Find the flux the the vector ar $F =$ outward across the given surfaces. Each surface ar is oriented, unless otherwise specified, with outward-pointing common pointing away from the origin.

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the top hemisphere that radius 2 centered at the origin.the cone $z = 2\sqrtx^2+y^2$, $z$ = 0 come 2 with exterior normal pointing upward  We have the right to make the flux calculation because that each surface straight by assessing the surface integral $\ \iint_S \ \tastecraftedmcd.combfF \cdot \tastecraftedmcd.combf\hatn \ \ dS \$ , and likewise by using the divergence Theorem as a check.

For (1), the "upper" hemisphere of radius 2 focused on the origin has actually 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" normal to the hemispherical surface, then continuing on to compute the flux integral:

$$g(x,y,z) \ = \ x^2 \ + \ y^2 \ + \ z^2 \ - \ 4 \ \ \Rightarrow \ \ \nabla g \ = \ \langle \ 2x \ , \ 2y \ , \ 2z \ \rangle$$

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

$$\Rightarrow \ \ \tastecraftedmcd.combf\hatn \ = \ \frac\nabla g \nabla g \ \ \ = \ \langle \ \fracx2 \ , \ \fracy2 \ , \ \fracz2 \ \rangle \quad \text$$

$$\Rightarrow \ \ \tastecraftedmcd.combfF \cdot \tastecraftedmcd.combf\hatn \ = \ \langle \ x^2 \ , \ y^2 \ , \ z^2 \ \rangle \cdot \ \langle \ \fracx2 \ , \ \fracy2 \ , \ \fracz2 \ \rangle \ = \ \fracx^3 \ + \ y^3 \ + \ z^32 \ \ .$$

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

$$\ \iint_S \ \tastecraftedmcd.combfF \cdot \tastecraftedmcd.combf\hatn \ \ dS$$

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

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

$$= \ \ \frac12 \cdot \ 2^5 \ \left< \ \int_0^2 \pi \int_0^\pi / 2 \ \sin^4 \phi \ ( \ \cos^3 \theta \ + \ \sin^3 \theta \ ) \ \ d\phi \ d\theta \quad + \ \ \int_0^2 \pi \int_0^\pi / 2 \ \cos^3 \phi \ \sin \ \phi \ \ d\phi \ d\theta \ \right>$$

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

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

b) utilizing the graph duty --

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

$$\Rightarrow \ \ 2 z \ \frac\partial g\partial x \ = \ - 2 x \ \ , \ \ 2 z \ \frac\partial g\partial y \ = \ - 2 y$$

$$\Rightarrow \ \ \frac\partial g\partial x \ = \ - \fracxz \ \ , \ \ \frac\partial g\partial y \ = \ - \fracyz$$

$$\Rightarrow \ \ \iint_S \ \tastecraftedmcd.combfF \cdot \tastecraftedmcd.combf\hatn \ \ dS \ \ = \ \ \iint_D \ -F_x \ \frac\partial g\partial x \ -F_y \ \frac\partial g\partial y \ + \ F_z \ \ dA$$

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

This surface integral is performed over the projected area that the hemispherical surface ar onto the $\ xy-$ plane, i m sorry is a decaying of radius 2 ; this lends itself fine to the use of polar coordinates:

$$\iint_S \ \tastecraftedmcd.combfF \cdot \tastecraftedmcd.combf\hatn \ \ dS$$

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

$$= \ \ 0 \ \ + \ \ \int_0^2 \pi d\theta \ \int_0^2 \ ( \ 4r \ - \ r^3 ) \ \ dr \ \ = \ 2 \pi \ \left( \ 2r^2 \ - \ \frac14r^4 \right) \vert_0^2$$

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

This climate is the "outward" flux v the hemispherical surface. We can apply the divergence Theorem over the volume that the hemisphere together a check:

$$\ \nabla \cdot \tastecraftedmcd.combfF \ = \ 2x \ + \ 2y \ + 2z$$

$$\Rightarrow \ \ \iiint_V \ \nabla \cdot \tastecraftedmcd.combfF \ \ dV \ \ = \ \ \iiint_V \ ( \ 2x \ + \ 2y \ + 2z \ ) \ \ dV$$

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

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

$$= \ \ 2 \ \left< \ \int_0^2 \pi \int_0^\pi / 2 \int_0^2 \ r^3 \ \sin^2 \phi \ < \ \cos \theta \ + \ \sin \theta \ > \ \ dr \ d\phi \ d\theta \quad + \ \ \int_0^2 \pi \int_0^\pi / 2 \int_0^2 \ r^3 \ \cos \phi \ \sin \phi \ \ dr \ d\phi \ d\theta \ \right>$$

$$= \ \ 2 \ \left< \ 0 \ \ + \ \ \int_0^2 \pi d\theta \ \int_0^\pi / 2 \cos \phi \ \sin \phi \ \ d\phi \ \int_0^2 \ r^3 \ \ dr \ \right>$$

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

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

The external flux v the hemispherical surface ar is equal to this amount less the flux v the basic of the hemisphere. But this circular basic (of radius 2) lies in the $\ xy-$ airplane ( $\ z \ = \ 0 \$ ) , so we have

$$\iint_B \ \tastecraftedmcd.combfF \cdot \tastecraftedmcd.combf\hatn \ \ dS \ \ = \ \ \iint_B \ \langle \ x^2 \ , \ y^2 \ , \ 0^2 \ \rangle \cdot \langle \ 0 \ , \ 0 \ , \ -1 \ \rangle \ \ dS \ = \ 0 \ \ .$$

As over there is no $\ z-$ component of the field $\ \tastecraftedmcd.combfF \$ in the $\ xy-$ plane, there is no flux with the base of the hemisphere. Hence, we confirm our result for the flux v the hemispherical surface.

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For (2), we address the "upper" nappe the 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 facets of this difficulty are similar to the first one, so us will elaborate less top top the details. One peculiarity below is the the "upward" common to the conical surface ar is chosen, which points "into" the volume that the nappe; thus, we take the negative of the standard definition for the typical vector.

$$g(x,y,z) \ = \ 4x^2 \ + \ 4y^2 \ - \ z^2 \ \ \Rightarrow \ \ \nabla g \ = \ \langle \ 8x \ , \ 8y \ , \ -2z \ \rangle$$

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

$$\Rightarrow \ \ \tastecraftedmcd.combf\hatn \ = \ -\frac\nabla g \ = \ -\frac1\sqrt5 \ \langle \ \frac4xz \ , \ \frac4yz \ , \ -1 \ \rangle \quad \text< "upward" unit normal>$$

$$\Rightarrow \ \ \tastecraftedmcd.combfF \cdot \tastecraftedmcd.combf\hatn \ = \ \frac1\sqrt5 \left( -\frac4x^3z \ - \ \frac4y^3z \ + \ z^2 \right)$$

$$\Rightarrow \ \ \iint_S \ \tastecraftedmcd.combfF \cdot \tastecraftedmcd.combf\hatn \ \ dS \ \ = \ \ \frac1\sqrt5 \ \int_0^2 \pi \int_0^1 \ \left< \ - \frac4r^3 \ (\cos^3 \theta \ + \ \sin^3 \theta) 2r \ + \ (2r)^2 \ \right> \ \ (\sqrt5) \ r \ dr \ d\theta$$

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

b) using the graph function --

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

$$\Rightarrow \ \ 2 z \ \frac\partial g\partial x \ = \ 8 x \ \ , \ \ 2 z \ \frac\partial g\partial y \ = \ 8 y \ \ \Rightarrow \ \ \frac\partial g\partial x \ = \ \frac4xz \ \ , \ \ \frac\partial g\partial y \ = \ \frac4yz$$

$$\Rightarrow \ \ \iint_S \ \tastecraftedmcd.combfF \cdot \tastecraftedmcd.combf\hatn \ \ dS \ \ = \ \ \iint_D \ -x^2 \left( \frac4xz\right) \ -y^2 \left( \frac4yz\right) \ + \ z^2 \ \ dA$$$$= \ \ \iint_D \ -\left(\frac4x^3 \ + \ 4y^3z\right) \ + \ z^2 \ \ dA \ \ ,$$

giving us the exact same integration end the disk of radius 1 top top the $\ xy-$ plane, i beg your pardon is the forecast of the conical surface, and also therefore the same result for the "upward" flux through the conical surface.

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Making a inspect using the divergence Theorem, we combine in cylindrical collaborates over the volume the the cone approximately $\ z \ = \ 2 \$ come obtain

$$\iiint_V \ \nabla \cdot \tastecraftedmcd.combfF \ \ dV \ \ = \ \ \iiint_V \ ( \ 2x \ + \ 2y \ + 2z \ ) \ \ dV$$

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

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

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

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

At the level $\ z \ = \ 2 \$ , the ar is $\ \tastecraftedmcd.combfF \ = \ \langle \ x^2 \ , \ y^2 \ , \ 2^2 \ \rangle \$ , thus for the top surface of the conical volume,

$$\tastecraftedmcd.combfF \cdot \tastecraftedmcd.combf\hatn \ = \ \langle \ x^2 \ , \ y^2 \ , \ 4 \ \rangle \cdot \ \langle \ 0 \ , \ 0 \ , \ 1 \ \rangle \ = \ 4 \ \ .$$

The "upward" flux v this surface, a circle of radius 1, is then

$$\iint_T \ \tastecraftedmcd.combfF \cdot \tastecraftedmcd.combf\hatn \ \ dS \ \ = \ 4 \ \cdot \ \pi \ \cdot \ 1^2 \ = \ 4 \pi \ \ .$$

With this lot of "upward" flux through the optimal of the conical volume and a net "upward" flux of $\ 2 \pi \$ with that volume, the "upward" flux through the cone "wall" have to be $\ 2 \pi \$ , as we have found from the flux integration.

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One additional comment might be made here. Through all of this discussion, us have uncovered the section of the flux integrations involving the azimuthal angle $\ \theta \$ to constantly be zero. The geometrical interpretation for this is the the contents of the field $\ \tastecraftedmcd.combfF \$ that are parallel to the $\ xy-$ plane are not only non-negative, but are symmetrical about the line $\ y = \ x \$ . Due to the fact that the "horizontal" cross-sections the both the hemisphere and the cone space circular and also centered ~ above the $\ z-$ axis, the flux entering the limits of these cross-sections top top one next of the line $\ y = -x \$ exactly matches the flux leaving these borders on the other side the the line (as might be seen in the graph below). Therefore only the $\ z-$ ingredient of the field makes any contribution to the network flux through the surfaces we have actually examined.