MATH 64: FINAL EXAM olutions Problem 1. Use a line integral to find the plane area enclosed by the curve C: r = a cos 3 t i + b sin 3 t j ( t π). olution: We assume a > b >. A = 1 π (xy yx )dt = 3ab π (sin 4 t cos t + sin t cos 4 t)dt = 3ab 8 π sin (t)dt = 3πab 8. Problem. Compute the outward flux of the vector field F = (e cos z + 3xy, 1/(1 + sin x) + 3x y, sin(e y ) + z 3 ) across the surface consisting of the cylinder x + y = 4, z, capped at the bottom by the disk D := {(x, y, z) : z =, x + y 4}, and capped at the top by the hemisphere {x + y + z = 4, z }. olution: Let R be the region in R 3 bounded by the surface. We use the Divergence Theorem. We have divf = 3(x + y + z ), so that F d = 3 (x + y + z )dv := I. D To compute the integral, we decompose the domain D into regions: D = D 1 D, where D 1 = {(x, y, z) : z, x + y 4}, and where D = {(x, y, z) : z, x + y + z 4}. Let I 1 = 3 D 1 (x + y + z )dv. To compute I 1 we switch to cylindrical coordinates. The integral I 1 can be computed as follows: I 1 = 3 z= r= π θ= (r + z )rdrdθdz = 6π z= ( r 4 dz 4 + r z ) = 8π. Let I = 3 D (x + y + z )dv. To compute I, we switch to spherical coordinates. We then have π/ π ( ) ρ I = 3 ρ ρ sin φdθdφdρ = 6π ( cos φ) π/ 5 = 19π 5 5. ρ= φ= θ= Finally, I = I 1 + I = 118.4π. Problem 3. Using tokes s Theorem, compute the integral (y z )dx + (z x )dy + (x y )dz, C 1
where C is the curve formed by the intersection of the cube [, 1] [, 1] [, 1] R 3 with the plane x + z = 1, oriented so that its projection into the (x, y)-plane has counterclockwise orientation. olution: We first remark that curl F = ( ) (y + z, x + z, x + y). The intersection of the unit cube with the plane x + z = 1 is a rectangle with vertices at (1,, ), (1,, 1), (, 1, ), (, 1, 1). It projects one-to-one onto the unit square [, 1] [, 1] in the (x, z)-plane, and the boundary curve is oriented counterclockwise. Accordingly, can be parametrized by {(x, y, 1 x) : x, y 1}. The normal N is pointing upward and is equal to (1,, 1). Thus, ( F) N = ( )(x + y + z) = ( )(y + 1). the flux integral is equal to Problem 4. Let ( F = ( ) x= x (x + y + z ) 3/ + 18xz, y= (y + 1)dydx = 4. ) y yx + (x + y + z ) 3/ + y3 3, z. (x + y + z ) 3/ Compute the outward flux of F through the boundary of the ellipsoid x /4 + y + 9z = 1. olution: We write F = F 1 + F, where ( ) x F 1 = (x + y + z ), y 3/ (x + y + z ), z = r 3/ (x + y + z ) 3/ r, 3 which has the singularity at r =, and F = (18xz, yx / + y 3 /3, ). Note that divf 1 =, and divf = 18z + x / + y. Let be the surface of the ellipse; then F d = F 1 d + F d := I 1 + I. ince the origin is contained in the ellipsoid R bounded by, to compute I 1, by applying the divergence theorem, we may let ( ) be a sphere with radius ɛ. Then, r I 1 = F 1 d = F 1 d = ( ) ( ) r r 3 r d = 1 ( ) r d 1 = d = 4π. ɛ ( )
To compute I, we again apply the Divergence Theorem. We have divf = 18z + x / + y. Then I = (x / + y + 18z )dxdydz. x /4+y +9z 1 We change coordinates: u = x/, v = y, w = 3z; the Jacobian is equal to /3. The integral becomes (u + v + w )dudvdw. 3 u +v +w 1 witching to spherical coordinates, we find that I = 4 3 ρ= π φ= π θ= ρ ρ sin φdρdφdθ = 8π 3 ( cos φ)π (ρ 5 /5) 1 = 16π 15. The total flux is equal to I 1 + I = 4π + 16π/15 = 76π/15. Problem 5. Given he following IBVP of heat conduction equation: u t = u 3 sin x ( < x < π, t > ) x u(, t) = 1, u(π, t) =, (t > ) u(x, ) = sin(3x) sin(5x), ( x π) (1) Find the equilibrium state solution for the above problem; () Find the unsteady solution for the above IBVP; (3) how that the above unsteady solutions approach to the equilibrium state solution as t. olution: (1).We look for steady solution of the inhomogeneous equation in the form u(x, t) = W (x), where d W/dx = 3 sin x, W () = 1, W (π) =. By integration, we find that W (x) = 1 x 3 sin x. π (). Let u(x, t) = ũ(x, t) + W (x). ubstituting into the equation, we find that ũ satisfies ũ t = ũ xx ; ũ(, t) = ũ(π, t) = ; ũ(x, ) = 1 + x + 3 sin x + sin(3x) sin(5x) = f(x). π To solve ũ, we use the ine Fourier series: ũ(x, t) = 3 sin xe t + sin(3x)e 3t sin(5x)e 5t + ã n sin nxe nt. From the IC, we derive ã n sin nx = x π 1, ( x π).
Thus, it follows that ã n = π π Therefore, we finally obtain: ( x π 1 ) sin nxdx = nπ cos nx ( x π 1 ) π = ( 1)n nπ. u(x, t) = 1 x π 3 sin x + 3 sin xe t + sin(3x)e 3t sin(5x)e 5 t ( 1) n + sin nxe nt. nπ (3). As t, u(x, t) W (x) = 1 x π 3 sin x. Problem 6. Given the following problem of the string vibration with external oscillatory force: u t u = cos πx cos πt, ( < x < 1, t > ) x u x (, t) = u x (1, t) =, (t > ) u(x, ) = f(x) = cos πx, u t (x, ) = g(x) = cos πx, ( x 1). (1) Use the method of separation of variables and Fourier series to find its solution; () Determine the behavior of the solution, as t : Is the solution bounded or unbounded as t? olution: 1. We first look for a particular solution of the form W (x, t) = cos(πx)φ(t). ubstituting into the equation we get cos(πx)φ (t) + 4π cos(πx)φ(t) = cos(πx) cos(πt). It follows that φ (t) + 4π φ(t) = cos(πt). From the theory of inhomogeneous second order ODE (method of undetermined coefficients), we know that solution will have the form φ(t) = A cos(πt)+b sin(πt)+ct cos(πt)+dt sin(πt). ubstituting into the ODE, we find that Taking the second derivative and equating the coefficients, we find that φ(t) = t sin(πt)/(4π), W (x, t) = t sin(πt)/(4π) cos(πx). We now set u(x, t) = ũ(x, t) + W (x, t). Then by substituting into the equation we have ũ t ũ =, ( < x < 1, t > ) x ũ x (, t) = ũ x (1, t) =, (t > ) ũ(x, ) = f(x) = cos πx, ũ t (x, ) = g(x) = cos πx, ( x 1).
One may make Cosine Fourier series for ũ(x, t), [ ] ũ(x, t) = A n cos(nπt) + B n sin(nπt) cos(nπx). n= To satisfy the IC s, we have ũ(x, ) = and It follows that Hence, Finally, we get ũ t (x, ) = n= A n cos(nπx) = f(x) = 1 + cos(πx). πb n cos(nπx) = g(x) = cos(πx). n= A = A = 1, B = 1 π, A n = B n =, otherwise. ũ(x, t) = 1 + [ 1 cos(πt) + 1 π sin(πt) ] cos(πx). u(x, t) = 1 [ 1 + cos(πt) + 1 ] π sin(πt) cos(πx) + t sin(πt) cos(πx). 4π