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aleksandr82 [10.1K]
3 years ago
10

PLEASE HELP ASAP 25 POINTS

Mathematics
2 answers:
Lena [83]3 years ago
4 0
The answer to the question is a)
Nookie1986 [14]3 years ago
4 0

Answer is A .


How?

See the horizontal line parallel to x-axis.Now with the help of graph,I can write some points on that horizontal line which are:(0,3),(2,3),(-2,3),etc.Just analyze these points.In these points you can see that y-coordinate is constant and is equal to 3.So the equation of this horizontal line is:

y=3 which is analogous to y<=3(Why not y>=3 because the line limits the value to 3 )

The slanted line can be represented by y>2x-1

Also see this image to know why this slanted line bounds the region y>2x-1





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Use the Divergence Theorem to evaluate S F · dS, where F(x, y, z) = z2xi + y3 3 + sin z j + (x2z + y2)k and S is the top half of
kifflom [539]

Looks like we have

\vec F(x,y,z)=z^2x\,\vec\imath+\left(\dfrac{y^3}3+\sin z\right)\,\vec\jmath+(x^2z+y^2)\,\vec k

which has divergence

\nabla\cdot\vec F(x,y,z)=\dfrac{\partial(z^2x)}{\partial x}+\dfrac{\partial\left(\frac{y^3}3+\sin z\right)}{\partial y}+\dfrac{\partial(x^2z+y^2)}{\partial z}=z^2+y^2+x^2

By the divergence theorem, the integral of \vec F across S is equal to the integral of \nabla\cdot\vec F over R, where R is the region enclosed by S. Of course, S is not a closed surface, but we can make it so by closing off the hemisphere S by attaching it to the disk x^2+y^2\le1 (call it D) so that R has boundary S\cup D.

Then by the divergence theorem,

\displaystyle\iint_{S\cup D}\vec F\cdot\mathrm d\vec S=\iiint_R(x^2+y^2+z^2)\,\mathrm dV

Compute the integral in spherical coordinates, setting

\begin{cases}x=\rho\cos\theta\sin\varphi\\y=\rho\sin\theta\sin\varphi\\z=\rho\cos\varphi\end{cases}\implies\mathrm dV=\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi

so that the integral is

\displaystyle\iiint_R(x^2+y^2+z^2)\,\mathrm dV=\int_0^{\pi/2}\int_0^{2\pi}\int_0^1\rho^4\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi=\frac{2\pi}5

The integral of \vec F across S\cup D is equal to the integral of \vec F across S plus the integral across D (without outward orientation, so that

\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=\frac{2\pi}5-\iint_D\vec F\cdot\mathrm d\vec S

Parameterize D by

\vec s(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec\jmath

with 0\le u\le1 and 0\le v\le2\pi. Take the normal vector to D to be

\dfrac{\partial\vec s}{\partial v}\times\dfrac{\partial\vec s}{\partial u}=-u\,\vec k

Then we have

\displaystyle\iint_D\vec F\cdot\mathrm d\vec S=\int_0^{2\pi}\int_0^1\left(\frac{u^3}3\sin^3v\,\vec\jmath+u^2\sin^2v\,\vec k\right)\times(-u\,\vec k)\,\mathrm du\,\mathrm dv

=\displaystyle-\int_0^{2\pi}\int_0^1u^3\sin^2v\,\mathrm du\,\mathrm dv=-\frac\pi4

Finally,

\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=\frac{2\pi}5-\left(-\frac\pi4\right)=\boxed{\frac{13\pi}{20}}

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