Where does the formula for the volume of a hemisphere come from, I have been using this The formula
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Welcome to MSE. Please read this text about how to ask a good question. – José Carlos Santos May 01 '20 at 06:55
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1See https://www.famousscientists.org/archimedes-makes-his-greatest-discovery/ – John Douma May 01 '20 at 07:30
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The circle centre $(0,0)$ with radius r is $x^2+y^2=r^2$. You want the volume of revolution generated by this curve from $r-h$ to $r$, which is $\pi\int_{r-h}^{r}{(r^2-x^2)dx}$. – Paul May 01 '20 at 07:45
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What I'm wondering, I have never seen this formula for spheres? What does $h$ mean? Anyway here's a calculation for the Volume of a sphere with radius $R$:
$$ \iiint_{\|\vec{x}\|=R}\mathrm{d}x\mathrm{d}y\mathrm{d}z\\ =\int_{\varphi=0}^{2\pi}\int_{\theta=0}^{\pi}\int_{r=0}^R\sin(\theta)*r^2\mathrm{d}r\mathrm{d}\theta\mathrm{d}\varphi\\ =2\pi*\int_{\theta=0}^\pi\sin(\theta)\int_{r=0}^Rr^2\mathrm{d}r\mathrm{d}\theta\\ =2\pi*\frac{R^3}{3}\int_{\theta=0}^\pi\sin(\theta)=\frac{4\pi R^3}{3} $$
If we want a hemisphere: $$ \frac{1}{2}*\frac{4\pi R^3}{3}=\frac{2\pi R^3}{3} $$
lphil
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It is a portion of the sphere of height h I should think. It can be done as a volume of revolution rather than a triple integral. – Paul May 01 '20 at 07:19
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@Paul your're right, rotation would've also been possible and thanks for the clarification for the height – lphil May 01 '20 at 07:24
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1We don't know if the person who asked this question has had Calculus and I don't believe Archimedes used a triple integral to solve this. – John Douma May 01 '20 at 07:32
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