HW13final - Solution for 16.7 2. 4. 10. 13. 14. 18. Problem...

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Unformatted text preview: Solution for 16.7 2. 4. 10. 13. 14. 18. Problem 16.7.19 a) Since F is conservative (test it) we get that curlF = 0 ⇒ F · ndσ = 0dσ = 0. S C F · dr = S × Problem 16.7.24 Yes: If × F = 0, then the circulation of F around the boundary of C of any oriented suface S in the domain of F is zero. The reason is this: By Stokes’ theorem, circulation = C F · dr = × F · ndσ = 0 · ndσ = 0. S S Problem 16.7.26 curlF = ( y 2 − x2 ∂M ∂P ∂N ∂M y 2 − x2 ∂P ∂N − )i+( − )j+( − )k = 0i+0j+ −2 2 + y 2 )2 ∂y ∂z ∂z ∂x ∂x ∂y (x (x + y 2 )2 k = 0. However, x2 + y 2 = 1 ⇒ r = (cos t)i + (sin t)j ⇒ dr = (− sin t)i + (cos t)j ⇒ F = dt 2π (− sin t)i + (cos t)j ⇒ F · dr = sin2 t + cos2 t = 1 ⇒ C F · dr = 0 1dt = 2π dt which is not zero. Problem 16.8.10 Using the divergence theorem we get that the outward flux across the surface of our region, is equal to the integral of the divergence · F over our region. We ∂ ∂ ∂ get that · F = ∂x (6x2 + 2xy ) + ∂y (2y + x2 z ) + ∂z (4x2 y 3 ) = 12x + 2y + 2 + 0 and that the outward flux is 3 π 2 2 Flux = D 3 0 0 π 2 (12x+2y +2)dV = 0 0 0 (12r cos(θ)+2r sin(θ)+2)rdrdθdz = 3 (32 cos(θ) + 16 sin(θ) + 4)dθdz = 3 (32 + 2π + 0 16 )dz = 112 + 6π 3 Problem 16.8.11 ∂ ∂ ∂ Here the divergence is · F = ∂x (2xz ) + ∂y (−xy ) + ∂z (−z 2 ) = −x When plugging z = 0 into the plane y + z = 4 we get y = 4 so our plane touches the boundary of the ellipse 4x2 + y 2 = 16 at (0, 4) in the xy -plane. We get 2 √ 16−4x2 0 4−y Flux = D 2 √ 16−4x2 (−x)dV = 0 2 0 (−x)dzdydx 1 ( x(16 − 4x2 ) − 4x 16 − 4x2 )dx 2 2 = 0 0 (xy − 4x)dydx = 0 3 1 1 = 4x2 − x4 + (16 − 4x2 ) 2 2 3 =− 0 40 3 1 Problem 16.8.19 a) div(g F) = = = b) × ( g F) = ∂ ∂ ∂ ∂ ∂ ∂ (gP ) − (gN ) i+ (gM ) − (gP ) j+ (gN ) − (gM ) k ∂y ∂z ∂z ∂x ∂x ∂y i+ M ∂g ∂m ∂g ∂P +g −P −g ∂z ∂z ∂x ∂x ∂g ∂g −P ∂z ∂x ×F+ j+ g j+ N ∂g ∂N ∂g ∂M +g −M −g ∂x ∂x ∂y ∂y j+ N ∂g ∂g −M ∂x ∂y k k M g · gF = ∂ ∂ ∂ (gM ) + (gN ) + (gP ) ∂x ∂y ∂z ∂N ∂g +N ∂y ∂y +g ∂P ∂g +P ∂z ∂z =g ·F+ g·F ∂g ∂M +M ∂x ∂x +g +g ∂g ∂g ∂g +N +P ∂x ∂y ∂z ∂M ∂N ∂P + + ∂x ∂y ∂z = = P P ∂P ∂g ∂N ∂g +g −N −g ∂y ∂y ∂z ∂z ∂g ∂g −N ∂y ∂z i+ g +g ∂N ∂P −g ∂y ∂z i+ M k=g ∂P ∂m −g ∂z ∂x ∂M ∂N −g ∂x ∂y g×F Problem 16.8.22 Yes, the outward flux through the top is 5. The reason is this: Since · F = · (xi − 2y j +(z +3)k) = 1 − 2+1 = 0, the outward flux across the closed cubelike surface is 0 by the Divergence Theorem. The flux across the top is therefore the negative of the flux across the sides and base. Routine calculations show that the sum of these latter fluxes is -5 (The flux across the sides that lie in the xz -plane and the yz -plane are 0, while the flux across the xy -plane is −3.) Therefore the flux across the top is 5. Problem 16.8.26 F=C⇒ · F = 0 ⇒ Flux = S F · ndσ = D · FdV = D 0dV = 0 2 ...
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