Vector calculus and divergence theorem properties

Question

I cannot seem to figure out why the following is true:

$\partial \over \partial r$ $\int_{B(x,r)}\triangle v(z) dz =$ $\int_{\partial B(x,r)}\triangle v(y) do(y)$

I attempted to apply divergence theorem to the LHS, but here's where I get stuck and don't know where to proceed:

$\partial \over \partial r$ $\int_{B(x,r)}\triangle v(z) dz = $ $\partial \over \partial r$ $\int_{B(x,r)}div(\nabla v(z)) dz =$ $\partial \over \partial r$ $\int_{\partial B(x,r)}\nabla v(y) \cdot \nu dy$

Where $\nu$ is an exterior normal to $\partial B(x,r)$.

I do not know when/how to pass the partial derivative into the integrand to get the result.

Thanks in advance,

score 1 · Answer 1 · answered Dec 04 '13 at 11:17

1

Hint: The thing you want to prove does not really depend on the integrand being a Laplacian. We have $$ \int_{B(x,r)} f(z)\, dz = \int_0^r \left(\int_{\partial B(x,s)} f(z)\, do(z)\right)\, ds. $$

answered Dec 04 '13 at 11:17

martini

84,101

What does $f(z) , do(z)$ mean? Is $o(z)$ meaning the outward normal? – Mark Fantini Dec 04 '13 at 12:08
As in your question, I denote by $do(z)$ the measure on the sphere. – martini Dec 05 '13 at 15:11
Sorry, I am not familiar with this notation. What is special about $do(z)$ instead of $dz$? Can you please point me to where it is commonplace? Thanks. – Mark Fantini Dec 05 '13 at 22:09

Vector calculus and divergence theorem properties

1 Answers1