2 years ago

I confused on this question

Mathematics

2 answers:

Fynjy0 [20]2 years ago

6 0

42 degrees. Comment down if you need the explanation why

zhenek [66]2 years ago

5 0

The answer is 42 degress

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Evaluate the surface integral ModifyingBelow Integral from nothing to nothing Integral from nothing to nothing With Upper S f (x

kykrilka [37]

Parameterize $S$ by

$\vec s(u,v)=6\cos u\sin v\,\vec\imath+6\sin u\sin v\,\vec\jmath+6\cos v\,\vec k$

with $0\le u\le2\pi$ and $0\le v\le\frac\pi2$ . Take a normal vector to $S$ ,

$\dfrac{\partial\vec s}{\partial v}\times\dfrac{\partial\vec s}{\partial u}=36\cos u\sin^2v\,\vec\imath+36\sin u\sin^2v\,\vec\jmath+36\cos v\sin v\,\vec k$

which has norm

$\left\|\dfrac{\partial\vec s}{\partial v}\times\dfrac{\partial\vec s}{\partial u}\right\|=36\sin v$

Then the integral of $f(x,y,z)=x^2+y^2$ over $S$ is

$\displaystyle\iint_Sx^2+y^2\,\mathrm dS=\iint_S\left((6\cos u\sin v)^2+(6\sin u\sin v)^2\right)\left\|\frac{\partial\vec s}{\partial v}\times\frac{\partial\vec s}{\partial u}\right\|\,\mathrm du\,\mathrm dv$