Surface integral

**Bruno J.** · April 17th 2010, 09:35 PM

Let $\vec F(x,y,z)=(P(x,y,z),Q(x,y,z),R(x,y,z))$ be a vector field on $\mathbb R^3$ with continuous partial derivatives, and $S$ a smooth orientable surface. Show that

$\iint_S \vec F \cdot \vec{dS} = \iint_S Pdydz + Qdzdx+Rdxdy$ .

**AllanCuz** · May 21st 2010, 12:51 PM

Originally Posted by Bruno J.

Let $\vec F(x,y,z)=(P(x,y,z),Q(x,y,z),R(x,y,z))$ be a vector field on $\mathbb R^3$ with continuous partial derivatives, and $S$ a smooth orientable surface. Show that

$\iint_S \vec F \cdot \vec{dS} = \iint_S Pdydz + Qdzdx+Rdxdy$ .

I'd go to the definition of the divergence theorem.

$\iint_S \vec F \cdot \vec{dS} = \iiint_V div F dV$

$\iiint_V ( \frac{ \partial P }{ \partial x } + \frac{ \partial Q }{ \partial y } + \frac{ \partial R }{ \partial z })dxdydz$

Let's multiply through and examine the first component

$\iiint_V ( \frac{ \partial P }{ \partial x } ) dxdydz$

Note that,

$\int \frac{ \partial P }{ \partial x } dx$

Represents the integral of the derivative of P with respect to x. Which means, we take the derivative and then integrate back to our original function. This means,

$\int \frac{ \partial P }{ \partial x } dx = P$

Thus,

$\iiint_V ( \frac{ \partial P }{ \partial x } ) dxdydz = \iint Pdydz$

**Bruno J.** · May 23rd 2010, 07:57 AM

The surface doesn't need to be closed, in which case you cannot use the divergence theorem!

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