution, and thus sums up to a constant value no matter how small we make the slices.

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This is more than faintly analogous to the old troll math meme: $\pi = 4$

But we know $\pi \neq 4$. The steps that are used in the image to approach the circle don't approximate it well enough to give you the length. Instead, you need the curves to not only approach the circle in distance, but you need the tangents of the curves to approach the tangents of the circle.

Analogously, you need to approximate the hemisphere with things that are are a little better than rings. The easiest way to go is to use little pieces of cones. If you wrote out the Riemann sums related to the technique you're using now, you'd be basically approximating the surface with rings. Try writing out the sums if you instead approximated with thin slices of a cone.

Alternatively, consider how the surface area of the sphere should relate to its volume as the radius changes. What is the derivative of the volume of the sphere wrt $r$ (this is a bit of cheat, and won't work with things like ellipses, but its okay for spheres).

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  • $\begingroup$ Haha, I've seen that meme :) I agree that approximating with pieces of hollow cones is better for a shell, but then how come we don't also use pieces of solid cones to approximate volume of a solid sphere? Archaick mentioned that the curvature of the outer rim of the disk doesn't matter that much, but does it really matter so little that there's 0 difference between approximations with cone pieces or disks? $\endgroup$
    – Raksha
    Aug 17 '15 at 1:03
  • $\begingroup$ Yeah, it makes no difference when you do the volume. Calculate the difference in volume of the cone pieces and the cylinders, and you'll see it goes to 0. Basically, the important part of the volume is all the stuff inside, the outer part doesn't contribute much. But, with surface area, its all about the outside part, and you can calculate the difference between approximations and see that its big enough to count. $\endgroup$
    – user24142
    Aug 17 '15 at 1:08
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Intuitively, when we consider the height of a super-thin disk, the curvature of the outer rim of the disk matters very little. However, if we want to consider the area of the rim of the disk, the fact that the top of the disk has a smaller radius than the bottom affects the area substantially (think a trapezoid whose sides have been glued together vs. a rectangle whose sides have been glued together). You may want to check out the Wikipedia article about Jacobian determinants which allow one to integrate using coordinates other than Cartesian ones: Jacobian Determinants. The relevant theorem is that if $\Phi: A \to B$ is a smooth change of coordinates, $f(\textbf{x}): A \to \mathbb{R}$ is also smooth, and $P \subset A$, then $$\int_P f(\textbf{x}) dV=\int_{\Phi(P)} (f\circ \phi^{-1})(\textbf{x}) \mathrm{det(J_{\Phi})} dV,$$ were $J_{\Phi}$ is the Jacobian matrix of $\Phi$.

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