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Question Number 82591 by Learner-123 last updated on 22/Feb/20

A closed surface is defined in spherical coordinates by 3<r<5 , 0.1π<θ<0.3π, 1.2π<φ<1.6π. Find the total surface area.

$${A}\:{closed}\:{surface}\:{is}\:{defined}\:{in}\:{spherical} \\ $$ $${coordinates}\:{by}\:\mathrm{3}<{r}<\mathrm{5}\:,\:\mathrm{0}.\mathrm{1}\pi<\theta<\mathrm{0}.\mathrm{3}\pi, \\ $$ $$\mathrm{1}.\mathrm{2}\pi<\phi<\mathrm{1}.\mathrm{6}\pi.\:\boldsymbol{{F}}{ind}\:{the}\:{total}\:{surface} \\ $$ $${area}. \\ $$

Commented byLearner-123 last updated on 22/Feb/20

Ans:36.79 square units.

$${Ans}:\mathrm{36}.\mathrm{79}\:{square}\:{units}. \\ $$

Commented byLearner-123 last updated on 23/Feb/20

Commented byLearner-123 last updated on 23/Feb/20

Sir, this is the explaination given. Can you tell how these 4 (actually 5)terms come..

$${Sir},\:{this}\:{is}\:{the}\:{explaination}\:{given}. \\ $$ $${Can}\:{you}\:{tell}\:{how}\:{these}\:\mathrm{4}\:\left({actually}\:\mathrm{5}\right){terms} \\ $$ $${come}.. \\ $$

Commented bymr W last updated on 23/Feb/20

Commented bymr W last updated on 23/Feb/20

in our case φ=0.1π..0.3π θ=1.2π..1.6π r=3..5 we have totally 6 faces: face 1: r=3, φ=0.1π...0.3π, θ=1.2π..1.6π A_1 =∫_(1.2π) ^(1.6π) ∫_(0.1π) ^(0.3π) r sin θ dφrdθ=3^2 ∫_(1.2π) ^(1.6π) ∫_(0.1π) ^(0.3π) sin θ dφdθ face 2: r=5, φ=0.1π...0.3π, θ=1.2π..1.6π A_2 =similarly=5^2 ∫_(1.2π) ^(1.6π) ∫_(0.1π) ^(0.3π) sin θ dφdθ face 3: φ=0.1π, θ=1.2π..1.6π, r=3..5 A_3 =(1/2)(5^2 −3^2 )(1.6π−1.2π) face 4: φ=0.3π, θ=1.2π..1.6π, r=3..5 A_4 =(1/2)(5^2 −3^2 )(1.6π−1.2π) face 5: φ=0.1π..0.3π, θ=1.2π, r=3..5 A_5 =∫_(0.1π) ^(0.3π) ∫_3 ^5 sin 1.2π dφrdr face 6: φ=0.1π..0.3π, θ=1.6π, r=3..5 A_6 =∫_(0.1π) ^(0.3π) ∫_3 ^5 sin 1.6π dφrdr A=ΣA =(3^2 +5^2 )∫_(1.2π) ^(1.6π) ∫_(0.1π) ^(0.3π) sin θ dφdθ +2×(1/2)(5^2 −3^2 )(1.6π−1.2π) +(sin 1.2π+sin 1.6π)(0.3π−0.1π)∫_3 ^5 rdφdr

$${in}\:{our}\:{case} \\ $$ $$\phi=\mathrm{0}.\mathrm{1}\pi..\mathrm{0}.\mathrm{3}\pi \\ $$ $$\theta=\mathrm{1}.\mathrm{2}\pi..\mathrm{1}.\mathrm{6}\pi \\ $$ $${r}=\mathrm{3}..\mathrm{5} \\ $$ $${we}\:{have}\:{totally}\:\mathrm{6}\:{faces}: \\ $$ $${face}\:\mathrm{1}:\:{r}=\mathrm{3},\:\phi=\mathrm{0}.\mathrm{1}\pi...\mathrm{0}.\mathrm{3}\pi,\:\theta=\mathrm{1}.\mathrm{2}\pi..\mathrm{1}.\mathrm{6}\pi \\ $$ $${A}_{\mathrm{1}} =\int_{\mathrm{1}.\mathrm{2}\pi} ^{\mathrm{1}.\mathrm{6}\pi} \int_{\mathrm{0}.\mathrm{1}\pi} ^{\mathrm{0}.\mathrm{3}\pi} {r}\:\mathrm{sin}\:\theta\:{d}\phi{rd}\theta=\mathrm{3}^{\mathrm{2}} \int_{\mathrm{1}.\mathrm{2}\pi} ^{\mathrm{1}.\mathrm{6}\pi} \int_{\mathrm{0}.\mathrm{1}\pi} ^{\mathrm{0}.\mathrm{3}\pi} \mathrm{sin}\:\theta\:{d}\phi{d}\theta \\ $$ $${face}\:\mathrm{2}:\:{r}=\mathrm{5},\:\phi=\mathrm{0}.\mathrm{1}\pi...\mathrm{0}.\mathrm{3}\pi,\:\theta=\mathrm{1}.\mathrm{2}\pi..\mathrm{1}.\mathrm{6}\pi \\ $$ $${A}_{\mathrm{2}} ={similarly}=\mathrm{5}^{\mathrm{2}} \int_{\mathrm{1}.\mathrm{2}\pi} ^{\mathrm{1}.\mathrm{6}\pi} \int_{\mathrm{0}.\mathrm{1}\pi} ^{\mathrm{0}.\mathrm{3}\pi} \mathrm{sin}\:\theta\:{d}\phi{d}\theta \\ $$ $${face}\:\mathrm{3}:\:\phi=\mathrm{0}.\mathrm{1}\pi,\:\theta=\mathrm{1}.\mathrm{2}\pi..\mathrm{1}.\mathrm{6}\pi,\:{r}=\mathrm{3}..\mathrm{5} \\ $$ $${A}_{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{5}^{\mathrm{2}} −\mathrm{3}^{\mathrm{2}} \right)\left(\mathrm{1}.\mathrm{6}\pi−\mathrm{1}.\mathrm{2}\pi\right) \\ $$ $${face}\:\mathrm{4}:\:\phi=\mathrm{0}.\mathrm{3}\pi,\:\theta=\mathrm{1}.\mathrm{2}\pi..\mathrm{1}.\mathrm{6}\pi,\:{r}=\mathrm{3}..\mathrm{5} \\ $$ $${A}_{\mathrm{4}} =\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{5}^{\mathrm{2}} −\mathrm{3}^{\mathrm{2}} \right)\left(\mathrm{1}.\mathrm{6}\pi−\mathrm{1}.\mathrm{2}\pi\right) \\ $$ $${face}\:\mathrm{5}:\:\phi=\mathrm{0}.\mathrm{1}\pi..\mathrm{0}.\mathrm{3}\pi,\:\theta=\mathrm{1}.\mathrm{2}\pi,\:{r}=\mathrm{3}..\mathrm{5} \\ $$ $${A}_{\mathrm{5}} =\int_{\mathrm{0}.\mathrm{1}\pi} ^{\mathrm{0}.\mathrm{3}\pi} \int_{\mathrm{3}} ^{\mathrm{5}} \mathrm{sin}\:\mathrm{1}.\mathrm{2}\pi\:{d}\phi{rdr} \\ $$ $${face}\:\mathrm{6}:\:\phi=\mathrm{0}.\mathrm{1}\pi..\mathrm{0}.\mathrm{3}\pi,\:\theta=\mathrm{1}.\mathrm{6}\pi,\:{r}=\mathrm{3}..\mathrm{5} \\ $$ $${A}_{\mathrm{6}} =\int_{\mathrm{0}.\mathrm{1}\pi} ^{\mathrm{0}.\mathrm{3}\pi} \int_{\mathrm{3}} ^{\mathrm{5}} \mathrm{sin}\:\mathrm{1}.\mathrm{6}\pi\:{d}\phi{rdr} \\ $$ $${A}=\Sigma{A} \\ $$ $$=\left(\mathrm{3}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} \right)\int_{\mathrm{1}.\mathrm{2}\pi} ^{\mathrm{1}.\mathrm{6}\pi} \int_{\mathrm{0}.\mathrm{1}\pi} ^{\mathrm{0}.\mathrm{3}\pi} \mathrm{sin}\:\theta\:{d}\phi{d}\theta \\ $$ $$+\mathrm{2}×\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{5}^{\mathrm{2}} −\mathrm{3}^{\mathrm{2}} \right)\left(\mathrm{1}.\mathrm{6}\pi−\mathrm{1}.\mathrm{2}\pi\right) \\ $$ $$+\left(\mathrm{sin}\:\mathrm{1}.\mathrm{2}\pi+\mathrm{sin}\:\mathrm{1}.\mathrm{6}\pi\right)\left(\mathrm{0}.\mathrm{3}\pi−\mathrm{0}.\mathrm{1}\pi\right)\int_{\mathrm{3}} ^{\mathrm{5}} {rd}\phi{dr} \\ $$

Commented byLearner-123 last updated on 24/Feb/20

Thank You Sir! Can you please solve my other 3 doubts?

$${Thank}\:{You}\:{Sir}! \\ $$ $${Can}\:{you}\:{please}\:{solve}\:{my}\:{other}\:\mathrm{3}\:{doubts}? \\ $$

Commented bymr W last updated on 24/Feb/20

i don′t know what are your doubts. you should not assume that i know all posts.

$${i}\:{don}'{t}\:{know}\:{what}\:{are}\:{your}\:{doubts}.\:{you} \\ $$ $${should}\:{not}\:{assume}\:{that}\:{i}\:{know}\:{all}\:{posts}. \\ $$

Commented byLearner-123 last updated on 24/Feb/20

Question no: 82627 , 82631,82647

$${Question}\:{no}:\:\mathrm{82627}\:,\:\mathrm{82631},\mathrm{82647} \\ $$

Commented byLearner-123 last updated on 24/Feb/20

I think there is some technical problem in this app, whenever i press ♮sort by recent activityε it gets me to same integral question everytime...(75986)

$${I}\:{think}\:{there}\:{is}\:{some}\:{technical}\:{problem} \\ $$ $${in}\:{this}\:{app},\:{whenever}\:{i}\:{press}\:\natural\boldsymbol{{s}}{ort}\:{by} \\ $$ $${recent}\:{activity}\varepsilon\:{it}\:{gets}\:{me}\:{to}\:{same} \\ $$ $${integral}\:{question}\:{everytime}...\left(\mathrm{75986}\right) \\ $$

Commented bymr W last updated on 24/Feb/20

i can′t help you.

$${i}\:{can}'{t}\:{help}\:{you}. \\ $$

Commented bymr W last updated on 26/Feb/20

both

$${both} \\ $$

Commented byLearner-123 last updated on 26/Feb/20

Are you referring to my doubt or this technical glitch?

$${Are}\:{you}\:{referring}\:{to}\:{my}\:{doubt}\:{or} \\ $$ $${this}\:{technical}\:{glitch}? \\ $$

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