# MATHS 267 Answers to Stokes Practice Dr. Jones

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1 MATH 267 Answers to tokes Practice Dr. Jones 1. Calculate the flux F d where is the hemisphere x2 + y 2 + z 2 1, z > and F (xz + e y2, yz, z 2 + 1). Note: the surface is open (doesn t include any of the plane z ) and the normal vector points upward. olution: Let be the unit disk in the xy-plane with the downward pointing normal. This makes where is the upper half of the unit ball. Using the divergence theorem, F d + F d F dv We ll then calculate the other two pieces of this equation and solve for the third one (the one we want). F dv 4z dv 1 1 x 2 1 x 2 y 2 4z dz dy dx 1 x 2 2π π/2 1 (4ρ cos ϕ)(ρ 2 sin ϕ) dρ dϕ dθ ( 2π ) ( ) π/2 ( 1 ) 4 dθ sin ϕ cos ϕ dϕ ρ 3 dρ 4(2π)(1/2)(1/4) π and the integral over can be integrated directly because is flat remember that we re using the downward pointing normal, and that z on so F d e y2,, 1,, da 1 1 x 2 1 x 2 dy dx π so that F d F dv F d π ( π) 2π 2. Evaluate x d where is the portion of the surface z x + y that is cut out by z x 2 + y 2 x + y. olution: This is probably easiest to simply evaluate directly. Let ϕ(u, v) u, v, u + v so that ϕ u (u, v) 1,, 1 and ϕ v (u, v), 1, 1 so that the integral in question is u ϕ u ϕ v du dv 3u du dv where is the region in the uv-plane where u + v u 2 + v 2 u + v or (u 1) 2 + v 2 1. ince this is a circle of radius 1 centered at (1, ), we can change variables by letting 1

2 u 1 + r cos θ, v r sin θ, yielding 3u du dv π ( 1 2π ( 1 r(1 + r cos θ) dθ dr r dθ dr + ) ( 2π r dr dθ 3(1/2)(2π) π 3 1 2π ) + ( 1 3 ) r 2 cos θ dθ dr ) ( 2π ) r 2 dr cos θ dθ 3. Let D be the boundary of the region D in the plane that is bounded by x + 2y 1, x + 2y 2, x y 1 and x y 3. Evaluate D 2y dx + x dy. olution: By tokes Theorem, the integral of 2y, x, around the boundary is the flux of the curl across the region. The curl is,, and the normal to the flat parametrization is,, 1 so 2y dx + x dy ()da D Now, either change coordinates by u x + 2y, v x y (x (u + 2v)/3, y (u v)/3, so the Jacobian is 1/3) or simply calculate the location of the vertices and the area of a parallelogram to see that ()da 2/3 D 4. Calculate the surface integral ( F ) d where is the portion of the surface x z 1 inside the cylinder x 2 + y 2 1 and F (2y, x z, y). olution: I didn t specify, but let s use the upward pointing normal. clearly applies here to give D ( F ) d F dr tokes Theorem so we ll parametrize the boundary of as γ(t) cos t, sin t, cos t 1 ( t 2π) and compute ( F ) d 2π 2π 2π F dr F (γ(t)) γ (t) dt 2 sin t, 1 2 cos t, sin t sin t, cos t, sin t dt 3 sin 2 t + cos t 2 cos 2 t dt 5π Alternatively, it s really not that hard to do directly let ϕ(u, v) u cos v, u sin v, u cos v 1 ( u 1, v 2π) 2

3 so that ϕ u (u, v) cos v, sin v, cos v, ϕ v (u, v) u sin v, u cos v, u sin v and ϕ u ϕ v u,, u (note that this normal points down, so we ll need to change the sign at the end). ince F 2,, 3, we are simply calculating ( F ) d 1 2π 5u dv du 5π 5. How would you calculate each of the following (i.e., which theorem(s) would you use if direct calculation is the best option, state this)? Which hypotheses, if any, are irrelevant? a. The flux of an irrotational field through a closed surface direct calculation O divergence theorem b. The flux of an incompressible field through a closed surface by divergence theorem c. The flux of an irrotational field through an open surface direct O use the cap off a region trick from 1. d. The flux of an incompressible field through an open surface tokes + un-curling e. The line integral of an irrotational field around a closed curve by tokes f. The line integral of an incompressible field around a closed curve direct calculation O tokes with a nice surface g. The line integral of an irrotational field along an open curve un-gradient h. The line integral of an incompressible field along an open curve direct calculation 6. Let be the portion of the surface (x 2 + y 2 + z 2 )(e x2 ) 1 + x 3 which intersects x. Find the flux through of the field F (, y 3 + e x2 +z 3, 3y 2 z + sin(x 2 y 2 )). olution: The surface looks horrible, but it intersects the yz-plane in a unit circle, so the boundary can be parametrized by γ(t), cos t, sin t ( t 2π). Furthermore, the divergence of F is, so F G for some vector field G. Furthermore, the basic un-curling formula I gave you has the third component, and the second component is an integral of the first component of F, which is. Therefore, G can be chosen to be of the form f(x, y, z),, where f is some (probably inutterably horrible) function. The line integral over the boundary, though, will be F d 2π f(x, y, z),, dr f(, cos t, sin t),,, sin t, cos t dt 7. Find the flux of the field F (x, y, z) y, zx, x through the surface where is the hemisphere of the unit sphere above the xy-plane (upward pointing normal - open surface). olution: ince F, this can be done by un-curling F and integrating around the unit circle in the xy-plane. It can also be done by the capping-off method of 1. I ll do it both ways. First, the un-curling formula gives z2 x 2 xy, yz, y, zx, x 3

4 and may be parametrized by γ(t) cos t, sin t, ( t 2π) so that F d z2 x 2 2π 2π xy, yz, dr cos t sin t,, sin t, cos t, dt cos 2 t sin t dt Alternatively, if is the unit disk in the xy-plane with the downward pointing normal, then is the boundary of the upper half of the unit ball, so that so that by the symmetry of F d + F d F dv F d F d F (x, y, ),, da y,, x,, da x da 8. Find the work done by the field F (x, y, z) yz y, xz x, xy in moving along the portion of the curve which is the intersection of the surface y x 2 with the plane z x + y between the points (, 1, ) and (1, 1, 2). olution: ince F, there is some function f(x, y, z) such that F f. Using the un-gradient formula yields f(x, y, z) xyz xy. The gradient theorem then says that the integral in question is C F dr f(1, 1, 2) f(, 1, ) (2 1) () 2 9. Let be the unit sphere {(x, y, z) x 2 + y 2 + z 2 1} and Calculate F (x, y, z) x y 2 sin z, x 2 + 1e z3 y, z + (1 + x 2 + y 2 ) 5/4 F d. olution: is the boundary of, where is the unit ball. The divergence theorem then implies that F d F dv (1) dv 4π/3 4

5 1. Let be the open portion of the surface x 2 + y 2 1 between z and z 1 and let F (x, y, z) x y 2 sin z, x 2 + 1e z3 y, z + (1 + x 2 + y 2 ) 5/4 Calculate F d. olution: Let be the unit disk in the xy-plane with the downward pointing normal and 1 be the unit disk in the plane z 1 with the upward pointing normal. Then 1 are, where is the region bounded by z, z 1 and x 2 + y 2 1. Consequently, and thus F d + F d + F d 1 (1) dv π We then calculate directly that F d π F d F d 1 F dv F d x, x y, (1 + x 2 + y 2 ) 5/4,, da 1 1 x 2 (1 + x 2 + y 2 ) 5/4 dy dx 1 x 2 and F d x y 2 sin(1), e x y, 1 + (1 + x 2 + y 2 ) 5/4,, 1 da x 2 π x 2 1 x (1 + x 2 + y 2 ) 5/4 dy dx 1 x 2 (1 + x 2 + y 2 ) 5/4 dy dx and thus, without even evaluating the ugly integral (which can be evaluated by changing to polar coordinates it s (4π/9)( )), we see that F d π F d F d 1 π π 5

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