evaluating integral over surface of the sphere

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This is what I've tried. Am I approaching this the right way? Also, where do I go next?



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    Looks good so far. The inner antiderivative suggests $u=cos theta$ and that's good since then $du=-sin theta$ and you have the $sin theta$ where it's needed.
    – coffeemath
    Feb 15 '14 at 0:40










  • Há um erro: exp(a*cos(x)) != exp(a)*exp(cos(x))
    – Rômulo Silva
    Aug 27 at 17:06














up vote
3
down vote

favorite
2












enter image description here



This is what I've tried. Am I approaching this the right way? Also, where do I go next?



enter image description here







share|cite|improve this question
















  • 1




    Looks good so far. The inner antiderivative suggests $u=cos theta$ and that's good since then $du=-sin theta$ and you have the $sin theta$ where it's needed.
    – coffeemath
    Feb 15 '14 at 0:40










  • Há um erro: exp(a*cos(x)) != exp(a)*exp(cos(x))
    – Rômulo Silva
    Aug 27 at 17:06












up vote
3
down vote

favorite
2









up vote
3
down vote

favorite
2






2





enter image description here



This is what I've tried. Am I approaching this the right way? Also, where do I go next?



enter image description here







share|cite|improve this question












enter image description here



This is what I've tried. Am I approaching this the right way? Also, where do I go next?



enter image description here









share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Feb 14 '14 at 22:48









user120227

510517




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  • 1




    Looks good so far. The inner antiderivative suggests $u=cos theta$ and that's good since then $du=-sin theta$ and you have the $sin theta$ where it's needed.
    – coffeemath
    Feb 15 '14 at 0:40










  • Há um erro: exp(a*cos(x)) != exp(a)*exp(cos(x))
    – Rômulo Silva
    Aug 27 at 17:06












  • 1




    Looks good so far. The inner antiderivative suggests $u=cos theta$ and that's good since then $du=-sin theta$ and you have the $sin theta$ where it's needed.
    – coffeemath
    Feb 15 '14 at 0:40










  • Há um erro: exp(a*cos(x)) != exp(a)*exp(cos(x))
    – Rômulo Silva
    Aug 27 at 17:06







1




1




Looks good so far. The inner antiderivative suggests $u=cos theta$ and that's good since then $du=-sin theta$ and you have the $sin theta$ where it's needed.
– coffeemath
Feb 15 '14 at 0:40




Looks good so far. The inner antiderivative suggests $u=cos theta$ and that's good since then $du=-sin theta$ and you have the $sin theta$ where it's needed.
– coffeemath
Feb 15 '14 at 0:40












Há um erro: exp(a*cos(x)) != exp(a)*exp(cos(x))
– Rômulo Silva
Aug 27 at 17:06




Há um erro: exp(a*cos(x)) != exp(a)*exp(cos(x))
– Rômulo Silva
Aug 27 at 17:06










1 Answer
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the next steps are the following:
$$beginmatrix
iint_Se^z,dS&=&int_0^2piint_0^pie^a,costhetaa^2sintheta,dtheta dphi\
&=&a^2e^aint_0^2piint_0^pi e^costhetasintheta,dtheta dphi \
&=&2pi a^2e^aint_0^pi d(-e^costheta)\
&=&2pi a^2e^aleft(e-frac1eright)
endmatrix$$






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    1 Answer
    1






    active

    oldest

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    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

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    active

    oldest

    votes








    up vote
    1
    down vote



    accepted










    the next steps are the following:
    $$beginmatrix
    iint_Se^z,dS&=&int_0^2piint_0^pie^a,costhetaa^2sintheta,dtheta dphi\
    &=&a^2e^aint_0^2piint_0^pi e^costhetasintheta,dtheta dphi \
    &=&2pi a^2e^aint_0^pi d(-e^costheta)\
    &=&2pi a^2e^aleft(e-frac1eright)
    endmatrix$$






    share|cite|improve this answer
























      up vote
      1
      down vote



      accepted










      the next steps are the following:
      $$beginmatrix
      iint_Se^z,dS&=&int_0^2piint_0^pie^a,costhetaa^2sintheta,dtheta dphi\
      &=&a^2e^aint_0^2piint_0^pi e^costhetasintheta,dtheta dphi \
      &=&2pi a^2e^aint_0^pi d(-e^costheta)\
      &=&2pi a^2e^aleft(e-frac1eright)
      endmatrix$$






      share|cite|improve this answer






















        up vote
        1
        down vote



        accepted







        up vote
        1
        down vote



        accepted






        the next steps are the following:
        $$beginmatrix
        iint_Se^z,dS&=&int_0^2piint_0^pie^a,costhetaa^2sintheta,dtheta dphi\
        &=&a^2e^aint_0^2piint_0^pi e^costhetasintheta,dtheta dphi \
        &=&2pi a^2e^aint_0^pi d(-e^costheta)\
        &=&2pi a^2e^aleft(e-frac1eright)
        endmatrix$$






        share|cite|improve this answer












        the next steps are the following:
        $$beginmatrix
        iint_Se^z,dS&=&int_0^2piint_0^pie^a,costhetaa^2sintheta,dtheta dphi\
        &=&a^2e^aint_0^2piint_0^pi e^costhetasintheta,dtheta dphi \
        &=&2pi a^2e^aint_0^pi d(-e^costheta)\
        &=&2pi a^2e^aleft(e-frac1eright)
        endmatrix$$







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        share|cite|improve this answer



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        answered Feb 15 '14 at 18:51









        DiegoMath

        1,9331021




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