Prove that $frac{d}{dr}frac{1}{4 pi r^2} iint_{S_r} u dS = frac{1}{4 pi r^2} iint_{S_r} nabla u cdot dvec{S}$
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How can I show that if $u$ is a $C^1$ function and $S_r = {(x, y, z) : x^2 + y^2 + z^2 = r}$, then for all $r$:
$$frac{d}{dr}frac{1}{4 pi r^2} iint_{S_r} u dS = frac{1}{4 pi r^2} iint_{S_r} nabla u cdot dvec{S}$$
multivariable-calculus
asked Nov 19 at 18:58
John
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How can I show that if $u$ is a $C^1$ function and $S_r = {(x, y, z) : x^2 + y^2 + z^2 = r}$, then for all $r$:
$$frac{d}{dr}frac{1}{4 pi r^2} iint_{S_r} u dS = frac{1}{4 pi r^2} iint_{S_r} nabla u cdot dvec{S}$$
multivariable-calculus
asked Nov 19 at 18:58
John
31
add a comment |
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0
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favorite
up vote
0
down vote
favorite
How can I show that if $u$ is a $C^1$ function and $S_r = {(x, y, z) : x^2 + y^2 + z^2 = r}$, then for all $r$:
$$frac{d}{dr}frac{1}{4 pi r^2} iint_{S_r} u dS = frac{1}{4 pi r^2} iint_{S_r} nabla u cdot dvec{S}$$
multivariable-calculus
asked Nov 19 at 18:58
John
31
How can I show that if $u$ is a $C^1$ function and $S_r = {(x, y, z) : x^2 + y^2 + z^2 = r}$, then for all $r$:
$$frac{d}{dr}frac{1}{4 pi r^2} iint_{S_r} u dS = frac{1}{4 pi r^2} iint_{S_r} nabla u cdot dvec{S}$$
multivariable-calculus
multivariable-calculus
asked Nov 19 at 18:58
John
31
asked Nov 19 at 18:58
John
31
asked Nov 19 at 18:58
John
31
asked Nov 19 at 18:58
John
31
asked Nov 19 at 18:58
John
31
31
add a comment |
add a comment |
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Get the "$r$" out of the limits of integration. With the substitution $rz = y$ you have
$r^2 dS(u) = dS(y)$ so that $$frac 1{4pi r^2} iint_{S_r} u(y) , dS(y) = frac 1{4pi} iint_{S_1} u(rz) , dS(z) = frac 1{4pi} iint_{S_1} u(rz_1,rz_2) , dS(z).$$
The derivative is calculated using the chain rule:
$$frac{d}{dr} u(rz_1,rz_2) = frac{partial u}{partial z_1}u(rz_1,rz_2) z_1 + frac{partial u}{partial z_2} u(rz_1,rz_2) z_2 = nabla u(rz_1,rz_2) cdot (z_1,z_2) = nabla u(rz) cdot z.$$ Re-apply the substitution $rz = y$ to obtain
$$ frac{d}{dr} frac 1{4pi} iint_{S_1} u(rz_1,rz_2) , dS(z) = frac 1{4pi} iint_{S_1} nabla u(rz) cdot z , dS(z) = frac 1{4pi r^2} iint_{S_r} nabla u(y) cdot frac yr , dS(y)$$
Finally observe that the exterior unit normal vector to the sphere $S_r$ at a point $y in S_r$ is just $dfrac yr$.
answered Nov 19 at 21:05
Umberto P.
38.3k13063
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1 Answer
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1 Answer
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active
oldest
votes
active
oldest
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active
oldest
votes
up vote
0
down vote
accepted
Get the "$r$" out of the limits of integration. With the substitution $rz = y$ you have
$r^2 dS(u) = dS(y)$ so that $$frac 1{4pi r^2} iint_{S_r} u(y) , dS(y) = frac 1{4pi} iint_{S_1} u(rz) , dS(z) = frac 1{4pi} iint_{S_1} u(rz_1,rz_2) , dS(z).$$
The derivative is calculated using the chain rule:
$$frac{d}{dr} u(rz_1,rz_2) = frac{partial u}{partial z_1}u(rz_1,rz_2) z_1 + frac{partial u}{partial z_2} u(rz_1,rz_2) z_2 = nabla u(rz_1,rz_2) cdot (z_1,z_2) = nabla u(rz) cdot z.$$ Re-apply the substitution $rz = y$ to obtain
$$ frac{d}{dr} frac 1{4pi} iint_{S_1} u(rz_1,rz_2) , dS(z) = frac 1{4pi} iint_{S_1} nabla u(rz) cdot z , dS(z) = frac 1{4pi r^2} iint_{S_r} nabla u(y) cdot frac yr , dS(y)$$
Finally observe that the exterior unit normal vector to the sphere $S_r$ at a point $y in S_r$ is just $dfrac yr$.
answered Nov 19 at 21:05
Umberto P.
38.3k13063
add a comment |
up vote
0
down vote
accepted
Get the "$r$" out of the limits of integration. With the substitution $rz = y$ you have
$r^2 dS(u) = dS(y)$ so that $$frac 1{4pi r^2} iint_{S_r} u(y) , dS(y) = frac 1{4pi} iint_{S_1} u(rz) , dS(z) = frac 1{4pi} iint_{S_1} u(rz_1,rz_2) , dS(z).$$
The derivative is calculated using the chain rule:
$$frac{d}{dr} u(rz_1,rz_2) = frac{partial u}{partial z_1}u(rz_1,rz_2) z_1 + frac{partial u}{partial z_2} u(rz_1,rz_2) z_2 = nabla u(rz_1,rz_2) cdot (z_1,z_2) = nabla u(rz) cdot z.$$ Re-apply the substitution $rz = y$ to obtain
$$ frac{d}{dr} frac 1{4pi} iint_{S_1} u(rz_1,rz_2) , dS(z) = frac 1{4pi} iint_{S_1} nabla u(rz) cdot z , dS(z) = frac 1{4pi r^2} iint_{S_r} nabla u(y) cdot frac yr , dS(y)$$
Finally observe that the exterior unit normal vector to the sphere $S_r$ at a point $y in S_r$ is just $dfrac yr$.
answered Nov 19 at 21:05
Umberto P.
38.3k13063
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Get the "$r$" out of the limits of integration. With the substitution $rz = y$ you have
$r^2 dS(u) = dS(y)$ so that $$frac 1{4pi r^2} iint_{S_r} u(y) , dS(y) = frac 1{4pi} iint_{S_1} u(rz) , dS(z) = frac 1{4pi} iint_{S_1} u(rz_1,rz_2) , dS(z).$$
The derivative is calculated using the chain rule:
$$frac{d}{dr} u(rz_1,rz_2) = frac{partial u}{partial z_1}u(rz_1,rz_2) z_1 + frac{partial u}{partial z_2} u(rz_1,rz_2) z_2 = nabla u(rz_1,rz_2) cdot (z_1,z_2) = nabla u(rz) cdot z.$$ Re-apply the substitution $rz = y$ to obtain
$$ frac{d}{dr} frac 1{4pi} iint_{S_1} u(rz_1,rz_2) , dS(z) = frac 1{4pi} iint_{S_1} nabla u(rz) cdot z , dS(z) = frac 1{4pi r^2} iint_{S_r} nabla u(y) cdot frac yr , dS(y)$$
Finally observe that the exterior unit normal vector to the sphere $S_r$ at a point $y in S_r$ is just $dfrac yr$.
answered Nov 19 at 21:05
Umberto P.
38.3k13063
Get the "$r$" out of the limits of integration. With the substitution $rz = y$ you have
$r^2 dS(u) = dS(y)$ so that $$frac 1{4pi r^2} iint_{S_r} u(y) , dS(y) = frac 1{4pi} iint_{S_1} u(rz) , dS(z) = frac 1{4pi} iint_{S_1} u(rz_1,rz_2) , dS(z).$$
The derivative is calculated using the chain rule:
$$frac{d}{dr} u(rz_1,rz_2) = frac{partial u}{partial z_1}u(rz_1,rz_2) z_1 + frac{partial u}{partial z_2} u(rz_1,rz_2) z_2 = nabla u(rz_1,rz_2) cdot (z_1,z_2) = nabla u(rz) cdot z.$$ Re-apply the substitution $rz = y$ to obtain
$$ frac{d}{dr} frac 1{4pi} iint_{S_1} u(rz_1,rz_2) , dS(z) = frac 1{4pi} iint_{S_1} nabla u(rz) cdot z , dS(z) = frac 1{4pi r^2} iint_{S_r} nabla u(y) cdot frac yr , dS(y)$$
Finally observe that the exterior unit normal vector to the sphere $S_r$ at a point $y in S_r$ is just $dfrac yr$.
answered Nov 19 at 21:05
Umberto P.
38.3k13063
answered Nov 19 at 21:05
Umberto P.
38.3k13063
answered Nov 19 at 21:05
Umberto P.
38.3k13063
answered Nov 19 at 21:05
Umberto P.
38.3k13063
38.3k13063
add a comment |
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