find the flux of the vector ar $F = $ outward across the given surfaces. Each surface is oriented, unless otherwise specified, with outward-pointing typical pointing far from the origin.

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the upper hemisphere the radius 2 focused at the origin.the cone $z = 2sqrtx^2+y^2$, $z$ = 0 come 2 with outward normal pointing upward
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We deserve to make the flux calculation for each surface directly by examining the surface ar integral $ iint_S inter-base.netbfF cdot inter-base.netbfhatn dS $ , and additionally by applying the divergence Theorem together a check.

For (1), the "upper" hemisphere the radius 2 centered on the origin has the equation $ z = sqrt4 - x^2 - y^2 $ . We may likewise write $ x^2 + y^2 + z^2 = 4 , z ge 0 $ to identify the unit "outward" regular to the hemispherical surface, then proceeding on come compute the flux integral:

a) using the gradient --

$$ g(x,y,z) = x^2 + y^2 + z^2 - 4 Rightarrow abla g = langle 2x , 2y , 2z angle $$

$$ Rightarrow | abla g | = sqrt(2x)^2 + (2y)^2 + (2z)^2 = 2 cdot 2 = 4 $$

$$ Rightarrow inter-base.netbfhatn = frac abla g abla g = langle fracx2 , fracy2 , fracz2 angle quad ext $$

$$ Rightarrow inter-base.netbfF cdot inter-base.netbfhatn = langle x^2 , y^2 , z^2 angle cdot langle fracx2 , fracy2 , fracz2 angle = fracx^3 + y^3 + z^32 . $$

This is well-suited to the use of spherical coordinates, therefore integrating over the hemispherical surface of radius $ R = 2 $ gives

$$ iint_S inter-base.netbfF cdot inter-base.netbfhatn dS $$

$$ = int_0^2 pi int_0^pi / 2 frac12 ( R^3 sin^3 phi cos^3 heta + R^3 sin^3 phi sin^3 heta + R^3 cos^3 phi ) R^2 sin phi dphi d heta $$

$$ = frac12R^5 int_0^2 pi int_0^pi / 2 ( sin^4 phi < cos^3 heta + sin^3 heta > + cos^3 phi sin phi ) dphi d heta $$

$$ = frac12 cdot 2^5 left< int_0^2 pi int_0^pi / 2 sin^4 phi ( cos^3 heta + sin^3 heta ) dphi d heta quad + int_0^2 pi int_0^pi / 2 cos^3 phi sin phi dphi d heta ight> $$

$$ = 16 left< 0 + int_0^2 pi d heta int_0^pi / 2 cos^3 phi sin phi dphi ight> = 16 cdot 2 pi cdot left( -frac14 cos^4 phi ight) vert_0^pi / 2 $$

$$ = 16 cdot 2 pi cdot frac14 = 8 pi . $$

b) utilizing the graph function --

$$ z = g(x,y) = sqrt4 - x^2 - y^2 Rightarrow z^2 = 4 - x^2 - y^2 $$

$$ Rightarrow 2 z fracpartial gpartial x = - 2 x , 2 z fracpartial gpartial y = - 2 y $$

$$ Rightarrow fracpartial gpartial x = - fracxz , fracpartial gpartial y = - fracyz $$

$$ Rightarrow iint_S inter-base.netbfF cdot inter-base.netbfhatn dS = iint_D -F_x fracpartial gpartial x -F_y fracpartial gpartial y + F_z dA $$

$$ = iint_D -x^2 left(- fracxz ight) -y^2 left(- fracyz ight) + z^2 dA = iint_D left(fracx^3 + y^3z ight) + z^2 dA . $$

This surface ar integral is performed over the projected area of the hemispherical surface onto the $ xy-$ plane, which is a disk of radius 2 ; this lends itself fine to the use of polar coordinates:

$$ iint_S inter-base.netbfF cdot inter-base.netbfhatn dS $$

$$ = int_0^2 pi int_0^2 fracr^3 ( cos^3 heta + sin^3 heta )sqrt4 - r^2 r dr d heta + int_0^2 pi int_0^2 ( 4 - r^2 ) r dr d heta $$

$$ = 0 + int_0^2 pi d heta int_0^2 ( 4r - r^3 ) dr = 2 pi left( 2r^2 - frac14r^4 ight) vert_0^2 $$

$$ = 2 pi ( 8 - 4 ) = 8 pi . $$

This then is the "outward" flux with the hemispherical surface. Us can apply the divergence Theorem over the volume that the hemisphere as a check:

$$ abla cdot inter-base.netbfF = 2x + 2y + 2z $$

$$ Rightarrow iiint_V abla cdot inter-base.netbfF dV = iiint_V ( 2x + 2y + 2z ) dV $$

$$ = 2 int_0^2 pi int_0^pi / 2 int_0^2 r ( sin phi cos heta + sin phi sin heta + cos phi ) r^2 dr sin phi dphi d heta $$

$$ = 2 int_0^2 pi int_0^pi / 2 int_0^2 r^3 ( sin^2 phi < cos heta + sin heta > + cos phi sin phi ) dr dphi d heta $$

$$ = 2 left< int_0^2 pi int_0^pi / 2 int_0^2 r^3 sin^2 phi < cos heta + sin heta > dr dphi d heta quad + int_0^2 pi int_0^pi / 2 int_0^2 r^3 cos phi sin phi dr dphi d heta ight> $$

$$ = 2 left< 0 + int_0^2 pi d heta int_0^pi / 2 cos phi sin phi dphi int_0^2 r^3 dr ight> $$

$$ = 2 cdot 2 pi int_0^pi / 2 frac12 sin 2 phi dphi int_0^2 r^3 dr = 2 cdot 2 pi left( -frac14 cos 2 phi ight) vert_0^pi / 2 cdot left( frac14r^4 ight) vert_0^2 $$

$$ = 2 cdot 2 pi left( -frac14 ight) ( <-1> - 1 ) cdot frac14 cdot 2^4 = 8 pi . $$

The exterior flux through the hemispherical surface ar is equal to this amount much less the flux v the base of the hemisphere. Yet this circular base (of radius 2) lies in the $ xy-$ plane ( $ z = 0 $ ) , so us have

$$ iint_B inter-base.netbfF cdot inter-base.netbfhatn dS = iint_B langle x^2 , y^2 , 0^2 angle cdot langle 0 , 0 , -1 angle dS = 0 . $$

As over there is no $ z-$ ingredient of the field $ inter-base.netbfF $ in the $ xy-$ plane, over there is no flux through the basic of the hemisphere. Hence, we confirm our an outcome for the flux through the hemispherical surface.

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For (2), we attend to the "upper" nappe that the cone having actually the equation $ z = 2 sqrtx^2 + y^2 $ , or $ z^2 = 4x^2 + 4y^2 , z ge 0 $ . Plenty of of the aspects of this problem are similar to the an initial one, so we will sophisticated less top top the details. One peculiarity below is the the "upward" normal to the conical surface ar is chosen, i beg your pardon points "into" the volume of the nappe; thus, us take the an unfavorable of the standard definition for the typical vector.

a) using the gradient --

$$ g(x,y,z) = 4x^2 + 4y^2 - z^2 Rightarrow abla g = langle 8x , 8y , -2z angle $$

$$ Rightarrow | abla g | = sqrt(8x)^2 + (8y)^2 + (-2z)^2 = sqrt64 (x^2 + y^2) + 4z^2 $$$$ = sqrt16z^2 + 4z^2 = 2 sqrt5 z $$

$$ Rightarrow inter-base.netbfhatn = -frac abla g = -frac1sqrt5 langle frac4xz , frac4yz , -1 angle quad ext< "upward" unit normal> $$

$$ Rightarrow inter-base.netbfF cdot inter-base.netbfhatn = frac1sqrt5 left( -frac4x^3z - frac4y^3z + z^2 ight) $$

$$ Rightarrow iint_S inter-base.netbfF cdot inter-base.netbfhatn dS = frac1sqrt5 int_0^2 pi int_0^1 left< - frac4r^3 (cos^3 heta + sin^3 heta) 2r + (2r)^2 ight> (sqrt5) r dr d heta $$

$$ = int_0^2 pi d heta int_0^1 4r^3 dr = 2 pi ( r^4 ) vert_0^1 = 2 pi . $$

b) making use of the graph role --

$$ z = g(x,y) = 2 sqrtx^2 + y^2 Rightarrow z^2 = 4 x^2 + 4 y^2 $$

$$ Rightarrow 2 z fracpartial gpartial x = 8 x , 2 z fracpartial gpartial y = 8 y Rightarrow fracpartial gpartial x = frac4xz , fracpartial gpartial y = frac4yz $$

$$ Rightarrow iint_S inter-base.netbfF cdot inter-base.netbfhatn dS = iint_D -x^2 left( frac4xz ight) -y^2 left( frac4yz ight) + z^2 dA $$$$ = iint_D -left(frac4x^3 + 4y^3z ight) + z^2 dA , $$

giving united state the very same integration over the disk of radius 1 on the $ xy-$ plane, which is the projection of the conical surface, and also therefore the same an outcome for the "upward" flux v the conical surface.

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Making a examine using the divergence Theorem, we integrate in cylindrical collaborates over the volume of the cone up to $ z = 2 $ to obtain

$$ iiint_V abla cdot inter-base.netbfF dV = iiint_V ( 2x + 2y + 2z ) dV $$

$$ = 2 int_0^2 pi int_0^1 int_0^2r ( r cos heta + r sin heta + z ) dz r dr d heta $$

$$ = 2 int_0^2 pi int_0^1 left( rz < cos heta + sin heta > + frac12z^2 ight) vert_0^2r r dr d heta $$

$$ = 2 int_0^2 pi int_0^1 2r^3 ( cos heta + sin heta ) + frac12 (2r)^2 cdot r dr d heta $$

$$ = 2 int_0^2 pi d heta int_0^1 2r^3 dr = 2 cdot 2 pi left(frac12r^4 ight) vert_0^1 = 2 cdot 2 pi frac12 = 2 pi . $$

At the level $ z = 2 $ , the ar is $ inter-base.netbfF = langle x^2 , y^2 , 2^2 angle $ , thus for the peak surface that the conical volume,

$$ inter-base.netbfF cdot inter-base.netbfhatn = langle x^2 , y^2 , 4 angle cdot langle 0 , 0 , 1 angle = 4 . $$

The "upward" flux through this surface, a one of radius 1, is then

$$ iint_T inter-base.netbfF cdot inter-base.netbfhatn dS = 4 cdot pi cdot 1^2 = 4 pi . $$

With this amount of "upward" flux v the peak of the conical volume and also a net "upward" flux of $ 2 pi $ v that volume, the "upward" flux v the cone "wall" must be $ 2 pi $ , as we have discovered from the flux integration.

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One extr comment might be do here. Through every one of this discussion, us have found the section of the flux integrations including the azimuthal angle $ heta $ to constantly be zero. The geometrical translate for this is the the materials of the ar $ inter-base.netbfF $ that space parallel come the $ xy-$ airplane are not just non-negative, however are symmetrical about the line $ y = x $ . Since the "horizontal" cross-sections of both the hemisphere and the cone are circular and centered top top the $ z-$ axis, the flux entering the borders of these cross-sections on one side of the line $ y = -x $ exactly matches the flux leave these borders on the various other side of the line (as might be viewed in the graph below). So only the $ z-$ ingredient of the ar makes any kind of contribution to the net flux v the surface we have actually examined.