Volume integral sphere [closed]

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Compute $$iiint_E frac ,dx ,dy ,dzsqrtx^2+ y^2 +(z-2)^2$$



where $E$, i.e. the domain of integration, is specified by $$x^2+y^2+z^2 le 1$$




I tried using spherical coordinates



But i end up getting an integral solving which is an onerous task



Can anyone suggest an alternatively easier method ?




multivariable-calculus surface-integrals




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closed as off-topic by StubbornAtom, Strants, Xander Henderson, José Carlos Santos, user223391 Jul 20 at 1:33


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – StubbornAtom, Strants, Xander Henderson, José Carlos Santos, Community
If this question can be reworded to fit the rules in the help center, please edit the question.












  • There isn't any $E$ in the statement. And your inequality gives a ball, not a sphere.
    – Gerry Myerson
    Jul 19 at 7:18










  • Yeah i just realised that .... E is the domain
    – Speedy
    Jul 19 at 7:21










  • E is the domain of integration
    – Speedy
    Jul 19 at 7:21














up vote
-2
down vote

favorite













Compute $$iiint_E frac ,dx ,dy ,dzsqrtx^2+ y^2 +(z-2)^2$$



where $E$, i.e. the domain of integration, is specified by $$x^2+y^2+z^2 le 1$$




I tried using spherical coordinates



But i end up getting an integral solving which is an onerous task



Can anyone suggest an alternatively easier method ?




multivariable-calculus surface-integrals




share|cite|improve this question













closed as off-topic by StubbornAtom, Strants, Xander Henderson, José Carlos Santos, user223391 Jul 20 at 1:33


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – StubbornAtom, Strants, Xander Henderson, José Carlos Santos, Community
If this question can be reworded to fit the rules in the help center, please edit the question.












  • There isn't any $E$ in the statement. And your inequality gives a ball, not a sphere.
    – Gerry Myerson
    Jul 19 at 7:18










  • Yeah i just realised that .... E is the domain
    – Speedy
    Jul 19 at 7:21










  • E is the domain of integration
    – Speedy
    Jul 19 at 7:21












up vote
-2
down vote

favorite









up vote
-2
down vote

favorite












Compute $$iiint_E frac ,dx ,dy ,dzsqrtx^2+ y^2 +(z-2)^2$$



where $E$, i.e. the domain of integration, is specified by $$x^2+y^2+z^2 le 1$$




I tried using spherical coordinates



But i end up getting an integral solving which is an onerous task



Can anyone suggest an alternatively easier method ?




multivariable-calculus surface-integrals




share|cite|improve this question














Compute $$iiint_E frac ,dx ,dy ,dzsqrtx^2+ y^2 +(z-2)^2$$



where $E$, i.e. the domain of integration, is specified by $$x^2+y^2+z^2 le 1$$




I tried using spherical coordinates



But i end up getting an integral solving which is an onerous task



Can anyone suggest an alternatively easier method ?





multivariable-calculus surface-integrals





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 19 at 8:23









StubbornAtom

3,78111134




3,78111134









asked Jul 19 at 7:13









Speedy

65




65




closed as off-topic by StubbornAtom, Strants, Xander Henderson, José Carlos Santos, user223391 Jul 20 at 1:33


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – StubbornAtom, Strants, Xander Henderson, José Carlos Santos, Community
If this question can be reworded to fit the rules in the help center, please edit the question.




closed as off-topic by StubbornAtom, Strants, Xander Henderson, José Carlos Santos, user223391 Jul 20 at 1:33


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – StubbornAtom, Strants, Xander Henderson, José Carlos Santos, Community
If this question can be reworded to fit the rules in the help center, please edit the question.











  • There isn't any $E$ in the statement. And your inequality gives a ball, not a sphere.
    – Gerry Myerson
    Jul 19 at 7:18










  • Yeah i just realised that .... E is the domain
    – Speedy
    Jul 19 at 7:21










  • E is the domain of integration
    – Speedy
    Jul 19 at 7:21
















  • There isn't any $E$ in the statement. And your inequality gives a ball, not a sphere.
    – Gerry Myerson
    Jul 19 at 7:18










  • Yeah i just realised that .... E is the domain
    – Speedy
    Jul 19 at 7:21










  • E is the domain of integration
    – Speedy
    Jul 19 at 7:21















There isn't any $E$ in the statement. And your inequality gives a ball, not a sphere.
– Gerry Myerson
Jul 19 at 7:18




There isn't any $E$ in the statement. And your inequality gives a ball, not a sphere.
– Gerry Myerson
Jul 19 at 7:18












Yeah i just realised that .... E is the domain
– Speedy
Jul 19 at 7:21




Yeah i just realised that .... E is the domain
– Speedy
Jul 19 at 7:21












E is the domain of integration
– Speedy
Jul 19 at 7:21




E is the domain of integration
– Speedy
Jul 19 at 7:21










1 Answer
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$$∫∫∫ dfracdxdydzsqrtx^2 + y^2 + (z-2)^2_-1^1dR=2piint_0^1R^2dR\=dfrac2pi3$$






share|cite|improve this answer




























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    0
    down vote



    accepted










    $$∫∫∫ dfracdxdydzsqrtx^2 + y^2 + (z-2)^2_-1^1dR=2piint_0^1R^2dR\=dfrac2pi3$$






    share|cite|improve this answer

























      up vote
      0
      down vote



      accepted










      $$∫∫∫ dfracdxdydzsqrtx^2 + y^2 + (z-2)^2_-1^1dR=2piint_0^1R^2dR\=dfrac2pi3$$






      share|cite|improve this answer























        up vote
        0
        down vote



        accepted







        up vote
        0
        down vote



        accepted






        $$∫∫∫ dfracdxdydzsqrtx^2 + y^2 + (z-2)^2_-1^1dR=2piint_0^1R^2dR\=dfrac2pi3$$






        share|cite|improve this answer













        $$∫∫∫ dfracdxdydzsqrtx^2 + y^2 + (z-2)^2_-1^1dR=2piint_0^1R^2dR\=dfrac2pi3$$







        share|cite|improve this answer













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        share|cite|improve this answer











        answered Jul 19 at 8:19









        Mostafa Ayaz

        8,6023630




        8,6023630












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