MTHE 227 Problem Set 10 Solutions. (1 y2 +z 2., 0, 0), y 2 + z 2 < 4 0, Otherwise.

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1 MTHE 7 Problem Set Solutions. (a) Sketch the cross-section of the (hollow) clinder + = in the -plane, as well as the vector field in this cross-section. ( +,, ), + < F(,, ) =, Otherwise. This is a simple model of water flowing through a pipe without turbulence (interestingl, the velocit goes to ero at the boundar!). (b) The disk S described b =, + (the region bounded b the cross-section of the pipe in the plane = ) ma be parametried b (u, v) (, u cos(v), u sin(v)), u [, ], v [, ]. Find the flu S F ds of the vector field F of part (a) through S, with the normal pointing along e (that is, along the positive direction). (c) The portion Σ of the clinder + = between = and = ma be parametried b (u, v) (v, cos(u), sin(u)), u [, ], v [, ]. The surface Σ has the same central ais as the clinder + =, and is contained within the clinder. Check that Σ F ds =, and briefl eplain. Solution. (a) In the -plane, we have =, and the vector field restricts to the field (, ), < < (, ), Otherwise. The velocit has a parabolic front. = =

2 To see the flow in three dimensions, imagine rotating the above picture about the central ais of the clinder (that is, about the line = ). (b) To begin, we need to find the normal vector to S. We have T u = (, cos(v), sin(v)) T v = (, u sin(v), u cos(v)) e e e N = T u T v = det cos(v) sin(v) = (u cos (v) + u sin (v),, ) = (u,, ). u sin(v) u cos(v) Notice this points along the positive -direction, as required. Then, we compute F N. Inside the clinder, we have Therefore, F(σ(u, v)) = ( u cos (v) + u sin (v),, ) = ( u,, ). F(σ(u, v)) N(u, v) = ( u ) u + + = u u. Finall, we integrate over the domain of the parametriation, obtaining S F ds = D F(σ(u, v)) N(u, v) da [ u =. u u dudv u 6 ] dv 6 6 dv dv Remark. If we computed the average flu, b dividing the total flu b the area of the disc, we would obtain Average Flu = π = Maimum speed of flow =. This is a general feature of such flows (called laminar flows) the average flu is equal to one half the maimum speed of the flow (this is not too hard to show b replacing the vector field above b the scaled field v ma F, where v ma is the ma speed). (c) Of course, we epect that this is true, since the flow is everwhere tangent to such a clinder. To check this formall, let s find the normal vector and check that the dot

3 product of the flow with the normal is equal to ero. For Σ, T u = (, sin(u), cos(u)) T v = (,, ) e e e N = T u T v = det sin(u) cos(u) = (, cos(u), sin(u)). Therefore, F(σ(u, v)) N(u, v) = (,, ) (, cos(u), sin(u)) = + + = and so Σ F ds =.. Sketch the volume of integration for the iterated integral epress it in the five other possible orders of integration. (You do not have to evaluate an of the integrals.) d d d, and Solution. The volume is a tetrahedron with vertices at (,, ), (,, ), (,, ) and (,, ), and bounded b the planes =, =, = and =. The projections or shadows to the three coordinate planes look like

4 Therefore, in the si orders of integration the integral is d d d, d d d, d d d, d d d, d d d, d d d.. Describe the volume of integration, convert to clindrical coordinates, and evaluate (a) (b) 9 9 Solution. + d d d, + d d d. (a) The volume of integration is bounded below b the cone = + and above b the plane =. Notice that the cone intersects the plane when = +. In clindrical coordinates, the cone is described b the equalit = r, so the triple integral (in one of the possible orders of integration) is (remembering the Jacobian factor of r in clindrical coordinates) r dr d dθ = 7 6 = 9π. [ r [ 6 ] ] d dθ d dθ (b) The volume of integration is bounded below b the paraboloid = + and above b the paraboloid. The shadow of the two paraboloids in the -plane is a circle of radius. In clindrical coordinates, the two paraboloids are described b = r (bottom) and = r (top). The integral is r r r d dr dθ dθ [ ] r r dr dθ =. r

5 . (a) Show that the center of mass of the solid + +, (the top half of the ball of radius ) of uniform densit δ has Cartesian coordinates (,, ). (Suggestion: Integrate in spherical coordinates.) (b) Now suppose that the densit δ of the solid in part (a) is given b Solution. δ(,, ) = γ for some number γ. (Interpretation: the ball is made of lighter material at the top than at the base. The upper bound on γ is made to avoid regions of negative densit.) Find the coordinates of the center of mass as a function of γ. For which γ is the center of mass at the point (,, )? (a) Denote the solid b R, and its densit b δ. B smmetr, the center of mass of the top half of the ball must lie on the -ais. (For instance, if we made a calculation that placed the center of mass in a point with a non-ero or -coordinate, we could spin the ball around the -ais, and repeat the calculation, leading to a contradiction.) To find the -coordinate, we need to compute δ dv δ dv. In the uniform densit case, the δ factors out both integrals, and cancels in the fraction, so it is enough to compute dv dv. To do this, we integrate in spherical coordinates. The function converts to ρ cos(φ), and R is described b the inequalities ρ, φ π, θ. Finall, we need to remember the Jacobian factor ρ sin(φ). We have dv [ ρ ] π/ ρ sin(φ) dφ dρ dθ ρ [ cos(φ)] π/ dρ dθ ρ dρ dθ = () ( ) = π. dθ 5

6 This makes sense, since we should be getting half of the volume of a unit sphere, which is equal to π. Now, the second integral is dv = = = π. π/ (ρ cos(φ)) ρ sin(φ) dφ dρ dθ ρ [ sin (φ)] [ ρ ] Therefore, the -coordinate of the center of mass is dθ dv dv = So that the center of mass is at the point (,, ). π/ π π =, dρ dθ (b) Because the new densit is a function of onl, the center of mass is again along the -ais b smmetr. Our task is to compute δ dv δ dv = γ dv γ dv. One of the cleaner was of doing this is to appl the linearit of the integral, and begin b computing From part (a), we know that dv, dv, dv. dv = π, dv = π. 6

7 The third integral is We then have dv = [ ρ5 5 ] = 5. γ dv γ dv π/ π/ (ρ cos(φ)) ρ sin(φ) dφ dρ dθ ρ cos (φ) sin(φ) dφ dρ dθ ρ [ cos (φ)] π/ dρ dθ = dv γ dv dv γ dv π/ γπ/5 = / γπ/ 5 γ = 5γ. This is equal to when (5 γ) = 5γ, or, rearranging, 5 = 9γ, so at γ = 5 9. / at γ = / at γ = 5/9 7/5 at γ = γ 7