Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

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 ✓

$${R}={a} \\ $$$${r}={b} \\ $$$${L}=\mathrm{2}\sqrt{{R}^{\mathrm{2}} −{r}^{\mathrm{2}} } \\ $$$${S}=\mathrm{4}\pi{R}^{\mathrm{2}} +\mathrm{2}\pi{rL}−\mathrm{2}×\mathrm{2}\pi{R}\left({R}−\sqrt{{R}^{\mathrm{2}} −{r}^{\mathrm{2}} }\right) \\ $$$$\:\:\:=\mathrm{4}\pi{r}\sqrt{{R}^{\mathrm{2}} −{r}^{\mathrm{2}} }+\mathrm{4}\pi{R}\sqrt{{R}^{\mathrm{2}} −{r}^{\mathrm{2}} } \\ $$$$\:\:\:=\mathrm{4}\pi\left({R}+{r}\right)\sqrt{{R}^{\mathrm{2}} −{r}^{\mathrm{2}} } \\ $$$$\:\:\:=\mathrm{4}\pi\left({a}+{b}\right)\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} } \\ $$$${check}: \\ $$$${b}=\mathrm{0}\:\Rightarrow{S}=\mathrm{4}\pi{a}^{\mathrm{2}} \:{ok}\:\checkmark \\ $$$${b}={a}\:\Rightarrow{S}=\mathrm{0}\:{ok}\:\checkmark \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com