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Question-215344

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Question Number 215344 by universe last updated on 03/Jan/25
Answered by MrGaster last updated on 04/Jan/25
Let Σ be the boundary of D. Σ=Σ_1 ∪Σ_2 ∪Σ_3 where Σ_1 ={(x,y,z)∈ R^3 :x^2 +y^2 +b^2 ,∣z∣≤(√(a^2 −b^2 ))} Σ_2 ={(x,y,z)∈ R^3 :x^2 +y^2 +z^2 =a^2 ,z≥0} Σ_3 ={(x,y,z)∈ R^3 :x^2 +y^2 +z^2 =a^2 ,z≤0} The surface area of Σ_1 is ∫∫_2 ds=2πb∙2(√(a^2 −b^2 ))=4πb(√(a^2 −b^2 )) The surface areas of Σ_2 andΣ_3 are each half the suface area of a sphere with radius a,i.e., (1/2)∙4πa^2 =2πa^2 Therefore,the total surface area of Σis ∫∫_Σ ds=∫∫_Σ_1 ds+∫∫_Σ_2 ds+∫∫_Σ_3 ds =4πb(√(a^2 −b^2 ))+2πa^2 +2πa^2 =4πb(√(a^2 −b^2 ))+4πa^2
$$\:\mathrm{Let}\:\Sigma\:\mathrm{be}\:\mathrm{the}\:\mathrm{boundary}\:\mathrm{of}\:{D}. \\ $$$$\Sigma=\sum_{\mathrm{1}} \cup\sum_{\mathrm{2}} \cup\sum_{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{where} \\ $$$$\sum_{\mathrm{1}} =\left\{\left({x},{y},{z}\right)\in\:\mathbb{R}^{\mathrm{3}} :{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{b}^{\mathrm{2}} ,\mid{z}\mid\leq\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }\right\} \\ $$$$\sum_{\mathrm{2}} =\left\{\left({x},{y},{z}\right)\in\:\mathbb{R}^{\mathrm{3}} :{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} ={a}^{\mathrm{2}} ,{z}\geq\mathrm{0}\right\} \\ $$$$\sum_{\mathrm{3}} =\left\{\left({x},{y},{z}\right)\in\:\mathbb{R}^{\mathrm{3}} :{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} ={a}^{\mathrm{2}} ,{z}\leq\mathrm{0}\right\} \\ $$$$\mathrm{The}\:\mathrm{surface}\:\mathrm{area}\:\mathrm{of}\:\sum_{\mathrm{1}} \mathrm{is}\:\int\int_{\mathrm{2}} {ds}=\mathrm{2}\pi{b}\centerdot\mathrm{2}\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }=\mathrm{4}\pi{b}\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} } \\ $$$$\mathrm{The}\:\mathrm{surface}\:\mathrm{areas}\:\mathrm{of}\:\sum_{\mathrm{2}} \mathrm{and}\sum_{\mathrm{3}} \mathrm{are}\:\mathrm{each}\:\mathrm{half} \\ $$$$\mathrm{the}\:\mathrm{suface}\:\mathrm{area}\:\mathrm{of}\:\mathrm{a}\:\mathrm{sphere}\:\mathrm{with}\:\mathrm{radius}\:{a},\mathrm{i}.\mathrm{e}., \\ $$$$\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}\centerdot\mathrm{4}\pi{a}^{\mathrm{2}} =\mathrm{2}\pi{a}^{\mathrm{2}} \\ $$$$\mathrm{Therefore},\mathrm{the}\:\mathrm{total}\:\mathrm{surface}\:\mathrm{area}\:\mathrm{of}\:\Sigma\mathrm{is} \\ $$$$\:\:\:\:\:\:\:\int\int_{\Sigma} {ds}=\int\int_{\sum_{\mathrm{1}} } {ds}+\int\int_{\sum_{\mathrm{2}} } {ds}+\int\int_{\sum_{\mathrm{3}} } {ds} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{4}\pi{b}\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }+\mathrm{2}\pi{a}^{\mathrm{2}} +\mathrm{2}\pi{a}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{4}\pi{b}\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }+\mathrm{4}\pi{a}^{\mathrm{2}} \\ $$
Answered by mr W last updated on 04/Jan/25
Commented by mr W last updated on 04/Jan/25
the solid D is a spherical ring (also called napkin ring).
$${the}\:{solid}\:{D}\:{is}\:{a}\:{spherical}\:{ring} \\ $$$$\left({also}\:{called}\:{napkin}\:{ring}\right). \\ $$
Commented by mr W last updated on 04/Jan/25
R=a r=b L=2(√(R^2 −r^2 )) S=4πR^2 +2πrL−2×2πR(R−(√(R^2 −r^2 ))) =4πr(√(R^2 −r^2 ))+4πR(√(R^2 −r^2 )) =4π(R+r)(√(R^2 −r^2 )) =4π(a+b)(√(a^2 −b^2 )) check: b=0 ⇒S=4πa^2 ok ✓ b=a ⇒S=0 ok ✓
$${R}={a} \\ $$$${r}={b} \\ $$$${L}=\mathrm{2}\sqrt{{R}^{\mathrm{2}} −{r}^{\mathrm{2}} } \\ $$$${S}=\mathrm{4}\pi{R}^{\mathrm{2}} +\mathrm{2}\pi{rL}−\mathrm{2}×\mathrm{2}\pi{R}\left({R}−\sqrt{{R}^{\mathrm{2}} −{r}^{\mathrm{2}} }\right) \\ $$$$\:\:\:=\mathrm{4}\pi{r}\sqrt{{R}^{\mathrm{2}} −{r}^{\mathrm{2}} }+\mathrm{4}\pi{R}\sqrt{{R}^{\mathrm{2}} −{r}^{\mathrm{2}} } \\ $$$$\:\:\:=\mathrm{4}\pi\left({R}+{r}\right)\sqrt{{R}^{\mathrm{2}} −{r}^{\mathrm{2}} } \\ $$$$\:\:\:=\mathrm{4}\pi\left({a}+{b}\right)\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} } \\ $$$${check}: \\ $$$${b}=\mathrm{0}\:\Rightarrow{S}=\mathrm{4}\pi{a}^{\mathrm{2}} \:{ok}\:\checkmark \\ $$$${b}={a}\:\Rightarrow{S}=\mathrm{0}\:{ok}\:\checkmark \\ $$

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