MATH 263 ASSIGNMENT 9 SOLUTIONS. F dv =

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1 MAH AIGNMEN 9 OLUION ) Let F = (x yz)î + (y + xz)ĵ + (z + xy)ˆk and let be the portion of the cylinder x + y = that lies inside the sphere x + y + z = 4 be the portion of the sphere x + y + z = 4 that lies outside the cylinder x + y = be the volume bounded by and ompute a) F ˆn d with ˆn pointing inward b) F d c) F ˆn d with ˆn pointing outward Use the divergence theorem to answer at least one of parts (a), (b) and (c). olution. Observe that F =. o F d = he horizontal cross-section of at height z is a washer with outer radius 4 z (determined by the equation of the sphere) and inner radius (determined by the equation of the cylinder). o the crosssection has area π ( 4 z ) π = π ( z ). On the intersection of the sphere and cylinder z = 4 = so F d = π ( z ) dz = π d On the cylindrical sur, using (surprise!) cylindrical coordinates, ˆn = ( îı + sin θĵj ) so d = dθ dz ( z ) dz = π ( ) / = 8 π F ˆn = ( z sin θ )( ) + ( sin θ + z )( sin θ ) = By the divergence theorem F ˆn d = F ˆn d = dz π dθ = 8 π F d F ˆn d = π ) Evaluate the integral F ˆn d, where F = (x, y, ) and is the sur z = x y, for x +y, by two methods. a) First, by direct computation of the sur integral. b) econd, by using the divergence theorem. olution. a) Let G(x, y, z) = x + y + z. hen witching to polar coordinates G ˆn d = xîı+yĵj+ˆk dx dy = G ˆk dx dy = (xîı + yĵj + ˆk) dx dy F ˆn d = xîı + yĵj + ˆk xîı + yĵj + ˆk dx dy = x + y + dx dy F ˆn d = dr r π dθ (r + ) = π r4 + r = π

2 b) all the solid z x y,. Let denote the bottom sur of. he disk has radius, area π, z = the outward normal ˆk, so that F ˆn d = F ˆk dx dy = dx dy = π As the divergence theorem gives F ˆn d = = π + F = x (x) + y (y) + z () = F d dz x +y z = π + π z z = π F ˆn d = dxdy = π + d ( π) dz π( z) a) By applying the divergence theorem to F = φ a, where a is an arbitrary constant vector, show that φ d = φˆn d b) how that the centroid ( x, ȳ, z) of a solid is given by ( x, ȳ, z) = vol( ) (x + y + z ) ˆn d olution. a) he divergence of φ a is φ a. o, by the divergence theorem, φ a ˆn d = φ a d = φˆn d φ d his is true for all vectors a. o b) By part a, with φ = x + y + z, vol( ) φˆn d (x + y + z ) ˆn d = vol( ) φ d = a = (xîı + yĵj + zˆk) d = ( x, ȳ, z) 4) Find the flux of F = (y + xz)îı+ (y + yz)ĵj (x + z )ˆk upward through the first octant part of the sphere x + y + z = a. olution. Let = { (x, y, z) x + y + z a, x, y, z }. he consists of an x =, a y =, a z = and the first octant part of the sphere. all the latter. hen F d = z + + z z d = d = 4 8 πa = πa a π/ a F ( îı) dy dz = ( y) dy dz = dr r dθ r sin θ = r dr = a x= y= z= F ( ĵj) dx dz = F ( ˆk) dx dy = x= z= (x) dx dy = a

3 By the divergence theorem F ˆn dx dy = F d s x= F ( îı) dy dz y= F ( ĵj) dx dz z= F ( ˆk) dx dy = π 5) Let E( r) be the electric field due to a charge configuration that has density ρ( r). Gauss law states that, if is any solid in IR with sur, then the electric flux E ˆn d = 4πQ where Q = ρ d is the total charge in. Here, as usual, ˆn is the outward pointing unit normal to. how that E( r) = 4πρ( r) for all r in IR. his is one of Maxwell s equations. Assume that E( r) and ρ( r) are well defined and continuous everywhere. a olution. By the divergence theorem E ˆn d = E d o by Gauss law E d = 4π ρ d E 4πρ d = his is true for all solids for which the divergence theorem applies. If there were some point in IR for which E 4πρ were, say, strictly bigger than zero, then, by continuity, we could find a ball B ɛ centered on that point with E 4πρ > everywhere on B ɛ. his would force B ɛ E 4πρ d >, which violates E 4πρ d = with set equal to Bɛ. Hence E 4πρ must be zero everywhere. ) Evaluate, both by direct integration and by tokes heorem, (z dx+x dy +y dz) where is the circle x + y + z =, x + y + z =. Orient so that its projection on the xy plane is counterclockwise. olution. he projection of on the xy plane is x + y + ( x y) = or x + xy + y = or (x + y) + (x y) =. Hence we may parametrize the curve using x+y =, x y = sin θ x(θ) = sin θ x (θ) = sin θ y(θ) = + sin θ and y (θ) = sin θ + z(θ) = z (θ) = sin θ he sign in x y = sin θ has been chosen to make the projected motion counterclockwise. he check this, observe that at θ =, (x, y) = (, ) and ( dx dθ, dy ) dθ = (, ), which is up and to the left. his integral is of the form F d r where F = zîı + xĵj + yˆk and is curve parametrized above. Hence π F d r = F ( r(θ) ) r (θ) dθ = = π π ( sin θ ) + ( sin θ )( sin θ + ) + ( + sin θ )( sin θ ) dθ cos θ + sin θ + ( + ) sin θ dθ = π = π

4 hoose as the portion of the plane x + y + z = interior to the sphere. hen ˆn = (îı + ĵj + ˆk) and F = îı + ĵj + ˆk so, by tokes heorem, F d r = F ˆn d = (îı + ĵj + ˆk) (îı + ĵj + ˆk) d = d = π since is a circle of radius. 7) Evaluate (x sin y y )dx + (x y cos y + x)dy where is the counterclockwise boundary of the trapezoid with vertices (, ), (, ), (, ) and (, ). olution. By Green s theorem (or tokes theorem) (x sin y y )dx + (x y cos y ( + x)dy = x (x y cos y + x) y (x sin y y ) ) dx dy y (, ) ( (, ) = xy cos y + xy cos y + y ) dx dy x ( ) (, ) = + y dx dy (, ) he integral of y vanishes because the domain of integration is invariant under y y. he other integral is times the area of the trapezoid, which is its width () times the average of its heights ( + 4). o (x sin y y )dx + (x y cos y + x)dy = 9. 8) Evaluate F d r where F = ye x îı + (x + e x )ĵj + z ˆk and is the curve r(t) = ( + cos t)îı + ( + sin t)ĵj + ( sin t cos t)ˆk olution. F = ( + e x e x)ˆk, so by tokes heorem F d r = ˆk d where is the intersection of x + y + z = with (x ) + (y ). Now ˆk d is the area of the projection of on the xy plane. his projection is the circle of radius centred on (, ), which has area π. o F d r = π. 9) Let be the intersection of x+y z = 7 and x x+4y = 5. he curve is oriented counterclockwise when viewed from high on the z axis. Let Evaluate F d r. F = ( x e x + yz ) îı + ( sin y y + sin z x ) ĵj + ( xy + y cos z )ˆk olution. By tokes heorem and the observation that F = xîı (z + x)ˆk F d r = F ˆn d where is the part of x + y z = 7 inside (x ) + 4y = = xîı (z + x)ˆk z= 7+x+y (,, ) dx dy = 7 4x y dx dy = (area of ellipse with semi axes a = 4, b = )7 4 x ȳ = π = 4π 4

5 ) onsider ( F ) ˆn d where is the portion of the sphere x +y +z = that obeys x+y +z, ˆn is the upward pointing normal to the sphere and F = (y z)îı + (z x)ĵj + (x y)ˆk. Find another sur with the property that ( F ) ˆn d = ( F ) ˆn d and evaluate ( F ) ˆn d. olution. Let be the portion of x + y + z = that is inside the sphere x + y + z =. hen =, so, by tokes heorem, (with ˆn always the upward pointing normal) ( F ) ˆn d = F d r = As F = (îı + ĵj + ˆk) and, on, ˆn = (îı + ĵj + ˆk) F d r = ( F ) ˆn d ( ( ) F ) ˆn d = d = Area( ) is a circular disk. It s center (x c, y c, z c ) has to obey x c + y c + z c =. By symmetry, x c = y c = z c, so x c = y c = z c =. Any point, like (,, ), which satisfies both x + y + z = and x + y + z = is on the boundary of. o the radius of is (,, ( ) (,, ) =,, ) =. o the area of is π and ( F ) ˆn d = Area( ) = 4 π 5