Problem Set 6 Math 213, Fall 2016

Transcription

1 Problem Set 6 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. Determine the area bounded by the plane curve corresponding to the polar equation r = 3sin(2θ). The curve is a four-petaled rose. To determine the total area, determine the area of one petal, θ π /2, then multiply by 4. Solution 1. Since r = 3sin(2θ), the curve can be parametrized in polar coordinates by γ(θ) = (3sin(2θ)cosθ,3sin(2θ)sinθ), where θ π /2. Using the formula in problem 2, with x(θ) = and y(θ) =, would ulimately yield the area of one petal. However, there is a shortcut for polar equations of the form r = f (θ). According to equation (4.22) in the course notes, the area is Hence, the area of one petal is π /2 1 f (θ) 2 dθ. 2 π /2 1 (3sin(2θ)) 2 dθ = 1 π /2 9sin 2 (2θ)dθ 2 2 π /2 = 9 (1 cos(4θ))dθ 4 (θ 1 ) 4 sin(4θ) π/2 = 9 4 = 9π 8. So, the area of all four petals is 9π 2.

2 P2. Let M be a region in R 2 with boundary, which can be parametrized by γ(t) = (x(t),y(t)), a t b. Apply Stokes theorem to the covector field ω = x dy (and/or ω = y dx) to give an alternate proof that the area bounded by M, A(M), satisfies A(M) = 1 2 b a (x(t)y (t) x (t)y(t)) dt. Solution 2. Let ω = y dx + x dy. Then, dω = dy dx + dx dy = 2dx dy. So, the right-hand side of Stokes theorem ω = is 2dx dy, M which is twice the area of M. Since γ(t) parametrizes the boundary of M, the left-hand side is the same as the line integral of ω over γ. Computing the pullback of ω by γ, we have ω M γ ω = y(t)d(x(t)) + x(t)d(y(t)) = y(t)x (t)dt + x(t)y (t)dt, hence b a ( y(t)x (t) + x(t)y (t))dt = 2 dx dy = 2A(M). M P3. Let M be the ellipsoid x 2 + y 2 + 2z 2 = 24, z 2, with boundary. Let X = (zx + z 2 y + x,z 2 yx + y,z 4 x 2 ). Compute the flux of curlx across M with respect to the outward pointing normal. Solution 3. Here, we use the classical Stokes theorem. The boundary of M corresponds the the points on the ellipsoid for which z = 2, hence x 2 + y 2 = 16. That is, the boundary is a circle of radius 4 in the z = 2 plane. The orientation of this circle should be counterclockwise, when viewed from above, hence we may parametrize by γ(t) = (4cost,4sint,2),

3 for t 2π. We have X(γ),γ = 48sint cost 64sin 2 t + 256sint cos 2 t + 16sint cost. Using u-substitution and the trigonometric identity sin 2 t = 1 /2(1 cos(2t)), we find 2π ( X(γ),γ dt = 16cos 2 t 32t + 16sin(2t) 256 ) 2π 3 cos3 t = 64π. P4. Let M be the surface 2z = x 2 + y 2 below z = 2. Compute the flux of curlx, where X = (3y, xz, yz 2 ), across M directly and by using Stokes theorem. Solution 4. (Stokes theorem) Similar to the last problem, the boundary of M is the set of points on the paraboloid for which z = 2, hence x 2 + y 2 = 4. The boundary of M is a circle of radius 2 in the z = 2 plane. Taking this circle to be parametrized counterclockwise, we have for t 2π. We have γ(t) = (2cost,2sint,2), X(γ),γ = 12sin 2 t 8cos 2 t = 4sin 2 t 8. Applying the same trig. identity as in the previous problem, we integrate 2π 2π X(γ),γ dt = ( 2(1 cos(2t)) 8)dt = ( 1t + sin(2t)) 2π = 2π. (Direct calculation) The surface M can be parametrized in cylindrical coordinates by F(r,θ) = (r cosθ,r sinθ, 1 2 r2 ),

4 where r 2 and θ 2π. The tangent vectors are and The normal vector is and which becomes r = (cosθ,sinθ,r) = ( r sinθ,r cosθ,). θ r θ = ( r2 cosθ, r 2 sinθ,r) curlx(f) = curlx = (x z 2,, z 3), ( r cosθ 1 4 r4,, 1 ) 2 r2 3 when evaluated at the parametrization F. The flux of curlx is thus 2π 2 curlx(f), r 2π 2 dr dθ = r 3 cos 2 θ + 1 θ 4 r6 cosθ 1 2 r3 3r dr dθ 2π = 4cos 2 θ + 32 cosθ 8dθ 7 = ( sin(2θ) + 32 ) 7 sinθ 1 2π = 2π. Note that the identity cos 2 θ = 1 /2(1 + cos(2θ)) was used. Also, since the z-component of the normal vector is positive, the normal is upward pointing, which agrees with the counterclockwise orientation of. P5. Compute the line integral of ω = (2xyz + sinx)dx + x 2 z dy + x 2 y dz over γ(t) = (cos 5 t,sin 2 t,t 4 ), t π, directly and by using Stokes theorem. For the direct computation, after computing the pullback, I suggest integrating the terms using Mathematica or similar software. However, one of the terms can be integrated using u-substitution. As an example, suppose you wanted to compute the indefinite integral of 3t 2 cos 6 t sin 2 t using Mathematica. The following command Integrate[3t^2 * Cos[t]^6 * Sin[t]^2,t] would output the indefinite integral, where the first argument of Integrate is the function to be integrated and the second argument is the variable to integrate with respect to. For the same term, if you want the definite integral for t π, you would enter the command

5 Solution 5. (Stokes theorem) Integrate[3t^2 * Cos[t]^6 * Sin[t]^2,{t,,Pi}]. The version of Stokes theorem that applies here is the Fundamental Theorem of Line Integrals. We must determine a potential function for ω. If f is a potential, then df = ω, hence f = 2xyz + sinx x f y = x2 z f z = x2 y. Integrating with respect to x,y and z, we find that f (x,y,z) = x 2 yz cosx. Indeed, one can quickly verify that the differential df is precisely ω. Thus, the line integral of ω over γ equals since cos( a) = cos(a). (Direct calculation) f (γ(π)) f (γ()) = f ( 1,,π 4 ) f (1,,) = ( cos( 1)) ( cos(1)) =, To directly compute the line integral, we first compute the pullback γ ω = (2t 4 sin 2 t cos 5 t + sin(cos 5 t))( 5cos 4 t sint)dt + t 4 cos 1 t(2sint cost)dt + sin 2 t cos 1 t(4t 3 )dt = ( 1t 4 sin 3 t cos 9 t 5sin(cos 5 t)cos 4 t sint + 2t 4 sint cos 11 t + 4t 3 sin 2 t cos 1 t ) dt. Now, we integrate the pullback for t π. The integral of the first, third and fourth terms equals. The second term can be integrated using u-substitution. If u = cos 5 t, then du = 5cos 4 t sint, hence π 5sin(cos 5 t)cos 4 t sint dt = sinu du = cos(cos 5 t) π = cos( 1) + cos(1) =.

6 P6. Let M be the lower hemisphere of the unit sphere oriented by the exterior (outward) normal vector. Compute y dx + x dy. Solution 6. Since the lower hemisphere is oriented by the exterior normal, the boundary should be oriented clockwise, when viewed from above, by the right-hand rule. The boundary of M is the set of points on the unit sphere for which z =, which is simply the unit circle in the z = plane. A clockwise parametrization for the unit circle is for t 2π. The pullback is γ(t) = (cost, sint,), γ ( y dx + x dy) = sint( sint)dt + cost( cost)dt = dt. Hence, 2π y dx + x dy = dt = 2π. We can also apply Stokes theorem and integrate dω over M. In this case, dω = 2dx dy. As always, if we choose to change coordinates, we must pullback before integrating. If F(ϕ,θ) = (sinϕ cosθ,sinϕ sinθ,cosϕ), where π /2 ϕ π and θ 2π, then F (2dx dy) = 2sinϕ cosθ dϕ dθ. The orientation specified by the normal vector ϕ θ agrees with the orientation of specified above. P7. By using Green s theorem, evaluate (x 3 y)dx + (cosy + 2x)dy,

7 where M is the plane region with boundary. Here, a and b are constants. {(x,y) a 2 x 2 + y 2 b 2 } Solution 7. If ω = (x 3 y)dx + (cosy + 2x)dy, we have dω = 3dx dy, which we need to integrate over M. The 2-surface M looks like a washer; it is the solid region bounded by the circles of radii a and b. We parametrize M in polar coordinates with F(r,θ) = (r cosθ,r sinθ), a r b and θ 2π. To integrate dω, we first compute the pullback F (3dx dy) = 3r dr dθ. Hence, 2π b a 3r dr dθ = 2π 3 2 (b2 a 2 )dθ = 3π(b 2 a 2 ). P8. Let M be the portion of the plane x + y + z = 2 in the first octant oriented by the downward normal. Compute the work done by X = (x + xz 2,x,y) on a particle moving around the boundary of M, with respect to the induced orientation on. Solution 8. (Answer only) 16 3 P9. Let M be the lower hemisphere of the unit sphere oriented by the exterior (outward) normal vector. Verify Stokes theorem for X = ( y,x,z). Solution 9. (Answer only) The flux of curlx out of M is 2π, which is the same as the line integral of X over the boundary of M. The flux of X out of M is 2π, which is the same as the volume/triple integral of divx. 3