MATH 55 Applied Honors alculus III Winter Final xam Review heet : omments and elected olutions Note: The final exam will cover % among topics in chain rule, linear approximation, maximum and minimum values, and lagrange multipliers. 8% among material covered after the second midterm exam. I just added one more problem in this review sheet.. True or False. etermine whether the following statements are true or false and provide an explanation for your answer (5 points each). (a) Let F(x, y, z) be a vector field whose components have continuous second partial derivatives in R 3, then ( F) = : True, it just states Theorem 7.5.. (b) If F = P, Q and P y = Q x in an open region, then F is conservative. : False, change the open region to the simply connected region. An good counter example is for vector field F = < y,x> discussed in the example 5 on page 3 in the text book. x +y (c) If f has continuous partial derivatives on R 3 and is any circle, then f dr =. : True. ince the vector field F = f, which must be conservative therefore with zero work done along closed type II line integral. (d) f(x, y) ds = f(x, y) ds. : False. This is type I line integral, whose physical meaning is the area of the fence therefore change direction will not change the integral value. (e) If is a sphere and F is a constant vector field, then F d =. : True. We can get F =, then by divergence theorem, the statement is true. Of course this is under the assumption that the conditions for applying divergence theorem are valid. (f) uppose a solid and its boundary surface satisfy the conditions of the divergence theorem, then (f g g f) n d = (f g g f) dv : True. Find the divergence of f g g f and you will get it as the integrand of the right hand side integral. To realize this you need to use a very use identity as (ff) = f F + f F. The notation g = g (g) f(x, y) dx = f(x, y) dx.
: True. This is type II line integral therefore the direction matters. (h) The integral π π r dz dr dθ represents the volume enclosed by the cone z = x + y and the plane z =. : False. There are two mistakes here. First the up limit of the integral with respect to r should be. econd, the integrand has missed an r. (i) If is a simple closed curve in R, n its unit normal vector, the region it encloses, and F a vector field whose component functions have continuous first partial derivatives, then F n ds = F da. : True. This is the second vector form of Green s theorem. You may find its explanation on page 3. ince we did not cover it in the class, you don t need to worry about it. (j) If is a smooth simple closed curve in the xy-plane, then the area of the domain inside is given by x dy where the integration is counterclockwise. : True. This is an type II line integral F =<, Q > and if you evaluate Q get. o by green s theorem, this line integral equals the area of the domain inside x P y, you will
. onsider the two dimensional potential fluid flow with velocity field, where Φ(x, y) is a smooth scalar function. v(x, y) = v (x, y), v (x, y) = Φ(x, y) (a) how that v = for any choice of the function Φ(x, y), and explain why this implies that the circulation, v dr, is equal to zero for any simple closed curve. By theorem 7.5.3, we know this is true. Then by toke s theorem, v dr = v d = (b) A fluid is called incompressible if v =. how that in that case the velocity potential satisfies laplace s equation, Φ x + Φ y =. This can be shown by the definition of divergence and gradient, replacing v by Φ. (c) ompute the velocity field of the potential flow with Φ(x, y) = x y, and show that the fluid is incompressible. v =< x, y > and then v = (d) onsider the motion of a fluid particle in the velocity field from part (c). Let r(t) be the position of the particle at time t, and r() =, 3. Write the system of differential equations describing the position of the particle. how that the fluid particle is moving on the hyperbola xy = 6. The system of equations are x (t) = xand y (t) = dy. olving for it, we can get x = ce t and y = ce t, then with initial condition r() =, 3, we have xy = 6. 3. ompute the flux, F d, of the vector field F(x, y, z) = x3 3 i + y3 3 j + 4k through the boundary of the domain bounded by the paraboloid z = x + y and the plane z =. F d = FdV = (x + y )dv = π = π r r rdzdrdθ = π r3 ( r )dr = π 6 or z r rdrdzdθ = π 4 z dz = π 6 4. Let be a simple closed surface with outward unit normal n, and let r = x, y, z be a parametric representation of. (a) how that the volume of the solid contained in the surface is given by the integral V = r n d. 3 3
We also know o r n d = r d = rdv 3 3 3 r = < x, y, z >= 3 r n d == 3dV = dv = V 3 3 (b) alculate the volume of a cone of height H and base area A by putting the vertex of the cone at the origin and using the formula found in part (a). The volume of cone is V = r n d = r n d + r n d 3 3 3 Here, represents the surface on the cone x + y = z, and represents the surface on base z = Height. It s obvious the normal vector n on the surface is always perpendicular to the position vector r, so r n d = 3 And the normal vector n on is equal to k, and position vector r =< x, y, Height >, so r n d = < x, y, Height > <,, > da = Height Height Area da = 3 3 3 3 Here, Height is the cone height, Area is base area of cone. 5. Let the vector field F in R 3 be defined by F(x, y, z) = z, yz, xz + y. (a) how that the vector field F is conservative, and find a scalar function f(x, y, z) such that F = f. In order to prove F is conservative, calculate F =. F =< y y, z + z, >= very conservative vector field can be written as the gradient of scaler potential function, so o the potential function f x = z ; f y = yz; f z = xz + y f(x, y, z) = z x + c(y, z); f y = y z + c(x, z); f z = xz + y z + c(x, y) f(x, y, z) = xz + y z + ; 4
(b) If F, as given above, is the force acting on a particle, find the work of this force in moving the particle from the initial (t = ) to the final (t = ) points of the curve For conservative vector field, r(t) = t, 3 t, 3 sin t, t [, ] π F r = f(r(b)) f(r(a)) Here, r(b) =<,, 3 sin( π ) > and r(a) =<,, ) > F r = 3 sin( π ) + 3 sin( π ) = 9 sin( π ) 6. valuate P dx + Q dy where P (x, y) = tan y,q(x, y) = 3x + x sec y and is the circle (x ) + (y 5) = 4. By Green s theorem, P dx + Q dy = ( Q x P y )da = 3dA = 3 π = π 7. Find the area of the portion of the plane x + y + 3z = that is inside the cylinder x + y = Using the formula = fx + fy + da to calculate surface area if we know the surface formula z = f(x, y). o for this problem, we know the surface is: z = 3 x 3 y + 3, then = ( 3 ) + ( 3 ) + da = 8. Find the area of the parametric surface given by 4 9 da = 4 9 π = π 3 r(u, v) = u cos v, u sin v, u, u, v π, and describe the surface. According to the parametric surface formula, we get for the surface: z = (x + y ), z, which is a cone. If we know the parametric surface formula r(u, v), we can use = r u r v da to calculate surface area. o we have = r u r v da = < cos v, sin v, > < sin v, cos v, > da = da = π dudv = π Note here, the domain for (u, v) is a rectangle, instead of a circle. 9. Find the area of the parametric surface given by r(u, v) = sin u cos v, sin u sin v, sin u, u π 4, v π, 4 5
and describe the surface. According to the parametric surface, we get the surface: x + y + z =, and z /. o the surface is part of a sphere. = r u r v da = < cos u cos v, cos u sin v, cos u > < sin u sin v, sin u cos v, > da = = π/4 ( (sin u cos u( + sin v)da = sin u ) du π ( + sin v)dv = π/4 π/4 (sin u) (cos u) du 4 =... = 5 3 π cos 4u du π π ( + (sin v) )dv ( + cos v )dv. Use the divergence theorem to compute the flux across of the vector field given by F = x, y, z, where is the hemisphere x + y + z = 4, z (without the bottom). aution: the divergence theorem is only part of what you need, you will still need to calculate a surface integral! efine: is the hemisphere x + y + z = 4, z (without the bottom) b is the bottom under the hemisphere closed = + b o closed is a closed surface, which is also the boundary of half solid ball. According to divergence theorem, we have And we also have o we get: F d = closed = FdV = 3dV = 3 4 3 π()3 = 6π F d = F d + F d closed b F d + < x, y, > <,, > d = b F d = 6π F d + Problems from the text, page 7:, 3, 5, 7, 9, 7, 9, 3, 37, 38, and 39 6