Problem Set 5 Math 213, Fall 216 Directions: Name: Show all your work. You are welcome and encouraged to use Mathematica, or similar software, to check your answers and aid in your understanding of the material. However, all calculations should be performed by you, using your internal software. You are also welcome to collaborate with other students, but the end result should be your own work. P1. Compute the surface area of the following surfaces. a) The torus (doughnut) parametrized by F(θ,ϕ) ((cosϕ + 2)cosθ,(cosϕ + 2)sinθ,sinϕ), for θ 2π and ϕ π. b) The portion of the cone x 2 + y 2 z 2 which is above z and inside the sphere x 2 + y 2 + z 2 4ax, where a > is a constant. Solution 1. Solution 1a) The tangent vectors are and The normal vector is ( sinϕ cosθ, sinϕ sinθ,cosϕ) ϕ ( (cosϕ + 2)sinθ, (cosϕ + 2)cosθ,). ϕ ( cosϕ(cosϕ + 2)cosθ, cosϕ(cosϕ + 2)sinθ, (cosϕ + 2)sinϕ),
which has norm ϕ ( cos 2 ϕ(cosϕ + 2) 2 cos 2 θ + cos 2 ϕ(cosϕ + 2) 2 sin 2 θ + (cosϕ + 2) 2 sin 2 ϕ ) ( (cosϕ + 2) 2 cos 2 ϕ + (cosϕ + 2) 2 sin 2 ϕ ) (cosϕ + 2) 2 cosϕ + 2. Note that we do not need to consider the absolute value of cosϕ + 2, since cosϕ + 2 1. Thus, the surface area is ϕ 2π π dϕ dθ (cosϕ + 2)dϕ dθ 2π (sinϕ + 2ϕ) π dθ 2π (2π) 2. 2π dθ Solution 1b) Using cylindrical coordinates, the equation for the cone becomes z 2 r 2. Since we want the portion above z (and since r cannot be negative), we have z r. With x r cosθ and y r sinθ, the cone is parametrized in cylindrical coordinates by ψ(r,θ) (r cosθ,r sinθ,r). Substituting these values of x,y and z into x 2 + y 2 + z 2 4ax, we have r 2 + r 2 4ar cosθ. Solving this last equation for r yields two solutions, r and r 2acosθ, hence the bounds for r are given by r 2acosθ. Now, the equation x 2 + y 2 + z 2 4ax is equivalent to (x 2a) 2 + y 2 + z 2 (2a) 2 ( after completing the square in x), thus the sphere has radius 2a and is centered at (2a,,). So, the values of x are nonnegative, hence cosθ and θ π /2. We have tangent vectors and r (cosθ,sinθ,1) ( r sinθ,r cosθ,).
The normal vector has norm r ( r cosθ, r sinθ,r) r r 2 cos 2 θ + r 2 sin 2 θ + r 2 2r. Given the bounds for r and θ, the surface area is given by 2acosθ r dr dθ 2r dr dθ π /2 2 2acosθ 2 r2 dθ 2 2a 2 cos 2 θ dθ π 2a 2 /2 (1 + cos(2θ))dθ 2a 2 ( θ + 1 2 sin(2θ) ) 2πa 2. Here, we used the identity cos 2 θ 1 (1 + cos(2θ)). 2 P2. Let the surface M be the graph of z f (x,y) x 4 + 6x 2 y 2 for values of (x,y) satisfying y x and x 2. Determine the flux of the vector field X (xy,2y 2,4yz) across M with respect to the upward-pointing normal vector. Solution 2. Given the bounds for y and x and since the surface M is the graph of a function z f (x,y), this problem works best in Cartesian coordinates. Let ψ(x,y) (x,y,x 4 + 6x 2 y 2 ) parametrize the surface M. We have tangent vectors and x (1,,4x3 + 12xy 2 ) y (,1,12x2 y),
so the normal vector is x y ( (4x3 + 12xy 2 ), 12x 2 y,1). Since we are computing the flux of a vector field, we do not need the norm of the normal vector. We evaluate the vector field X at ψ by plugging in the x,y and z components of ψ, so X(ψ) (xy,2y 2,4y(x 4 + 6x 2 y 2 )). Then, So, the flux is given by X(ψ), x xy(4x 3 + 12xy 2 ) 24x 2 y 3 + 4x 4 y + 24x 2 y 3 y M 12x 2 y 3. X(ψ), x dx dy y 2 x 2 2 3 5 x5 2 96 5. 12x 2 y 3 dy dx 3x 2 y 4 x dx 3x 4 dx
P3. Integrate the 3-form ω xy dx dy dz over the portion of the unit ball in the first octant. Solution 3. The unit ball is the solid unit sphere, given by x 2 + y 2 + z 2 1 in Cartesian coordinates. We can parametrize the unit ball using standard spherical coordinates, ψ(ρ,ϕ,θ) (ρ sinϕ cosθ,ρ sinϕ sinθ,ρ cosϕ), where ρ 1. Since we only want the portion in the first octant, we have ϕ π /2 and θ π /2. Recall that hence the pullback of ω is Hence, we integrate π /2 π /2 1 ψ (dx dy dz) ρ 2 sinϕ dρ dϕ dθ, ψ ω (ρ sinϕ cosθ)(ρ sinϕ sinθ)ρ 2 sinϕ dρ dϕ dθ ρ 4 sin 3 ϕ sinθ cosθ dρ dϕ dθ ρ 4 sin 3 ϕ sinθ cosθ dρ dϕ dθ 1 5 1 5 1 5 π /2 π /2 π /2 π /2 sin 3 ϕ sinθ cosθ dϕ dθ (sinϕ sinϕ cos 2 ϕ)sinθ cosθ dϕ dθ ( cosϕ + 1 ) 3 cos3 ϕ sinθ cosθ dθ 2 sinθ cosθ dθ 15 2 1 15 2 sin2 θ 1 15. P4. Suppose the position of a particle at time t is given by the parametrized curve γ(t) (e t cost,e t sint). If ω x dx + y dy, determine the work done by the vector force field ω on the particle as it traverses from γ() to γ( π /2).
Solution 4. The pullback of ω by γ is γ ω e t costd(e t cost) + e t sintd(e t sint) e t cost(e t cost e t sint)dt + e t sint(e t sint + e t cost)dt e 2t (cos 2 t cost sint + sin 2 t + sint cost)dt e 2t dt. Thus, the work done by the vector field ω (x,y) is e 2t dt 1 2 e2t 1 2 (eπ 1). P5. Let γ(t) (cos 4 t,sin 2 t + cos 3 t,t), t π, and ω x dx + y dy + z dz. Compute the line integral ω and interpret your answer in terms of work done on a particle. γ Solution 5. The pullback of ω by γ is γ ω cos 4 t d(cos 4 t) + (sin 2 t + cos 3 t)d(sin 2 t + cos 3 t) + t dt cos 4 t( 4cos 3 t sint)dt + (sin 2 t + cos 3 t)(2sint cost 3cos 2 t sint)dt + t dt ( 4cos 7 t sint + 2sin 3 t cost 3sint cos 2 t + 5sint cos 4 t 3sint cos 5 t + t ) dt. All but the last term can be integrating via u-substitution and all but the last term vanish. The line integral is π ω t dt 1 2 π2, γ which is the amount of work done by the vector field X ω (x,y,z) on a particle as it traverses from γ() (1, 1, ) to γ(π) (1, 1, π).