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 F across S. For closed surfaces, use the positive (outward) orientation. F(x, y, z) = xy i + yz j + zx k S is the part of the paraboloid z = 8 − x2 − y2 that lies above the square 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, and has upward orientation

Respuesta :

Parameterize [tex]S{/tex] by

[tex]\vec s(u,v)=u\,\vec\imath+v\,\vec\jmath+(8-u^2-v^2)\,\vec k[/tex]

with [tex]0\le u\le1[/tex] and [tex]0\le v\le1[/tex].

Take the normal vector to [tex]S[/tex] to be

[tex]\vec s_u\times\vec s_v=2u\,\vec\imath+2v\,\vec\jmath+\vec k[/tex]

Then the flux of [tex]\vec F[/tex] across [tex]S[/tex] is

[tex]\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=\int_0^1\int_0^1\vec F(x(u,v),y(u,v),z(u,v))\cdot(\vec s_u\times\vec s_v)\,\mathrm du\,\mathrm dv[/tex]

[tex]=\displaystyle\int_0^1\int_0^1(uv\,\vec\imath+v(8-u^2-v^2)\,\vec\jmath+u(8-u^2-v^2)\,\vec k)\cdot(2u\,\vec\imath+2v\,\vec\jmath+\vec k)\,\mathrm du\,\mathrm dv[/tex]

[tex]=\displaystyle\int_0^1\int_0^1\bigg(2u^2v+(u+2v^2)(8-u^2-v^2)\bigg)\,\mathrm du\,\mathrm dv=\boxed{\frac{1553}{180}}[/tex]

Answer:

The flux of F across S is 8.627.

Step-by-step explanation:

Given,

F(x, y, z) = xy i + yz j + zx k

or F=(xy, yz, zx)

S is the part of the paraboloid [tex]z=8-x^{2} -y^{2}[/tex] above the square 0 ≤ x ≤ 1, 0 ≤ y ≤ 1

By differentiating z with respect to x we get:

fx=-2x

By differentiating z with respect to y we get:

fy=-2y

So, Surface integral is given by:

[tex]=\int\limits^1_0\int\limits^1_0( {(xy)*(2x)+y(8-x^{2} -y^{2} *(2y)+x(8-x^{2} -y^{2})} \, )dx \, dy[/tex][tex]=\int\limits^1_0\int\limits^1_0 (2x^{2} y+16y^{2} -2x^{2} y^{2} -2y^{4}+8x-x^{3}-xy^{2} } \,) dx \, dy[/tex]

Integrating with respect y:

[tex]=\int\limits^1_0(x^{2} y^{2} +\frac{16}{3} y^{3} -\frac{2}{3} x^{2} y^{3} -\frac{2}{5} y^{5}+8xy-x^{3}y-\frac{1}{3} xy^{3} } \, )dx \,[/tex]

After Substituting limits of y, we get:

[tex]=\int\limits^1_0(x^{2} +\frac{16}{3} -\frac{2}{3} x^{2} -\frac{2}{5}+8x-x^{3}-\frac{1}{3} x } \,) dx \,[/tex]

Integrating with respect x:

[tex]=(\frac{1}{3} x^{3} +\frac{16}{3}x -\frac{2}{9} x^{3} -\frac{2}{5}x+4 x^{2} -\frac{1}{4} x^{4}-\frac{1}{6} x^{2} } \,)[/tex]

After Substituting limits of x, we get:

[tex]=(\frac{1}{3} +\frac{16}{3} -\frac{2}{9} -\frac{2}{5}+4 -\frac{1}{4} -\frac{1}{6} } \,)\\\\=\frac{1553}{180}[/tex]

[tex]= 8.627[/tex]

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