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Question Number 215344 by universe last updated on 03/Jan/25

Answered by MrGaster last updated on 04/Jan/25

$$\mathrm{Let}\mathrm{\Sigma }\mathrm{be}\mathrm{the}\mathrm{boundary}\mathrm{of}D.$$$$\mathrm{\Sigma }=\sum _1\cup \sum _2\cup \sum _3$$$$\mathrm{where}$$$$\sum _1=\{(x,y,z)\in {\mathbb{R}}^3:x^2+y^2+b^2,\mid z\mid \le \sqrt{a^2-b^2}\}$$$$\sum _2=\{(x,y,z)\in {\mathbb{R}}^3:x^2+y^2+z^2=a^2,z\ge 0\}$$$$\sum _3=\{(x,y,z)\in {\mathbb{R}}^3:x^2+y^2+z^2=a^2,z\le 0\}$$$$\mathrm{The}\mathrm{surface}\mathrm{area}\mathrm{of}\sum _1\mathrm{is}\int {\int }_{2}ds=2\pi b\cdot 2\sqrt{a^2-b^2}=4\pi b\sqrt{a^2-b^2}$$$$\mathrm{The}\mathrm{surface}\mathrm{areas}\mathrm{of}\sum _2\mathrm{and}\sum _3\mathrm{are}\mathrm{each}\mathrm{half}$$$$\mathrm{the}\mathrm{suface}\mathrm{area}\mathrm{of}\mathrm{a}\mathrm{sphere}\mathrm{with}\mathrm{radius}a,\mathrm{i}.\mathrm{e}.,$$$$\frac{1}{2}\cdot 4\pi a^2=2\pi a^2$$$$\mathrm{Therefore},\mathrm{the}\mathrm{total}\mathrm{surface}\mathrm{area}\mathrm{of}\mathrm{\Sigma }\mathrm{is}$$$$\int {\int }_{\mathrm{\Sigma }}ds=\int {\int }_{\sum _1}ds+\int {\int }_{\sum _2}ds+\int {\int }_{\sum _3}ds$$$$=4\pi b\sqrt{a^2-b^2}+2\pi a^2+2\pi a^2$$$$=4\pi b\sqrt{a^2-b^2}+4\pi a^2$$

Answered by mr W last updated on 04/Jan/25

Commented by mr W last updated on 04/Jan/25

$$thesolidDisasphericalring$$$$\left(alsocallednapkinring\right).$$

Commented by mr W last updated on 04/Jan/25

$$R=a$$$$r=b$$$$L=2\sqrt{R^2-r^2}$$$$S=4\pi R^2+2\pi rL-2\times 2\pi R(R-\sqrt{R^2-r^2})$$$$=4\pi r\sqrt{R^2-r^2}+4\pi R\sqrt{R^2-r^2}$$$$=4\pi (R+r)\sqrt{R^2-r^2}$$$$=4\pi (a+b)\sqrt{a^2-b^2}$$$$check:$$$$b=0\Rightarrow S=4\pi a^{2}ok✓$$$$b=a\Rightarrow S=0ok✓$$

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