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Calculating a surface integral between a cone and a hemisphere
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Let $S$ be the surface formed by the portions of the hemisphere $z=sqrt1-x^2-y^2$ and of the cone $z =sqrtx^2+y^2$ with $x^2+y^2lefrac12$. Calculate $int_SvecFcdot dvecS$ (with the normal outside) where
$$vecF=(xz+e^ysin z,2yz+cos xz,-z^2+e^xcos y)$$
I'm using the divergence theorem, which would have $$int_SvecFcdot dvecS=iiint_D divvecFdV $$
Changing the original variables to spherical coordinates:
$$x=fracsqrt22sinvarphicostheta,$$
$$y=fracsqrt22sinvarphisintheta,$$
$$z=fracsqrt22cosvarphi.$$
And
$$div(vecF)=z+2z-2z=z.$$
Is it appropriate for you to use the divergence theorem?
How would the limits of integration in $iiint_D divvecFdV$?
multivariable-calculus
asked Jul 14 at 22:58
Alex Pozo
488214
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0
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Let $S$ be the surface formed by the portions of the hemisphere $z=sqrt1-x^2-y^2$ and of the cone $z =sqrtx^2+y^2$ with $x^2+y^2lefrac12$. Calculate $int_SvecFcdot dvecS$ (with the normal outside) where
$$vecF=(xz+e^ysin z,2yz+cos xz,-z^2+e^xcos y)$$
I'm using the divergence theorem, which would have $$int_SvecFcdot dvecS=iiint_D divvecFdV $$
Changing the original variables to spherical coordinates:
$$x=fracsqrt22sinvarphicostheta,$$
$$y=fracsqrt22sinvarphisintheta,$$
$$z=fracsqrt22cosvarphi.$$
And
$$div(vecF)=z+2z-2z=z.$$
Is it appropriate for you to use the divergence theorem?
How would the limits of integration in $iiint_D divvecFdV$?
multivariable-calculus
asked Jul 14 at 22:58
Alex Pozo
488214
1
It would be quite a bit easier if you used cylindrical co-ordinates, then $$rho in [0, frac 1 sqrt 2] \ theta in [0,2 pi] \ rho < z < sqrt1-rho^2 $$
– WW1
Jul 15 at 2:49
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $S$ be the surface formed by the portions of the hemisphere $z=sqrt1-x^2-y^2$ and of the cone $z =sqrtx^2+y^2$ with $x^2+y^2lefrac12$. Calculate $int_SvecFcdot dvecS$ (with the normal outside) where
$$vecF=(xz+e^ysin z,2yz+cos xz,-z^2+e^xcos y)$$
I'm using the divergence theorem, which would have $$int_SvecFcdot dvecS=iiint_D divvecFdV $$
Changing the original variables to spherical coordinates:
$$x=fracsqrt22sinvarphicostheta,$$
$$y=fracsqrt22sinvarphisintheta,$$
$$z=fracsqrt22cosvarphi.$$
And
$$div(vecF)=z+2z-2z=z.$$
Is it appropriate for you to use the divergence theorem?
How would the limits of integration in $iiint_D divvecFdV$?
multivariable-calculus
asked Jul 14 at 22:58
Alex Pozo
488214
Let $S$ be the surface formed by the portions of the hemisphere $z=sqrt1-x^2-y^2$ and of the cone $z =sqrtx^2+y^2$ with $x^2+y^2lefrac12$. Calculate $int_SvecFcdot dvecS$ (with the normal outside) where
$$vecF=(xz+e^ysin z,2yz+cos xz,-z^2+e^xcos y)$$
I'm using the divergence theorem, which would have $$int_SvecFcdot dvecS=iiint_D divvecFdV $$
Changing the original variables to spherical coordinates:
$$x=fracsqrt22sinvarphicostheta,$$
$$y=fracsqrt22sinvarphisintheta,$$
$$z=fracsqrt22cosvarphi.$$
And
$$div(vecF)=z+2z-2z=z.$$
Is it appropriate for you to use the divergence theorem?
How would the limits of integration in $iiint_D divvecFdV$?
multivariable-calculus
asked Jul 14 at 22:58
Alex Pozo
488214
asked Jul 14 at 22:58
Alex Pozo
488214
asked Jul 14 at 22:58
Alex Pozo
488214
asked Jul 14 at 22:58
Alex Pozo
488214
488214
1
It would be quite a bit easier if you used cylindrical co-ordinates, then $$rho in [0, frac 1 sqrt 2] \ theta in [0,2 pi] \ rho < z < sqrt1-rho^2 $$
– WW1
Jul 15 at 2:49
add a comment |Â
1
It would be quite a bit easier if you used cylindrical co-ordinates, then $$rho in [0, frac 1 sqrt 2] \ theta in [0,2 pi] \ rho < z < sqrt1-rho^2 $$
– WW1
Jul 15 at 2:49
1
1
It would be quite a bit easier if you used cylindrical co-ordinates, then $$rho in [0, frac 1 sqrt 2] \ theta in [0,2 pi] \ rho < z < sqrt1-rho^2 $$
– WW1
Jul 15 at 2:49
It would be quite a bit easier if you used cylindrical co-ordinates, then $$rho in [0, frac 1 sqrt 2] \ theta in [0,2 pi] \ rho < z < sqrt1-rho^2 $$
– WW1
Jul 15 at 2:49
add a comment |Â
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1
It would be quite a bit easier if you used cylindrical co-ordinates, then $$rho in [0, frac 1 sqrt 2] \ theta in [0,2 pi] \ rho < z < sqrt1-rho^2 $$
– WW1
Jul 15 at 2:49