The value of the integral -r-n-dS over the closed surface S bounding a volume V- where r-x-y- -zk- is the position vector and n- is the normal to the surface S- is

∵ nbsp; r⃗=xî+yĵ+3k̂ then div (r⃗) nbsp; =∂/∂ x ( x )+∂/∂ y ( y )+∂/∂ z ( z ) =1+1+1 = 3 ∵ Surface is closed with volume V so by Gauss Divergence theorem ...

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Gauss Divergence Theorem

The value of ...

Question

The value of the integral $∯ \to r . \to n d S$ over the closed surface S bounding a volume V, where $\to r = x^i + y^j + z^k$ is the position vector and $\to n$ is the normal to the surface S, is

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Solution

The correct option is C 3V
$∵$ $\to r = x^i + y^j + 3^k$ then
div $(\to r$ ) $= \frac{\partial}{\partial x} (x) + \frac{\partial}{\partial y} (y) + \frac{\partial}{\partial z} (z)$
=1+1+1 = 3
$∵$ Surface is closed with volume V so by Gauss-Divergence theorem

$\iint_{s} (f .^n) d s = ∭_{v} (d i v \to δ) d v$
$\Rightarrow \int \oint (\to r . \to n) d s = ∭_{v} (3) d v = 3 v$

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The value of the integral ∯→r.→ndS over the closed surface S bounding a volume V, where →r=x^i+y^j+z^k is the position vector and →n is the normal to the surface S, is

The correct option is C 3V∵ →r=x^i+y^j+3^k then div (→r) =∂∂x(x)+∂∂y(y)+∂∂z(z) =1+1+1 = 3 ∵ Surface is closed with volume V so by Gauss-Divergence theorem ∬s(f.^n)ds=∭v(div →δ)dv ⇒∫∮(→r.→n)ds=∭v(3)dv=3v

The value of the integral $∯ \to r . \to n d S$ over the closed surface S bounding a volume V, where $\to r = x^i + y^j + z^k$ is the position vector and $\to n$ is the normal to the surface S, is