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3 years ago

हवा से भी तेज चलने वाला कौन है

Mathematics

1 answer:

nevsk [136]3 years ago

3 0

Step-by-step explanation:

हवा से भी तेज चलने वाली रोशनी है।

light travels faster than air

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3 years ago

Evaluate the surface integral S F · dS for the given vector field F and the oriented surface S. In other words, find the flux of

Tomtit [17]

Apparently my answer was unclear the first time?

The flux of F across S is given by the surface integral,

$\displaystyle\iint_S\mathbf F\cdot\mathrm d\mathbf S$

Parameterize S by the vector-valued function r(u, v) defined by

$\mathbf r(u,v)=7\cos u\sin v\,\mathbf i+7\sin u\sin v\,\mathbf j+7\cos v\,\mathbf k$

with 0 ≤ u ≤ π/2 and 0 ≤ v ≤ π/2. Then the surface element is

dS = n • dS

where n is the normal vector to the surface. Take it to be

$\mathbf n=\dfrac{\frac{\partial\mathbf r}{\partial v}\times\frac{\partial\mathbf r}{\partial u}}{\left\|\frac{\partial\mathbf r}{\partial v}\times\frac{\partial\mathbf r}{\partial u}\right\|}$

The surface element reduces to

$\mathrm d\mathbf S=\mathbf n\,\mathrm dS=\mathbf n\left\|\dfrac{\partial\mathbf r}{\partial u}\times\dfrac{\partial\mathbf r}{\partial v}\right\|\,\mathrm du\,\mathrm dv$

$\implies\mathbf n\,\mathrm dS=-49(\cos u\sin^2v\,\mathbf i+\sin u\sin^2v\,\mathbf j+\cos v\sin v\,\mathbf k)\,\mathrm du\,\mathrm dv$

so that it points toward the origin at any point on S.

Then the integral with respect to u and v is

$\displaystyle\iint_S\mathbf F\cdot\mathrm d\mathbf S=\int_0^{\pi/2}\int_0^{\pi/2}\mathbf F(x(u,v),y(u,v),z(u,v))\cdot\mathbf n\,\mathrm dS$

$=\displaystyle-49\int_0^{\pi/2}\int_0^{\pi/2}(7\cos u\sin v\,\mathbf i-7\cos v\,\mathbf j+7\sin u\sin v\,\mathbf )\cdot\mathbf n\,\mathrm dS$

$=-343\displaystyle\int_0^{\pi/2}\int_0^{\pi/2}\cos^2u\sin^3v\,\mathrm du\,\mathrm dv=\boxed{-\frac{343\pi}6}$

4 0

3 years ago

हवा से भी तेज चलने वाला कौन है​

हवा से भी तेज चलने वाला कौन है