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Fofino [41]
2 years ago
7

At midnight the temperature was ➖ 5 degrees Celsius. At noon, the temperature was 19 degrees Celsius. Which expression represent

s the change in temperature?
Mathematics
1 answer:
Yanka [14]2 years ago
7 0

Answer:

cold and hot Is the answer

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Let f(x,y,z) = ztan-1(y2) i + z3ln(x2 + 1) j + z k. find the flux of f across the part of the paraboloid x2 + y2 + z = 3 that li
Sophie [7]
Consider the closed region V bounded simultaneously by the paraboloid and plane, jointly denoted S. By the divergence theorem,

\displaystyle\iint_S\mathbf f(x,y,z)\cdot\mathrm dS=\iiint_V\nabla\cdot\mathbf f(x,y,z)\,\mathrm dV

And since we have

\nabla\cdot\mathbf f(x,y,z)=1

the volume integral will be much easier to compute. Converting to cylindrical coordinates, we have

\displaystyle\iiint_V\nabla\cdot\mathbf f(x,y,z)\,\mathrm dV=\iiint_V\mathrm dV
=\displaystyle\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=1}\int_{z=2}^{z=3-r^2}r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta
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Then the integral over the paraboloid would be the difference of the integral over the total surface and the integral over the disk. Denoting the disk by D, we have

\displaystyle\iint_{S-D}\mathbf f\cdot\mathrm dS=\frac\pi2-\iint_D\mathbf f\cdot\mathrm dS

Parameterize D by

\mathbf s(u,v)=u\cos v\,\mathbf i+u\sin v\,\mathbf j+2\,\mathbf k
\implies\mathbf s_u\times\mathbf s_v=u\,\mathbf k

which would give a unit normal vector of \mathbf k. However, the divergence theorem requires that the closed surface S be oriented with outward-pointing normal vectors, which means we should instead use \mathbf s_v\times\mathbf s_u=-u\,\mathbf k.

Now,

\displaystyle\iint_D\mathbf f\cdot\mathrm dS=\int_{u=0}^{u=1}\int_{v=0}^{v=2\pi}\mathbf f(x(u,v),y(u,v),z(u,v))\cdot(-u\,\mathbf k)\,\mathrm dv\,\mathrm du
=\displaystyle-4\pi\int_{u=0}^{u=1}u\,\mathrm du
=-2\pi

So, the flux over the paraboloid alone is

\displaystyle\iint_{S-D}\mathbf f\cdot\mathrm dS=\frac\pi2-(-2\pi)=\dfrac{5\pi}2
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