MATH 255 Applied Honors Calculus III Winter Homework 11. Due: Monday, April 18, 2011

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1 MATH 255 Applied Honors Calculus III Winter 211 Homework 11 ue: Monday, April 18, 211 ection 17.7, pg. 1155: 5, 13, 19, 24. ection 17.8, pg. 1161: 3, 7, 13, 17 ection 17.9, pg. 1168: 3, 7, 19, # 5 z = g(x,y) = 1+2x+3y. x 2 yd = 3 2 = 14 3 = x2 yz 1+g 2 x +g2 y da x2 y(1+2x+3y) 1+4+9dydx 2 (x2 y +2x 3 y +3x 2 y 2 )dydx = 14 3 (x2 y 2 /2+x 3 y 2 +3x 2 y 3 ) 2 dx = 14 3 (1x2 +4x 3 )dx = # 13 (x2 z +y 2 z)d. is a hemisphere, so spherical coordinates seem reasonable. ρ = 2, φ π/2, θ 2π. r(u,v) = 2sinφcosφ,2sinφsinθ,2cosφ. In spherical coordinates, r φ r θ = ρ 2 sinφ = 4sinφ. (x 2 z +y 2 z)d = (4sin2 φcos 2 θ2cosφ+4sin 2 φcos 2 θ2cosφ)4sinφdφdθ = 32 2π π/2 sin 3 φcosφdφdθ = 64π π/2 sin 3 φcosφdφ = 32π 1 udu = 16π The substitution was u = sin 2 φ so du = 2sinφcosφ # 19 : z = 4 x 2 y 2 over x 1 and y 1. The paraboloid begins to suggest cylindrical coordinates, but the domain does not have cylindrical symmetry. o we ll keep rectangular coordinates. F = P,Q,R = xy,yz,zx. 1

2 F d = 1 1 = g ( P x Q g +R)dA y = 1 1 [ (xy)( 2x) (yz)( 2y)+(zx)]dxdy [ (xy)( 2x) y(4 x2 y 2 )( 2y)+(4 x 2 y 2 )x]dxdy = 1 1 [4x x3 +2x 2 y +8y 2 xy 2 2x 2 y 2 2y 4 ]dxdy = 713/18 ection 17.7, #24 Evaluate the surface integral F d for the vector field F(x,y,z) = xz i+x j +y k and the oriented surface given by the hemisphere x 2 +y 2 +z 2 = 25, y, oriented in the direction of the positive y-axis. In other words, find the flux of F across. olution. As in problem #13, we need to start by choosing a parametrization of, and any one will do. Given a parametrization r = r(u,v) of, with (u,v) R 2, the formula is F d = ± F( r(u,v)) [ r u (u,v) r v (u,v)] da(u,v), where the sign is chosen so that ± r u r v points in the direction of desired orientation. One parametrization of the hemisphere is to take x = u, z = v, and y = 25 u 2 v 2, where the domain is = {(u,v) such that u 2 +v 2 25}. Then, r(u,v) = u, 25 u 2 v 2,v, so u v r u (u,v) r v (u,v) = 1, 25 u2 v 2,, 25 u2 v 2,1 u v = 25 u2 v 2, 1,. 25 u2 v 2 It is clear by examination of the y-component that this normal vector to points in the opposite direction desired; therefore we select the minus sign and obtain F d = F( r(u,v)) [ r u (u,v) r v (u,v)] da(u,v) u v = F( r(u,v)) 25 u2 v 2,1, da(u,v). 25 u2 v 2 Finally, using the parametrization of we have F( r(u,v)) = uv,u, 25 u 2 v 2, so evaluating the dot product, we arrive at the double integral formula [ ] F d u 2 v = 25 u2 v +u+v da(u,v). 2 This double integral is going to turn out to be zero, basically because each term in the integrand is an odd function of either u or v, and the region is symmetrical with respect to u u and v v. But to confirm this we should just evaluate the integral by iterated integration as follows: u 2 [ F d u v 2 = ]vdvdu+ 25 u 2 25 u2 v +1 ududv =, 2 25 v 2 5 5

3 since in each case, the inner integral is identically zero because the integrand is an odd function of the integration variable while the interval of integration is symmetric about zero. Alternatively, we might choose to parametrize by spherical coordinates in which we take the north-south axis to be the y-axis instead of the z-axis. Thus we have x = 5sin(v)cos(u), y = 5cos(v), and z = 5sin(v)sin(u), with longitude angle u [,2π] and latitude angle v [,π/2], so the domain in the (u,v)-plane is the rectangle = [,2π] [,π/2]. Then, r(u,v) = 5sin(v)cos(u),5cos(v),5sin(v)sin(u), so r u (u,v) = 5sin(v)sin(u),,5sin(v)cos(u) and r v (u,v) = 5cos(v)cos(u), 5sin(v),5cos(v)sin(u), and therefore a normal vector to is given by r u (u,v) r v (u,v) = 25 sin 2 (v)cos(u),sin(v)cos(v)cos 2 (u)+sin(v)cos(v)sin 2 (u),sin 2 (v)sin(u) = 25 sin(v) sin(v) cos(u), cos(v), sin(v) sin(u). ince v π/2, we see that the y-component of this normal vector is nonnegative, so the vector is pointing in the desired direction for the orientation of. Therefore, for this parametrization we need to take the plus sign in the formula: F d r = F( r(u,v)) [ r u (u,v) r v (u,v)] da(u,v) = 25 sin(v) F( r(u,v)) sin(v)cos(u),cos(v),sin(v)sin(u) da(u,v). Finally, we note that in this parametrization, F( r(u,v)) = 25sin 2 (v)sin(u)cos(u),5sin(v)cos(u),5cos(v), o computing the dot product in the integrand gives F d r = 125 = 125 = 125 = 125 =. [ 5sin 4 (v)sin(u)cos 2 (u)+sin 2 (v)cos(v)cos(u)+sin 2 (v)cos(v)sin(u) ] da(u,v) π/2 2π π/2 π/2 [ sin 4 (v)cos 2 (u)[ sin(u)]+sin 2 (v)cos(v)cos(u)+sin 2 (v)cos(v)sin(u) ] dudv [ 1 3 sin4 (v)cos 3 (u)+sin 2 (v)cos(v)sin(u) sin 2 (v)cos(v)cos(u) dv 17.8 # 3 Using tokes s theorem means we want F d r, F = x 2 e yz,y 2 e xz,z 2 e xy. is the top hemisphere of radius 2, so the boundary is the circle of radius 2 in the xy plane. r(t) = 2cost,2sint,. r (t) = 2sint,2cost,. ] u=2π u= dv

4 ( F) d = F d r = 2π F r (t)dt = 2 2π ( x(t)2 e y(t)z(t) sint+y(t) 2 e z(t)x(t) cost)dt = 8 2π ( cost2 e sint+sint 2 e cost)dt = 8 2π ( cost2 sint+sint 2 cost)dt = [ cos 3 t/3+sin 3 t/3 ] 2π = 17.8 # 7 To use tokes s theorem, we identify the surface, which is the part of a plane in the 1st octant. The plane goes through (1,,), but most importantly, has normal vector n = 1/ 3 1,1,1. This is clear, since the x-, y-, and z-directions are all equivalent. The curl is F = 2 z,x,y. F d r = 1 = F d 1 x = 1 F 1,1,1 dx dy = 1 1 x ( 2z 2x 2y) dx dy 1 x ( 2(1 x y) 2x 2y) dx dy = 1 1 x ( 2) dx dy = 2+1 = # 13 We want to confirm tokes s theorem for a paraboloid z = x 2 +y 2, z 1 and F = y 2,x,z 2. a) F d F = î ĵ ˆk x x x y 2 x z 2 = î() ĵ()+ˆk(1 2y) F =,,1 2y = P,Q,R

5 The surface is a graph of z = g(x,y) = x 2 +y 2, so we can use the following approach: F d = ( g xp g y Q+R)dA = ( g x g y +1 2y)dA = 2π = 2π = (1 2y)dA 1 (1 2rcosθ)r dr dθ 1 (r 2r2 cosθ)dr dθ = 2π (1/2 2/3cosθ)dθ = 2π/2 = π b) F d r The boundary of the surface is the curve around the open end. That is r(t) = 1 cost,1 sint,1, θ 2π. o r (t) = sint,cost,. F d r = b a F( r(t)) r (t)dt = 2π y(t)2,x(t),z(t) 2 sint,cost, dt = 2π sin2 t,cost,1 sint,cost, dt = 2π sin 3 t+cos 2 tdt = [ 1/3(2+sin 2 t)cost ] 2π +[t/2+1/4sin2t]2π = +π + = π The two match, ( F) d = F d r 17.8 # 17 The work around those edges is C F d r, where C is a closed curve. Clearly, given this F, we can t simply integrate. We use tokes s theorem. The curl is F = 2y,2z,2x. We can parameterize the surface as r(θ,φ) = 2sinφcosθ,2sinφsinθ,2cosφ. This leads to: r φ r θ = 4cosθsin 2 φ,4sinθsin 2 φ,4sinφcosφ.

6 F d r = π/2 π/2 = π/2 = π/2 = F d = π/2 π/2 ( F) ( r φ r θ )da π/2 2y,2z,2x 4cosθsin 2 φ,4sinθsin 2 φ,4sinφcosφ da 4sinφsinθ,4cosφ,4sinφcosθ 4cosθsin 2 φ,4sinθsin 2 φ,4sinφcosφ da π/2 (16cosθsinθsin 3 φ+16cosφsin 2 φsinθ +16cosθsin 2 φcosφ)dθ dφ = π/2 π/2 (8sin2θsin 3 φ+32cosφsin 2 φsinθdθ dφ = π/2 (8sin 3 φ+32cosφsin 2 φ dφ = [ (8( 3/4cosφ+1/12cos3φ)+32(sin 3 φ/3) ] π/2 = [ (8( 3/4cosφ+1/12cos3φ)+32(sin 3 φ/3) ] π/2 = 16 The table in the back of the book was used for integrating sin 3 φ, and sin 2 φcosφ is recognized as the derivative of sin 3 φ/ # 3 Calculate in both ways. a) FdV F = 3+x+2x = 3+3x E FdV = (3+3x) dy dz dx = 1 (3+3x) dx = 3+3/2 = 9/2 b) F d The boundary of the box is the union of the six surfaces. We need a normal vector for each, but this is easy. E.g. top has n = ˆk to point upwards.

7 F d = F ˆkd + F ( ˆk)d + F îd top bottom front + F ( î)d + F ( ĵ)d + F ĵd back left right F d = 2xzd 2xzd + 3xd top bottom front 2xd xyd + xyd back left right F d = 2x 1d 2x d + 3 1d top bottom front 2 d x d + x 1d back left right F d = 3+3/2 = 9/2 The same! 17.9 # 7 The divergence is F = e x siny e x siny +2yz = 2yz. The flux is: F d = 2 = E FdV 1 1 2yz dx dy dz = # 19 The trick here is to note that the divergence theorem relates F d = V FdV. But, in this case, = 2. We want F d = 2 F d 1 F d. The second term is easy to compute, but the first would be hard without the divergence theorem. F = z 2 x,y 3 /3+tanz,x 2 z +y 2. Thus F = z 2 +y 2 +x 2. F d = V FdV = V (z2 +y 2 +x 2 )dv Given the hemisphere solid and the sum of square terms, spherical coordinates seem like a good choice. F d = V (z2 +y 2 +x 2 )dv = π/2 2π 1 (ρ)2 ρ 2 sinφ dρ dθ dφ = π/2 sinφdφ 2π dθ 1 (ρ)4 = 2π/5

8 Now for: 1 F d. The normal vector is clearly ˆk. Recall that z = on 1. 1 F d = 1 F ˆkd = 1 x 2 z +y 2 d = 1 y 2 d = 1 2π r 2 sin 2 θdθdr = π/4dr F d = 2 F d 1 F d = 2π/5 ( π/4) = 13π/ # 25 is the boundary of V. That is = V. Thus is a closed surface. ( F) d = ( F) d = ( F)dV = dv =. V V V Using the identity ( F) =.