How to Find the integral $int_{S}−(xy^2)~dydz + (2x ^2 y)~dzdx − (zy^2)~dxdy$ where is the portion of the...












2












$begingroup$



Find the integral
$$
int_{S}−(xy^2)~dydz + (2x^2y)~dzdx − (zy^2)~dxdy
$$
where $S$ is the portion of the sphere $x^2 + y^2 + z^2 = 1$ above the plane $z=frac{1}{2}$. Choose the direction of the normal to be outside the sphere.




Can i get some help? I got that the flux is
$$
huge phi = huge 0.
$$
and I am not sure if thats right.
begin{split}
vec F &= (-xy^2,2x^2y,-zy^2)\
vec n_mathrm{sphere} &= (2x,2y,2z)\
vec{F}cdot{}vec{n} = 2y^2(x^2 - z^2) &=^{{~(z^2 = 1-x^2-y^2)}}~4x^2y +2y^3 -2y\
intint 4x^2y +2y^3 - 2ydydx &=
end{split}



$$int_{r=0}^{r=sqrt{frac{3}{4}}}int_{theta=0}^{theta=2pi}4r^4cos^2theta sintheta + 2r^4sin^3theta - 2r^2sintheta dtheta dr = boxed{0}\
$$






multivariable-calculus vector-analysis divergence







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  • $begingroup$
    Anyone ? I need help with the question
    $endgroup$
    – Mather
    Jan 23 at 6:11
















2












$begingroup$



Find the integral
$$
int_{S}−(xy^2)~dydz + (2x^2y)~dzdx − (zy^2)~dxdy
$$
where $S$ is the portion of the sphere $x^2 + y^2 + z^2 = 1$ above the plane $z=frac{1}{2}$. Choose the direction of the normal to be outside the sphere.




Can i get some help? I got that the flux is
$$
huge phi = huge 0.
$$
and I am not sure if thats right.
begin{split}
vec F &= (-xy^2,2x^2y,-zy^2)\
vec n_mathrm{sphere} &= (2x,2y,2z)\
vec{F}cdot{}vec{n} = 2y^2(x^2 - z^2) &=^{{~(z^2 = 1-x^2-y^2)}}~4x^2y +2y^3 -2y\
intint 4x^2y +2y^3 - 2ydydx &=
end{split}



$$int_{r=0}^{r=sqrt{frac{3}{4}}}int_{theta=0}^{theta=2pi}4r^4cos^2theta sintheta + 2r^4sin^3theta - 2r^2sintheta dtheta dr = boxed{0}\
$$






multivariable-calculus vector-analysis divergence







share|cite|improve this question











$endgroup$












  • $begingroup$
    Anyone ? I need help with the question
    $endgroup$
    – Mather
    Jan 23 at 6:11














2












2








2


0



$begingroup$



Find the integral
$$
int_{S}−(xy^2)~dydz + (2x^2y)~dzdx − (zy^2)~dxdy
$$
where $S$ is the portion of the sphere $x^2 + y^2 + z^2 = 1$ above the plane $z=frac{1}{2}$. Choose the direction of the normal to be outside the sphere.




Can i get some help? I got that the flux is
$$
huge phi = huge 0.
$$
and I am not sure if thats right.
begin{split}
vec F &= (-xy^2,2x^2y,-zy^2)\
vec n_mathrm{sphere} &= (2x,2y,2z)\
vec{F}cdot{}vec{n} = 2y^2(x^2 - z^2) &=^{{~(z^2 = 1-x^2-y^2)}}~4x^2y +2y^3 -2y\
intint 4x^2y +2y^3 - 2ydydx &=
end{split}



$$int_{r=0}^{r=sqrt{frac{3}{4}}}int_{theta=0}^{theta=2pi}4r^4cos^2theta sintheta + 2r^4sin^3theta - 2r^2sintheta dtheta dr = boxed{0}\
$$






multivariable-calculus vector-analysis divergence







share|cite|improve this question











$endgroup$





Find the integral
$$
int_{S}−(xy^2)~dydz + (2x^2y)~dzdx − (zy^2)~dxdy
$$
where $S$ is the portion of the sphere $x^2 + y^2 + z^2 = 1$ above the plane $z=frac{1}{2}$. Choose the direction of the normal to be outside the sphere.




Can i get some help? I got that the flux is
$$
huge phi = huge 0.
$$
and I am not sure if thats right.
begin{split}
vec F &= (-xy^2,2x^2y,-zy^2)\
vec n_mathrm{sphere} &= (2x,2y,2z)\
vec{F}cdot{}vec{n} = 2y^2(x^2 - z^2) &=^{{~(z^2 = 1-x^2-y^2)}}~4x^2y +2y^3 -2y\
intint 4x^2y +2y^3 - 2ydydx &=
end{split}



$$int_{r=0}^{r=sqrt{frac{3}{4}}}int_{theta=0}^{theta=2pi}4r^4cos^2theta sintheta + 2r^4sin^3theta - 2r^2sintheta dtheta dr = boxed{0}\
$$






multivariable-calculus vector-analysis divergence




multivariable-calculus vector-analysis divergence






share|cite|improve this question















share|cite|improve this question













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edited Jan 23 at 6:09







Mather

















asked Jan 22 at 21:40









Mather Mather

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  • $begingroup$
    Anyone ? I need help with the question
    $endgroup$
    – Mather
    Jan 23 at 6:11


















  • $begingroup$
    Anyone ? I need help with the question
    $endgroup$
    – Mather
    Jan 23 at 6:11
















$begingroup$
Anyone ? I need help with the question
$endgroup$
– Mather
Jan 23 at 6:11




$begingroup$
Anyone ? I need help with the question
$endgroup$
– Mather
Jan 23 at 6:11










1 Answer
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$begingroup$

Let $$
Omega: frac{1}{2}le zle sqrt{1-x^2-y^2}
$$ be the region enclosed by $Sbigcup S_1$ where $S_1 :z=frac{1}{2}, x^2+y^2lefrac{3}{4}$. We obtain
$$
iiint_Omega 2(x^2-y^2) mathrm{d}xmathrm{d}ymathrm{d}z =iint_{Sbigcup S_1}vec{F}cdot vec{n} mathrm{d}S
$$ by divergence theorem. Note that
$$
iiint_Omega x^2 mathrm{d}xmathrm{d}ymathrm{d}z=iiint_Omega y^2 mathrm{d}xmathrm{d}ymathrm{d}z
$$ by the rotational symmetry, hence the left-hand side equals $0$. This gives
$$iint_{Sbigcup S_1}vec{F}cdot vec{n} mathrm{d}S=0.
$$ Since outward unit normal vector $vec{n}=(0,0,-1)^T$ on $S_1$, it follows
$$begin{eqnarray}
iint_{S}vec{F}cdot vec{n} mathrm{d}S&=&-iint_{S_1}vec{F}cdot vec{n} mathrm{d}S\&=&-frac{1}{2}iint_{x^2+y^2le frac{3}{4}} y^2 mathrm{d}xmathrm{d}y\
&=&-frac{1}{4}iint_{x^2+y^2le frac{3}{4}} (x^2+y^2) mathrm{d}xmathrm{d}y\
&=&-frac{1}{4}int_0^{2pi}int_{0}^{frac{sqrt{3}}{2}}r^3 mathrm{d}rmathrm{d}theta=-frac{9pi}{128}.
end{eqnarray}$$






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    1 Answer
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    1 Answer
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    active

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    1












    $begingroup$

    Let $$
    Omega: frac{1}{2}le zle sqrt{1-x^2-y^2}
    $$     be the region enclosed by $Sbigcup S_1$ where $S_1 :z=frac{1}{2}, x^2+y^2lefrac{3}{4}$. We obtain
    $$
    iiint_Omega 2(x^2-y^2) mathrm{d}xmathrm{d}ymathrm{d}z =iint_{Sbigcup S_1}vec{F}cdot vec{n} mathrm{d}S
    $$     by divergence theorem. Note that
    $$
    iiint_Omega x^2 mathrm{d}xmathrm{d}ymathrm{d}z=iiint_Omega y^2 mathrm{d}xmathrm{d}ymathrm{d}z
    $$     by the rotational symmetry, hence the left-hand side equals $0$. This gives
    $$iint_{Sbigcup S_1}vec{F}cdot vec{n} mathrm{d}S=0.
    $$     Since outward unit normal vector $vec{n}=(0,0,-1)^T$ on $S_1$, it follows
    $$begin{eqnarray}
    iint_{S}vec{F}cdot vec{n} mathrm{d}S&=&-iint_{S_1}vec{F}cdot vec{n} mathrm{d}S\&=&-frac{1}{2}iint_{x^2+y^2le frac{3}{4}} y^2 mathrm{d}xmathrm{d}y\
    &=&-frac{1}{4}iint_{x^2+y^2le frac{3}{4}} (x^2+y^2) mathrm{d}xmathrm{d}y\
    &=&-frac{1}{4}int_0^{2pi}int_{0}^{frac{sqrt{3}}{2}}r^3 mathrm{d}rmathrm{d}theta=-frac{9pi}{128}.
    end{eqnarray}$$






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      Let $$
      Omega: frac{1}{2}le zle sqrt{1-x^2-y^2}
      $$       be the region enclosed by $Sbigcup S_1$ where $S_1 :z=frac{1}{2}, x^2+y^2lefrac{3}{4}$. We obtain
      $$
      iiint_Omega 2(x^2-y^2) mathrm{d}xmathrm{d}ymathrm{d}z =iint_{Sbigcup S_1}vec{F}cdot vec{n} mathrm{d}S
      $$       by divergence theorem. Note that
      $$
      iiint_Omega x^2 mathrm{d}xmathrm{d}ymathrm{d}z=iiint_Omega y^2 mathrm{d}xmathrm{d}ymathrm{d}z
      $$       by the rotational symmetry, hence the left-hand side equals $0$. This gives
      $$iint_{Sbigcup S_1}vec{F}cdot vec{n} mathrm{d}S=0.
      $$       Since outward unit normal vector $vec{n}=(0,0,-1)^T$ on $S_1$, it follows
      $$begin{eqnarray}
      iint_{S}vec{F}cdot vec{n} mathrm{d}S&=&-iint_{S_1}vec{F}cdot vec{n} mathrm{d}S\&=&-frac{1}{2}iint_{x^2+y^2le frac{3}{4}} y^2 mathrm{d}xmathrm{d}y\
      &=&-frac{1}{4}iint_{x^2+y^2le frac{3}{4}} (x^2+y^2) mathrm{d}xmathrm{d}y\
      &=&-frac{1}{4}int_0^{2pi}int_{0}^{frac{sqrt{3}}{2}}r^3 mathrm{d}rmathrm{d}theta=-frac{9pi}{128}.
      end{eqnarray}$$






      share|cite|improve this answer











      $endgroup$
















        1












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        1





        $begingroup$

        Let $$
        Omega: frac{1}{2}le zle sqrt{1-x^2-y^2}
        $$         be the region enclosed by $Sbigcup S_1$ where $S_1 :z=frac{1}{2}, x^2+y^2lefrac{3}{4}$. We obtain
        $$
        iiint_Omega 2(x^2-y^2) mathrm{d}xmathrm{d}ymathrm{d}z =iint_{Sbigcup S_1}vec{F}cdot vec{n} mathrm{d}S
        $$         by divergence theorem. Note that
        $$
        iiint_Omega x^2 mathrm{d}xmathrm{d}ymathrm{d}z=iiint_Omega y^2 mathrm{d}xmathrm{d}ymathrm{d}z
        $$         by the rotational symmetry, hence the left-hand side equals $0$. This gives
        $$iint_{Sbigcup S_1}vec{F}cdot vec{n} mathrm{d}S=0.
        $$         Since outward unit normal vector $vec{n}=(0,0,-1)^T$ on $S_1$, it follows
        $$begin{eqnarray}
        iint_{S}vec{F}cdot vec{n} mathrm{d}S&=&-iint_{S_1}vec{F}cdot vec{n} mathrm{d}S\&=&-frac{1}{2}iint_{x^2+y^2le frac{3}{4}} y^2 mathrm{d}xmathrm{d}y\
        &=&-frac{1}{4}iint_{x^2+y^2le frac{3}{4}} (x^2+y^2) mathrm{d}xmathrm{d}y\
        &=&-frac{1}{4}int_0^{2pi}int_{0}^{frac{sqrt{3}}{2}}r^3 mathrm{d}rmathrm{d}theta=-frac{9pi}{128}.
        end{eqnarray}$$






        share|cite|improve this answer











        $endgroup$



        Let $$
        Omega: frac{1}{2}le zle sqrt{1-x^2-y^2}
        $$         be the region enclosed by $Sbigcup S_1$ where $S_1 :z=frac{1}{2}, x^2+y^2lefrac{3}{4}$. We obtain
        $$
        iiint_Omega 2(x^2-y^2) mathrm{d}xmathrm{d}ymathrm{d}z =iint_{Sbigcup S_1}vec{F}cdot vec{n} mathrm{d}S
        $$         by divergence theorem. Note that
        $$
        iiint_Omega x^2 mathrm{d}xmathrm{d}ymathrm{d}z=iiint_Omega y^2 mathrm{d}xmathrm{d}ymathrm{d}z
        $$         by the rotational symmetry, hence the left-hand side equals $0$. This gives
        $$iint_{Sbigcup S_1}vec{F}cdot vec{n} mathrm{d}S=0.
        $$         Since outward unit normal vector $vec{n}=(0,0,-1)^T$ on $S_1$, it follows
        $$begin{eqnarray}
        iint_{S}vec{F}cdot vec{n} mathrm{d}S&=&-iint_{S_1}vec{F}cdot vec{n} mathrm{d}S\&=&-frac{1}{2}iint_{x^2+y^2le frac{3}{4}} y^2 mathrm{d}xmathrm{d}y\
        &=&-frac{1}{4}iint_{x^2+y^2le frac{3}{4}} (x^2+y^2) mathrm{d}xmathrm{d}y\
        &=&-frac{1}{4}int_0^{2pi}int_{0}^{frac{sqrt{3}}{2}}r^3 mathrm{d}rmathrm{d}theta=-frac{9pi}{128}.
        end{eqnarray}$$







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        edited Jan 23 at 8:39

























        answered Jan 23 at 7:20









        SongSong

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