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Standard XII

Physics

Can Area Be a Vector?

If r̅ is the ...

Question

If $¯ r$ is the position vector of any point on a closed surface S that encloses the volume V then
$\iint_{s} (¯ r . d ¯ s)$ is equal to.

\frac{1}{2}

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Solution

The correct option is D 3V
we know $\to r = x^i + y^j + z^k$
Then div $\to r$ = 1 + 1 + 1 = 3
By Gauss-Divergence theorem
$\iint_{s} \to r . d \to s = ∭_{v} (d i v \to r) d V$
= $3 ∭_{v} (1) d V$
=3V

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If ¯r is the position vector of any point on a closed surface S that encloses the volume V then ∬s(¯r.d¯s) is equal to.

The correct option is D 3Vwe know →r=x^i+y^j+z^k Then div →r = 1 + 1 + 1 = 3 By Gauss-Divergence theorem ∬s→r.d→s=∭v(div→r)dV =3∭v(1)dV =3V

If $¯ r$ is the position vector of any point on a closed surface S that encloses the volume V then
$\iint_{s} (¯ r . d ¯ s)$ is equal to.

The correct option is D 3V
we know $\to r = x^i + y^j + z^k$
Then div $\to r$ = 1 + 1 + 1 = 3
By Gauss-Divergence theorem
$\iint_{s} \to r . d \to s = ∭_{v} (d i v \to r) d V$
= $3 ∭_{v} (1) d V$
=3V