# Now φ θ 3 sin θ 3 cos θ 0 φ z 0 1 and φ

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Unformatted text preview: Now φ θ = (- 3 sin θ, 3 cos θ, 0), φ z = (0 , , 1) and φ θ × φ z = (3 cos θ, 3 sin θ, 0). Hence φ θ × φ z (0 , z ) = (3 , , 0) which is outward pointing. integraldisplay S F · n dS = integraldisplay θ,z F · φ θ × φ z dz dθ = integraldisplay 2 π integraldisplay 3+3 cos θ (3 cos θ, 3 sin θ, 3 z sin θ ) · (3 cos θ, 3 sin θ, 0) dz dθ = integraldisplay 2 π integraldisplay 3+3 cos θ (9 cos 2 θ + 9 sin 2 θ ) dz dθ = 9 integraldisplay 2 π integraldisplay 3+3 cos θ dz dθ = 9 integraldisplay 2 π parenleftBig a24 a24 a24 a24 a58 = 0 3 cos θ + 3 parenrightBig dθ-2 2 4 x-2 2 y 2 4 6 z n z Equal z Equal x Plus 3-2 2 4 x = 27 (2 π ) = 54 π . 7. If γ ( t ) = ( r cos t, r sin t ) bounds the disk, γ prime ( t ) = (- r sin t, r cos t ) will be tangent to the disk. Put v = ( r cos t, r sin t ). Then v · γ prime ( t ) = 0 and v points away from the disk; hence the required normal is n = v bardbl v bardbl = (cos t, sin t ). Now F ( γ ( t )) = parenleftbigg r cos t r 2 , r sin t r 2 parenrightbigg = 1 r (cos t, sin t ) and F · n = 1 r (cos t, sin t ) · (cos t, sin t ) = 1 r . Also bardbl γ prime ( t ) bardbl = radicalBig r 2 (cos 2 t + sin 2 t ) = | r | = r . Hence integraldisplay 2 π F ( γ ( t )) · n ( γ ( t )) bardbl γ prime ( t ) bardbl dt = integraldisplay 2 π parenleftbigg 1 r parenrightbigg ( r ) dt = 2 π . Computing the divergence, we have div F = x 2 + y 2- 2 x 2 ( x 2 + y 2 ) 2 + x 2 + y 2- 2 y 2 ( x 2 + y 2 ) 2 = 0, so integraldisplay γ F · n ds = 2 π negationslash = 0 = integraldisplay interior of γ div F dA . This fact does not contradict Stokes’ Theorem since Stokes’ Theorem does not apply ( F is not defined at = (0 , 0)). The region D between the circle and the ellipse does not contain the origin, so F is defined and of class C 1 in all of D . Here we can apply Stokes’ Theorem giving integraldisplay ∂D F · n ds = integraldisplay D div F dS = 0. MATB42H Solutions # 11 page 4 8. The intersection of the plane and the ellipsoid is the ellipse γ as shown below. The plane containing γ is orthogonal to the z –axis, so we can think of γ as the boundary of an oriented surface S- the piece of the plane enclosed by γ ....
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• Winter '10
• EricMoore

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