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Let's say we have a hollow hemisphere with a surface mass density given as $\sigma$ and radius $R$, We want to find the mass of this hemisphere in terms of $\sigma$ and $R$.

As usual, when trying to solve using single variable we'll take a ring of mass $dm$ at an angle $\theta$ (Which is from the horizontal), then

$$\int dm = \int_{0}^{\pi/2}\sigma (2 \pi R\cos\theta)Rd\theta$$

Where $Rd\theta$ is the thickness of the ring.

And we'll get the answer as,

$$M = \sigma(2\pi R^2)$$

But then I tried taking the mass $dm$ at a distance $x$ from the center of the base of hemisphere and perpendicular to it.

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The radius of the ring would be, by Pythagoras theorem $(R^2-x^2)^{1/2}$ and thickness dx.

By writing the expression for $dm$ we get,

$$\int dm = \int_0^R \sigma(2\pi (R^2-x^2)^{1/2}) dx$$

Both the answers after evaluating are different. My question is where am I wrong? Why are both the answers different even though the logic behind both the integrations are correct?

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Tony
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    The first answer is correct. For te second, the differential $dx$ is wrong. It should be $ds=\sqrt{dx^2+dr^2}$ – herb steinberg Sep 16 '19 at 19:08
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    The error you’re making is of a type with the one in the proof that $\pi=4$. In a nutshell, approximating a sphere with a series of cylinders isn’t “good enough” (which has a precise mathematical definition). – amd Sep 16 '19 at 23:23
  • But then in the case of a solid hemisphere, the second method (approximating the sphere with cylinders) does give me the correct answer. Turns out this is the easiest method to find its mass. Why does this method work there and not here? – Tony Sep 17 '19 at 01:05
  • Is anybody gonna say something? – Tony Sep 17 '19 at 18:19
  • The short, glib answer is that when approximating a volume, approximating it with nested cylinders is “good enough.” To get more technical than that requires analyzing the behavior of the respective Riemann sums. – amd Sep 18 '19 at 00:09
  • Incidentally, if you want people to see your comments, address them by prepending an @ sign to the user name, e.g., @amd. – amd Sep 18 '19 at 00:10
  • @amd I wasn't addressing any specific person. I just posted it publicly hoping somebody would see it. :) – Tony Sep 19 '19 at 04:30

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In the second method, the thickness is not $dx$. It is instead the length of curve within $dx$ and is given by,

$$\sqrt{1+(y_x')^2}dx$$

where $x^2+y^2=R^2$ and $y_x' = -x/y$.

As a result, the integral reads,

$$\int dm = \int_0^R \sigma2\pi (R^2-x^2)^{1/2} \sqrt{1+(y_x')^2}dx$$

$$=\int_0^R \sigma2\pi R dx= \sigma(2\pi R^2)$$

which is same as that of the first method.

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