B Integrating to find surface area/volume of hemisphere

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1. Apr 14, 2017

user240

To find the surface area of a hemisphere of radius $R$, we can do so by summing up rings of height $Rd\theta$ (arc length) and radius $r=Rcos(\theta)$. So the surface area is then $S=\int_0^{\frac{\pi}{2}}2\pi (Rcos(\theta))Rd\theta=2\pi R^2\int_0^{\frac{\pi}{2}}cos(\theta)d\theta=2\pi R^2$.

However, to find the volume, if you were to use disks and of height to be $Rd\theta$, you miss a factor of $cos(\theta)$.. The edge of the each disk in this case cannot be 'slanted'.

My question is - why not? And why can we not use rings with a 'straight' edge like we do for disks when finding the surface area?

2. Apr 14, 2017

Staff: Mentor

The edge region of the disks/rings has a volume fraction that goes to zero for thin disks.

The fraction of the surface in this region doesn't go to zero - all the surface is this tilted region.

3. Apr 18, 2017

user240

Could you please explain/rephrase that part? I'm not sure I understand why it doesn't go to zero but the volume does (just so we're on the same page, you mean the blue part in the picture I attached, right?).

Can we show this rigorously or more quantitatively?

Also, why does using $dV=Rd\theta$ in the integral to find volume give a slightly larger result? Isn't it more 'better' to use this than $dV=Rcos(\theta)d\theta$, which has 'gaps' (albeit, which go to zero)?

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4. Apr 18, 2017

Staff: Mentor

It is possible to derive it rigorously, but that needs much more mathematics to define smooth surfaces, integrals on them and so on.

Compare the red part (the error of the finite disk approximation) to the white part (the correctly assigned volume) in the disks: it is tiny already. Now replace every disk by two disks with half the height. You reduce your error - by about a factor 2. Replace every new disk by 2 disks with half the new height. Again you reduce your error by a factor of about 2. In the limit of infinite disks, the error goes to zero. With the disk height $\Delta h$, the disks have a volume of $\pi (R \cos \theta)^2 \Delta h$. That expression now depends on both the angle and h, but thanks to $h=R \sin\theta$ where h is the height above the center, we can write $R^2 \cos^2 \theta = (R^2-h^2)$, and our volume expression simplifies to $\pi (R^2-h^2) \Delta h$, which can be converted to an integral and solved.

Now try to repeat the same with the surface: If you would approximate the surface area corresponding to a disk by $2 \pi \cos \theta \Delta h$ with the disk height $\Delta h$, you would only count the outer surfaces of the disks. But the actual surface is tilted - you underestimate the area by a factor given by the tilt (and only the tilt). Making the disks smaller does not help, because you get the same factor between estimate and the actual surface in every estimate no matter how small the disks get. That is not what we want. Instead of $\Delta h$, we should use the actual length of the line segment: $R \Delta \theta$. In the limit of infinitely small disks, this leads to the integral you have in post 1.
I don't understand that part of your post.