Integration over the ellipsoid $M:={(x,y,z)mid x^2+frac{y^2}{4}+frac{z^2}{9}=1}$ using the divergence theorem...












0












$begingroup$


I have troubles with the following question:



Given is the following vector field :
$$f(x,y,z)= begin{pmatrix}x+sin(y)cos(z)\y+e^{x^2+y^2}\-z + log(1+x^2+y^2)end{pmatrix}$$



and the set $M$ is defined by
$$M:={(x,y,z)mid x^2+frac{y^2}{4}+frac{z^2}{9}=1}$$



Calculate the following integral :
$$int_Mlangle f, v rangle, dS$$
where $v$ is the unit normal field (german : Einheitsnormalenfeld) toward the exterior.





So I thought about using the divergence theorem as usual given that $M$ is just the border of an ellipsoid. I get the triple integral over $1+2ye^{x^2+y^2}$ . The 3 integrals over 1 give us the volume of a normal ellipsoid, in this case $8pi$, but how do I integrate $iiint_{text{Ellipsoid}} 2ye^{x^2+y^2} dx,dy,dz$ ?

Thanks for your help.






real-analysis integration







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    0












    $begingroup$


    I have troubles with the following question:



    Given is the following vector field :
    $$f(x,y,z)= begin{pmatrix}x+sin(y)cos(z)\y+e^{x^2+y^2}\-z + log(1+x^2+y^2)end{pmatrix}$$



    and the set $M$ is defined by
    $$M:={(x,y,z)mid x^2+frac{y^2}{4}+frac{z^2}{9}=1}$$



    Calculate the following integral :
    $$int_Mlangle f, v rangle, dS$$
    where $v$ is the unit normal field (german : Einheitsnormalenfeld) toward the exterior.





    So I thought about using the divergence theorem as usual given that $M$ is just the border of an ellipsoid. I get the triple integral over $1+2ye^{x^2+y^2}$ . The 3 integrals over 1 give us the volume of a normal ellipsoid, in this case $8pi$, but how do I integrate $iiint_{text{Ellipsoid}} 2ye^{x^2+y^2} dx,dy,dz$ ?

    Thanks for your help.






    real-analysis integration







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      I have troubles with the following question:



      Given is the following vector field :
      $$f(x,y,z)= begin{pmatrix}x+sin(y)cos(z)\y+e^{x^2+y^2}\-z + log(1+x^2+y^2)end{pmatrix}$$



      and the set $M$ is defined by
      $$M:={(x,y,z)mid x^2+frac{y^2}{4}+frac{z^2}{9}=1}$$



      Calculate the following integral :
      $$int_Mlangle f, v rangle, dS$$
      where $v$ is the unit normal field (german : Einheitsnormalenfeld) toward the exterior.





      So I thought about using the divergence theorem as usual given that $M$ is just the border of an ellipsoid. I get the triple integral over $1+2ye^{x^2+y^2}$ . The 3 integrals over 1 give us the volume of a normal ellipsoid, in this case $8pi$, but how do I integrate $iiint_{text{Ellipsoid}} 2ye^{x^2+y^2} dx,dy,dz$ ?

      Thanks for your help.






      real-analysis integration







      share|cite|improve this question











      $endgroup$




      I have troubles with the following question:



      Given is the following vector field :
      $$f(x,y,z)= begin{pmatrix}x+sin(y)cos(z)\y+e^{x^2+y^2}\-z + log(1+x^2+y^2)end{pmatrix}$$



      and the set $M$ is defined by
      $$M:={(x,y,z)mid x^2+frac{y^2}{4}+frac{z^2}{9}=1}$$



      Calculate the following integral :
      $$int_Mlangle f, v rangle, dS$$
      where $v$ is the unit normal field (german : Einheitsnormalenfeld) toward the exterior.





      So I thought about using the divergence theorem as usual given that $M$ is just the border of an ellipsoid. I get the triple integral over $1+2ye^{x^2+y^2}$ . The 3 integrals over 1 give us the volume of a normal ellipsoid, in this case $8pi$, but how do I integrate $iiint_{text{Ellipsoid}} 2ye^{x^2+y^2} dx,dy,dz$ ?

      Thanks for your help.






      real-analysis integration




      real-analysis integration






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 7 at 19:33







      Poujh

















      asked Jul 18 '18 at 19:18









      PoujhPoujh

      602516




      602516






















          1 Answer
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          $begingroup$

          So you are asking how to set up a triple integral?



          The ellipsoid $x^2+ frac{y^2}{4}+ frac{z^2}{9}= 1$ can be covered by taking $x$ from $-1$ to $1$; $y$, for each $x$, from $-2sqrt{1-x^2}$ to $2sqrt{1-x^2}$; and $z$, for each $x$ and $y$, from $-3sqrt{1-x^2-frac{y^2}{4}}$ to $3sqrt{1-x^2-frac{y^2}{4}}$.



          So the integral will be $$int_{x=-1}^1int_{y=-2sqrt{1- x^2}}^{2sqrt{1- x^2}}int_{z=-3sqrt{1-x^2-frac{y^2}{4}}}^{3sqrt{1-x^2-frac{y^2}{4}}}left(1+ 2ye^{x^2+y^2}right)dz,dy,dx$$



          Since the integrand is independent of $z$, the first integration gives
          $$-6int_{x=-1}^1int_{y=-2sqrt{1- x^2}}^{2sqrt{1- x^2}}sqrt{1- x^2}left(1+ 2ye^{x^2+y^2}right)dy,dx$$






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            Thanks for your help, but isn't that really "complicated" ? I mean, the integral over 1 is easy, and the divergence of the vector field is also a lot easier than in "first" appearence, so do we really need to integrate $2ye^{x^2+y^2}$ in that way ? Isn't there a trick, or couldn't we for example use cylindrical/ spherical coordinates to make things easier ? I found this question in an old exam, and given that time in exam is often limited, the questions are generally not "that" hard, so I'm just a bit surprised.
            $endgroup$
            – Poujh
            Jul 18 '18 at 19:42








          • 1




            $begingroup$
            The first integral is an integral of an odd function over a symmetric interval, thus the entire calculation yields zero.
            $endgroup$
            – zokomoko
            Jul 18 '18 at 20:07













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          1 Answer
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          active

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          2












          $begingroup$

          So you are asking how to set up a triple integral?



          The ellipsoid $x^2+ frac{y^2}{4}+ frac{z^2}{9}= 1$ can be covered by taking $x$ from $-1$ to $1$; $y$, for each $x$, from $-2sqrt{1-x^2}$ to $2sqrt{1-x^2}$; and $z$, for each $x$ and $y$, from $-3sqrt{1-x^2-frac{y^2}{4}}$ to $3sqrt{1-x^2-frac{y^2}{4}}$.



          So the integral will be $$int_{x=-1}^1int_{y=-2sqrt{1- x^2}}^{2sqrt{1- x^2}}int_{z=-3sqrt{1-x^2-frac{y^2}{4}}}^{3sqrt{1-x^2-frac{y^2}{4}}}left(1+ 2ye^{x^2+y^2}right)dz,dy,dx$$



          Since the integrand is independent of $z$, the first integration gives
          $$-6int_{x=-1}^1int_{y=-2sqrt{1- x^2}}^{2sqrt{1- x^2}}sqrt{1- x^2}left(1+ 2ye^{x^2+y^2}right)dy,dx$$






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            Thanks for your help, but isn't that really "complicated" ? I mean, the integral over 1 is easy, and the divergence of the vector field is also a lot easier than in "first" appearence, so do we really need to integrate $2ye^{x^2+y^2}$ in that way ? Isn't there a trick, or couldn't we for example use cylindrical/ spherical coordinates to make things easier ? I found this question in an old exam, and given that time in exam is often limited, the questions are generally not "that" hard, so I'm just a bit surprised.
            $endgroup$
            – Poujh
            Jul 18 '18 at 19:42








          • 1




            $begingroup$
            The first integral is an integral of an odd function over a symmetric interval, thus the entire calculation yields zero.
            $endgroup$
            – zokomoko
            Jul 18 '18 at 20:07


















          2












          $begingroup$

          So you are asking how to set up a triple integral?



          The ellipsoid $x^2+ frac{y^2}{4}+ frac{z^2}{9}= 1$ can be covered by taking $x$ from $-1$ to $1$; $y$, for each $x$, from $-2sqrt{1-x^2}$ to $2sqrt{1-x^2}$; and $z$, for each $x$ and $y$, from $-3sqrt{1-x^2-frac{y^2}{4}}$ to $3sqrt{1-x^2-frac{y^2}{4}}$.



          So the integral will be $$int_{x=-1}^1int_{y=-2sqrt{1- x^2}}^{2sqrt{1- x^2}}int_{z=-3sqrt{1-x^2-frac{y^2}{4}}}^{3sqrt{1-x^2-frac{y^2}{4}}}left(1+ 2ye^{x^2+y^2}right)dz,dy,dx$$



          Since the integrand is independent of $z$, the first integration gives
          $$-6int_{x=-1}^1int_{y=-2sqrt{1- x^2}}^{2sqrt{1- x^2}}sqrt{1- x^2}left(1+ 2ye^{x^2+y^2}right)dy,dx$$






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            Thanks for your help, but isn't that really "complicated" ? I mean, the integral over 1 is easy, and the divergence of the vector field is also a lot easier than in "first" appearence, so do we really need to integrate $2ye^{x^2+y^2}$ in that way ? Isn't there a trick, or couldn't we for example use cylindrical/ spherical coordinates to make things easier ? I found this question in an old exam, and given that time in exam is often limited, the questions are generally not "that" hard, so I'm just a bit surprised.
            $endgroup$
            – Poujh
            Jul 18 '18 at 19:42








          • 1




            $begingroup$
            The first integral is an integral of an odd function over a symmetric interval, thus the entire calculation yields zero.
            $endgroup$
            – zokomoko
            Jul 18 '18 at 20:07
















          2












          2








          2





          $begingroup$

          So you are asking how to set up a triple integral?



          The ellipsoid $x^2+ frac{y^2}{4}+ frac{z^2}{9}= 1$ can be covered by taking $x$ from $-1$ to $1$; $y$, for each $x$, from $-2sqrt{1-x^2}$ to $2sqrt{1-x^2}$; and $z$, for each $x$ and $y$, from $-3sqrt{1-x^2-frac{y^2}{4}}$ to $3sqrt{1-x^2-frac{y^2}{4}}$.



          So the integral will be $$int_{x=-1}^1int_{y=-2sqrt{1- x^2}}^{2sqrt{1- x^2}}int_{z=-3sqrt{1-x^2-frac{y^2}{4}}}^{3sqrt{1-x^2-frac{y^2}{4}}}left(1+ 2ye^{x^2+y^2}right)dz,dy,dx$$



          Since the integrand is independent of $z$, the first integration gives
          $$-6int_{x=-1}^1int_{y=-2sqrt{1- x^2}}^{2sqrt{1- x^2}}sqrt{1- x^2}left(1+ 2ye^{x^2+y^2}right)dy,dx$$






          share|cite|improve this answer











          $endgroup$



          So you are asking how to set up a triple integral?



          The ellipsoid $x^2+ frac{y^2}{4}+ frac{z^2}{9}= 1$ can be covered by taking $x$ from $-1$ to $1$; $y$, for each $x$, from $-2sqrt{1-x^2}$ to $2sqrt{1-x^2}$; and $z$, for each $x$ and $y$, from $-3sqrt{1-x^2-frac{y^2}{4}}$ to $3sqrt{1-x^2-frac{y^2}{4}}$.



          So the integral will be $$int_{x=-1}^1int_{y=-2sqrt{1- x^2}}^{2sqrt{1- x^2}}int_{z=-3sqrt{1-x^2-frac{y^2}{4}}}^{3sqrt{1-x^2-frac{y^2}{4}}}left(1+ 2ye^{x^2+y^2}right)dz,dy,dx$$



          Since the integrand is independent of $z$, the first integration gives
          $$-6int_{x=-1}^1int_{y=-2sqrt{1- x^2}}^{2sqrt{1- x^2}}sqrt{1- x^2}left(1+ 2ye^{x^2+y^2}right)dy,dx$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jul 18 '18 at 20:02









          Robert Howard

          1,9161822




          1,9161822










          answered Jul 18 '18 at 19:36









          user247327user247327

          10.6k1515




          10.6k1515








          • 1




            $begingroup$
            Thanks for your help, but isn't that really "complicated" ? I mean, the integral over 1 is easy, and the divergence of the vector field is also a lot easier than in "first" appearence, so do we really need to integrate $2ye^{x^2+y^2}$ in that way ? Isn't there a trick, or couldn't we for example use cylindrical/ spherical coordinates to make things easier ? I found this question in an old exam, and given that time in exam is often limited, the questions are generally not "that" hard, so I'm just a bit surprised.
            $endgroup$
            – Poujh
            Jul 18 '18 at 19:42








          • 1




            $begingroup$
            The first integral is an integral of an odd function over a symmetric interval, thus the entire calculation yields zero.
            $endgroup$
            – zokomoko
            Jul 18 '18 at 20:07
















          • 1




            $begingroup$
            Thanks for your help, but isn't that really "complicated" ? I mean, the integral over 1 is easy, and the divergence of the vector field is also a lot easier than in "first" appearence, so do we really need to integrate $2ye^{x^2+y^2}$ in that way ? Isn't there a trick, or couldn't we for example use cylindrical/ spherical coordinates to make things easier ? I found this question in an old exam, and given that time in exam is often limited, the questions are generally not "that" hard, so I'm just a bit surprised.
            $endgroup$
            – Poujh
            Jul 18 '18 at 19:42








          • 1




            $begingroup$
            The first integral is an integral of an odd function over a symmetric interval, thus the entire calculation yields zero.
            $endgroup$
            – zokomoko
            Jul 18 '18 at 20:07










          1




          1




          $begingroup$
          Thanks for your help, but isn't that really "complicated" ? I mean, the integral over 1 is easy, and the divergence of the vector field is also a lot easier than in "first" appearence, so do we really need to integrate $2ye^{x^2+y^2}$ in that way ? Isn't there a trick, or couldn't we for example use cylindrical/ spherical coordinates to make things easier ? I found this question in an old exam, and given that time in exam is often limited, the questions are generally not "that" hard, so I'm just a bit surprised.
          $endgroup$
          – Poujh
          Jul 18 '18 at 19:42






          $begingroup$
          Thanks for your help, but isn't that really "complicated" ? I mean, the integral over 1 is easy, and the divergence of the vector field is also a lot easier than in "first" appearence, so do we really need to integrate $2ye^{x^2+y^2}$ in that way ? Isn't there a trick, or couldn't we for example use cylindrical/ spherical coordinates to make things easier ? I found this question in an old exam, and given that time in exam is often limited, the questions are generally not "that" hard, so I'm just a bit surprised.
          $endgroup$
          – Poujh
          Jul 18 '18 at 19:42






          1




          1




          $begingroup$
          The first integral is an integral of an odd function over a symmetric interval, thus the entire calculation yields zero.
          $endgroup$
          – zokomoko
          Jul 18 '18 at 20:07






          $begingroup$
          The first integral is an integral of an odd function over a symmetric interval, thus the entire calculation yields zero.
          $endgroup$
          – zokomoko
          Jul 18 '18 at 20:07




















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