Show a Coarea-formular-like identity
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I want to show that
$displaystyle u(x)=frac{1}{|B_r(x)|}int_{B_r(x)}u,mathrm{d}x = frac{1}{|partial B_r(x)|}int_{partial B_r(x)}u,mathrm{d}S$
where $B_r(x)$ is the ball with radius $r$ centered around $x$. Note that $u(x)inmathbb{R}^2$ so that $displaystyle |B_r(x)| = r^2pi, |partial B_r(x)|=2pi r$.
Using the coarea formular I would get
$displaystyle int_{B_r(x)}u,mathrm{d}x = int_0^r int_{partial B_r(x)}u,mathrm{d}S,mathrm{d}r$
and form this point I am stuck, since the curve is also dependent on $r$. Is this the right approach for this problem?
integration contour-integration
asked Nov 16 at 16:38
EpsilonDelta
5851515
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up vote
0
down vote
favorite
I want to show that
$displaystyle u(x)=frac{1}{|B_r(x)|}int_{B_r(x)}u,mathrm{d}x = frac{1}{|partial B_r(x)|}int_{partial B_r(x)}u,mathrm{d}S$
where $B_r(x)$ is the ball with radius $r$ centered around $x$. Note that $u(x)inmathbb{R}^2$ so that $displaystyle |B_r(x)| = r^2pi, |partial B_r(x)|=2pi r$.
Using the coarea formular I would get
$displaystyle int_{B_r(x)}u,mathrm{d}x = int_0^r int_{partial B_r(x)}u,mathrm{d}S,mathrm{d}r$
and form this point I am stuck, since the curve is also dependent on $r$. Is this the right approach for this problem?
integration contour-integration
asked Nov 16 at 16:38
EpsilonDelta
5851515
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I want to show that
$displaystyle u(x)=frac{1}{|B_r(x)|}int_{B_r(x)}u,mathrm{d}x = frac{1}{|partial B_r(x)|}int_{partial B_r(x)}u,mathrm{d}S$
where $B_r(x)$ is the ball with radius $r$ centered around $x$. Note that $u(x)inmathbb{R}^2$ so that $displaystyle |B_r(x)| = r^2pi, |partial B_r(x)|=2pi r$.
Using the coarea formular I would get
$displaystyle int_{B_r(x)}u,mathrm{d}x = int_0^r int_{partial B_r(x)}u,mathrm{d}S,mathrm{d}r$
and form this point I am stuck, since the curve is also dependent on $r$. Is this the right approach for this problem?
integration contour-integration
asked Nov 16 at 16:38
EpsilonDelta
5851515
I want to show that
$displaystyle u(x)=frac{1}{|B_r(x)|}int_{B_r(x)}u,mathrm{d}x = frac{1}{|partial B_r(x)|}int_{partial B_r(x)}u,mathrm{d}S$
where $B_r(x)$ is the ball with radius $r$ centered around $x$. Note that $u(x)inmathbb{R}^2$ so that $displaystyle |B_r(x)| = r^2pi, |partial B_r(x)|=2pi r$.
Using the coarea formular I would get
$displaystyle int_{B_r(x)}u,mathrm{d}x = int_0^r int_{partial B_r(x)}u,mathrm{d}S,mathrm{d}r$
and form this point I am stuck, since the curve is also dependent on $r$. Is this the right approach for this problem?
integration contour-integration
integration contour-integration
asked Nov 16 at 16:38
EpsilonDelta
5851515
asked Nov 16 at 16:38
EpsilonDelta
5851515
asked Nov 16 at 16:38
EpsilonDelta
5851515
asked Nov 16 at 16:38
EpsilonDelta
5851515
asked Nov 16 at 16:38
EpsilonDelta
5851515
5851515
add a comment |
add a comment |
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