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zmey [24]
3 years ago
9

Find the area of the figure below.

Mathematics
1 answer:
beks73 [17]3 years ago
6 0

Answer:

i think the answer is a 106 sq units

Step-by-step explanation:

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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
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Apparently my answer was unclear the first time?

The flux of <em>F</em> across <em>S</em> is given by the surface integral,

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

Parameterize <em>S</em> by the vector-valued function <em>r</em>(<em>u</em>, <em>v</em>) 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 ≤ <em>u</em> ≤ π/2 and 0 ≤ <em>v</em> ≤ π/2. Then the surface element is

d<em>S</em> = <em>n</em> • d<em>S</em>

where <em>n</em> 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 <em>S</em>.

Then the integral with respect to <em>u</em> and <em>v</em> 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}

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3 years ago
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1 is congruent 2 is neither and 3 is similar

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