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Hoochie [10]
3 years ago
5

What is the solution to the absolute value inequality?

Mathematics
1 answer:
Sloan [31]3 years ago
4 0
Answer -  c </= 4 or c >/= -3
2c - 1 </ 7
2c - 1 + 1 </= 7 +1
2c </= 8
2c/2  </=   8/2
c </=  4
2c - 1 >/= - 7
2c - 1 + 1 >/=  -7 + 1
2c  >/=  -6
2c/2  >/=  -6/2
c  >/=  -3
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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
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Close off the hemisphere S by attaching to it the disk D of radius 3 centered at the origin in the plane z=0. By the divergence theorem, we have

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

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Compute the integral of the divergence over R. Easily done by converting to cylindrical or spherical coordinates. I'll do the latter:

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From this we need to subtract the contribution of

\displaystyle\iint_D\vec F(x,y,z)\cdot\mathrm d\vec S

that is, the integral of \vec F over the disk, oriented downward. Since z=0 in D, we have

\vec F(x,y,0)=\dfrac{y^3}3\,\vec\jmath+y^2\,\vec k

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

where 0\le u\le 3 and 0\le v\le2\pi. Take the normal vector to be

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

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\vec F(x(u,v),y(u,v),0)\cdot(-u\,\vec k)=-y(u,v)^2u=-u^3\sin^2v

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(integral of divergence of <em>F</em>) - (integral over <em>D</em>) = integral over <em>S</em>

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m < b = 76

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Because b is directly across from the 76 degree angle, that makes them vertical angles, which means they are the same degree I believe.

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