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My idea to calculate the volume of the hemisphere is to sum up the area of circles of all radii up to the radius of the hemisphere we are interested in: $$\int_0^r \pi x^2 dx$$ This gives $\frac{1}{3}\pi r^3 \neq \frac{2}{3}\pi r^3$, so deviating by a factor of $2$ from the known formula for the volume of a hemisphere. Can you explain where my reasoning is wrong and why it deviates by a factor of $2$?

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    $\begingroup$ Remembering having once experienced the same misapprehension myself, I'd like to differently summarize the answers you've gotten in case it proves helpful: while $\pi x^2$ does range over the correct set of values for cross-sectional area as $x$ varies between $0$ and $r$, their density $dx$ with respect to the $x$-coordinate differs from what the correct expression $\pi (r^2-x^2)$ for the hemisphere would give, and both 'width' and 'height' matter in integration. Specifically, you had the radii $x$ of the disks decreasing linearly as $x$ goes from $r$ to $0$, actually resulting in a cone... $\endgroup$ Feb 21 at 2:28
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    $\begingroup$ ...whereas if you inspect a hemisphere at equally spaced $x$-coordinates then you see it shrinks slower than linearly as you move away from the equator and remains disproportionately wider than the cone would be throughout, only catching up by tapering off rapidly at the pole -- all as reflected in the radius-function $\sqrt{r^2-x^2}$ as $x$ goes from $0$ to $r$ (yes, here $x=0$ at the widest part whereas for the radius-function $x$, it corresponds to the narrowest, but this reversal of the $x$-coordinate turns out to be unimportant -- because it doesn't distort densities). $\endgroup$ Feb 21 at 2:28
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    $\begingroup$ Funny that the volume of a hemisphere is exactly twice the volume of a cone with the same height and ground area! I wonder if there's simple a geometric argument that illustrates this. $\endgroup$ Feb 21 at 9:44
  • $\begingroup$ I could never have got that far, but isn't this re-inventing the wheel? Is there not an established formula that gives a better solution? $\endgroup$ Feb 21 at 20:13
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    $\begingroup$ @RobbieGoodwin. Of course there is, but someone had to come up with that formula, which indicates that there is value in learning to derive things of that nature. $\endgroup$ Feb 21 at 22:55

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You're computing the volume of a cone, not a hemisphere!

More precisely, your integral gives the volume obtained by rotating the triangle $0 \le y \le x \le r$ about the $x$-axis.

The hemisphere, on the other hand, is obtained by rotating the region $0 \le y \le \sqrt{r^2-x^2}$, $0 \le x \le r$ about the $x$-axis, so to get the volume you need to write $$ \int_0^r \pi \Bigl( \sqrt{r^2-x^2} \Bigr)^2 \, dx $$ instead.

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  • $\begingroup$ You are right, I originally misread the question. $\endgroup$ Feb 20 at 12:27
  • $\begingroup$ I see. I did not think about it as a rotation but rather as piling up disks of different sizes. Could you give me some intuition why integrating from 0 to r can be seen as a rotation? It's hard for me to grasp it, because intuitively it seems to me we are not rotating around the full circle, but only for a segment of the circle of length r $\endgroup$
    – timtam
    Feb 20 at 16:31
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    $\begingroup$ You are adding discs together: they are the discs of radius $\sqrt{r^2-x^2}$ as $ x $ goes from $0$ to $r $, not the discs of radius $x$. $\endgroup$ Feb 20 at 18:42
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    $\begingroup$ You can make a cone by piling discs whose radii linearly transition from 0 up to the radius of the base of the cone. A hemisphere’s disks’ radii do not transition linearly because the profile of a hemisphere is curved not straight as is the profile of a cone. $\endgroup$
    – bob
    Feb 21 at 14:32
  • $\begingroup$ @HansLundmark Jiminy Christmas, you'd think I hadn't literally just been teaching volumes of revolution yesterday. I'll turn my math degrees back in now.... Or maybe just take a beat to sanity-check before commenting =D $\endgroup$
    – nitsua60
    Feb 22 at 13:15
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The thing to keep in mind with disk integration is the fact that the variable of integration represents the thickness of each disk.

You are trying to integrate the volume of a half-sphere, using: $$ \int_0^r \pi x^2 dx $$ with the idea being that a bunch of circular disks (each of area $A=\pi x^2$) should add up to make a half-sphere. So if we investigate, things seem to make sense at the endpoints of $x=0$ and $x=r$ where the areas are $A=\pi 0^2$ and $A=\pi r^2$ respectively as expected. But because the overall answer is incorrect we clearly have to investigate further!

What occurs at the half-way point? At $x=r/2$ the area is $A=\pi (r/2)^2$... but that is much too narrow a disc! A hemisphere is still pretty wide midway up. Doing a simple Pythagorean Theorem check, the hypotenuse being $r$ and the side being $r/2$ would mean the disk radius should be $\sqrt{3/4}r\approx 0.866r$ not $0.5r$!

Do you see the issue now? Setting the area of each disk to $A=\pi x^2$ inherently implies that the disk's radius at each point $x$ is exactly equal to $x$... but that $x$ relationship is one of a straight line, not a half-circle profile. So, as others have mentioned, what you end up with is a stack of disks with linearly increasing radius... a cone! (The reason you're off by a factor of 2 is coincidence due to $V_{cone}=\pi r^2 h/3=\pi r^3/3$ versus $V_{hemisphere}=V_{sphere}/2=(4\pi r^3/3)/2=2\pi r^3/3$ and not because you forgot a 2 somewhere.)

To fix the issue you have to make sure that each disk (which has thickness $dx$ at point $x$) has the correct radius to match the circle for that $x$ location. This means you need to use the Pythagorean Theorem to find a relationship between each slice's $x$ position and it's radius.

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Your problem seems to be that you don't have a clear concept of how to integrate the volume of an object by stacking disks on each other.

In order to compute a volume by stacking disks, you need the distance between disks to be proportional to how much the variable of integration (in your case $x$) has increased. A way to look at this is that if you take a stack of disks for which $x$ increases by $\Delta x$ for each new disk, the "thickness" of that disk needs to be $\Delta x.$

A classic puzzle that is constructed with disks of uniform thickness that increase in radius by a constant amount for each disk is the Towers of Hanoi. Usually the disks are stacked from the largest to the smallest rather than smallest to largest, but you would just have to turn the stack upside down to fix that. And the stack of disks looks like a cone, never like a hemisphere:

towers of hanoi with all disks on one peg

The correct way to do what you are trying to do is called the disk method. It works when you are computing the volume of an object that has a rotational axis of symmetry such that the space occupied by the object never changes when you spin it around this axis. Such an object can be sliced into disks perpendicular to the axis. (For some objects, some of the disks will have circular holes in the middle, but this does not happen for a hemisphere.)

To do the disk method you have to start at one end of the object's axis and integrate along the axis. At each distance $x$ from the end of the axis, you have to determine what radius of disk fits in the object on that particular plane.

For a hemisphere you can choose to integrate from the flat end to the round end or from the round end to the flat end. But in neither case will the radius of each disk be equal to the distance from the end you started at. The only object for which that is true is a cone, which is why the integral you wrote will give you the volume of a cone of height $r$ and circular radius $r$ every time.

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