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Question Number 183216 by mr W last updated on 23/Dec/22
Commented by mr W last updated on 23/Dec/22
Commented by mr W last updated on 23/Dec/22
the general case for sphere (red) cut by a cylinder (green)
$${the}\:{general}\:{case}\:{for}\:{sphere}\:\left({red}\right)\:{cut} \\ $$$${by}\:{a}\:{cylinder}\:\left({green}\right) \\ $$
Answered by mr W last updated on 23/Dec/22
given: R, r, a find the volume of the portion of the sphere cut by the cylinder b^2 =a^2 +r^2 −2ar cos θ z=(√(R^2 −b^2 ))=(√(R^2 −a^2 −r^2 +2ar cos θ)) a sin θ=(a−r cos θ) tan ϕ ⇒ϕ=tan^(−1) (((a sin θ)/(a−r cos θ))) 2bdb=2ar sin θ dθ dS=2ϕbdb=2ar tan^(−1) (((a sin θ)/(a−r cos θ))) sin θ dθ V=2∫_0 ^π zdS V=4arR∫_0 ^π (√(1−((a/R))^2 −((r/R))^2 +((2ar)/R^2 ) cos θ)) tan^(−1) (((sin θ)/(1−(r/a) cos θ))) sin θ dθ let λ=(a/R), μ=(r/R) ⇒V=4μλR^3 ∫_0 ^π (√(1−μ^2 −λ^2 +2μλ cos θ)) tan^(−1) (((sin θ)/(1−(μ/λ) cos θ))) sin θ dθ
$${given}:\:{R},\:{r},\:{a} \\ $$$${find}\:{the}\:{volume}\:{of}\:{the}\:{portion}\:{of} \\ $$$${the}\:{sphere}\:{cut}\:{by}\:{the}\:{cylinder} \\ $$$${b}^{\mathrm{2}} ={a}^{\mathrm{2}} +{r}^{\mathrm{2}} −\mathrm{2}{ar}\:\mathrm{cos}\:\theta \\ $$$${z}=\sqrt{{R}^{\mathrm{2}} −{b}^{\mathrm{2}} }=\sqrt{{R}^{\mathrm{2}} −{a}^{\mathrm{2}} −{r}^{\mathrm{2}} +\mathrm{2}{ar}\:\mathrm{cos}\:\theta} \\ $$$${a}\:\mathrm{sin}\:\theta=\left({a}−{r}\:\mathrm{cos}\:\theta\right)\:\mathrm{tan}\:\varphi \\ $$$$\Rightarrow\varphi=\mathrm{tan}^{−\mathrm{1}} \left(\frac{{a}\:\mathrm{sin}\:\theta}{{a}−{r}\:\mathrm{cos}\:\theta}\right) \\ $$$$\mathrm{2}{bdb}=\mathrm{2}{ar}\:\mathrm{sin}\:\theta\:{d}\theta \\ $$$${dS}=\mathrm{2}\varphi{bdb}=\mathrm{2}{ar}\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{{a}\:\mathrm{sin}\:\theta}{{a}−{r}\:\mathrm{cos}\:\theta}\right)\:\mathrm{sin}\:\theta\:{d}\theta \\ $$$${V}=\mathrm{2}\int_{\mathrm{0}} ^{\pi} {zdS} \\ $$$${V}=\mathrm{4}{arR}\int_{\mathrm{0}} ^{\pi} \sqrt{\mathrm{1}−\left(\frac{{a}}{{R}}\right)^{\mathrm{2}} −\left(\frac{{r}}{{R}}\right)^{\mathrm{2}} +\frac{\mathrm{2}{ar}}{{R}^{\mathrm{2}} }\:\mathrm{cos}\:\theta}\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{sin}\:\theta}{\mathrm{1}−\frac{{r}}{{a}}\:\mathrm{cos}\:\theta}\right)\:\mathrm{sin}\:\theta\:{d}\theta \\ $$$${let}\:\lambda=\frac{{a}}{{R}},\:\mu=\frac{{r}}{{R}} \\ $$$$\Rightarrow{V}=\mathrm{4}\mu\lambda{R}^{\mathrm{3}} \int_{\mathrm{0}} ^{\pi} \sqrt{\mathrm{1}−\mu^{\mathrm{2}} −\lambda^{\mathrm{2}} +\mathrm{2}\mu\lambda\:\mathrm{cos}\:\theta}\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{sin}\:\theta}{\mathrm{1}−\frac{\mu}{\lambda}\:\mathrm{cos}\:\theta}\right)\:\mathrm{sin}\:\theta\:{d}\theta \\ $$
Commented by manxsol last updated on 23/Dec/22
Thanks, Sir W,applause for your effort, very clear
$${Thanks},\:{Sir}\:{W},{applause} \\ $$$$\:{for}\:{your}\:{effort},\:{very}\:{clear} \\ $$$$ \\ $$
Commented by mr W last updated on 23/Dec/22
thank you too!
$${thank}\:{you}\:{too}! \\ $$

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