M273Q Multivariable Calculus Spring 2017 Review Problems for Exam 3


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1 M7Q Multivariable alculus Spring 7 Review Problems for Exam Exam covers material from Sections and and 7.. As you prepare, note well that the Fall 6 Exam posted online did not cover exactly the same material. The formulas in the box will be provided on the exam. da = r dr dθ dv = r dz dr dθ dv = ρ sin φ dρ dφ dθ. Let F be a vector field in R and f a scalar function of three variables. For each of the following, state whether the operations shown produce a vector field, a scalar function, or whether they cannot be computed, in which case the statement is nonsense. Vector Scalar Nonsense F Vector Scalar Nonsense F Vector Scalar Nonsense F Vector Scalar Nonsense f F Vector Scalar Nonsense F F. alculate x da where D is the triangular region with vertices,,,, and,. D D x da = x x dy dx = x xy y=x y= x dx = x dx = x =. Evaluate e /x xye xy dy dx. We use integration by parts with u = y, du = dy, dv = xe xy dy, v = e xy to start the integration. e /x [ e /x ] /x [ e ] xye xy dy dx = ye xy e xy e dy dx = x exy /x e dx x dx = x = 4. Evaluate π π x/ sin y dy dx. We change the order of integration and evaluate. π π sin y dy dx = π y sin y dx dy = π π y sin y dy = cos y = x/ 5. Let W = { x, y, z x + y + z 4, x, y, z }. Evaluate W xz dv.
2 M7Q Multivariable alculus Review Problems for Exam  Page of 4 x 4 x y xz dz dy dx = 4 x x 4 x y dy dx = [ 4 x x 4 x 4 x ] / dx = x 4 x / dx = 4 x 5/ 5 = 5 6. alculate y x + y dx dy by changing to polar coordinates. y π/4 r r dr dθ = π/4 4 r4 r= r= dθ = π/4 dθ = π/4. 7. a Sketch the region of integration for x + y / 5 dydx. b Switch the order of integration for the integral in part a. y y + y / 5 dxdy c alculate the integral from part b. = y y + y / 6 = y / 5 dxdy = + y / 5 x= y x 6 = 4. x= y dy = + y / 5 y / dy 8. Set up but do not evaluate iterated triple integrals, with appropriate limits, for find the volume of the solid under by z = x + y and z = 8 x y in: a rectangular coordinates
3 M7Q Multivariable alculus Review Problems for Exam  Page of Volume = 4 x 8 x y 4 x x +y dz dy dx. b cylindrical coordinates Volume = π 8 r r r dzdrdθ 9. Set up an iterated triple integral, with the appropriate limits in the coordinates of your choice, for finding the volume of the region that lies between the spheres x + y + z = and x + y + z = 9, and inside the cone z = x + y. That is, z x + y. DO NOT EVALUATE. Volume of S = dv. In spherical coordinates: Volume = dv = S π π/4 S ρ sin φ dρ dφ dθ.. onvert order. 4 4 x x/4 4 x fx, y, z dz dy dx to an integral in dy dx dz order and to integrals in dx dz dy
4 M7Q Multivariable alculus Review Problems for Exam  Page 4 of z x y 4 4z 4 x In dxdzdy order two integrals are required. y /4 4 y 4 x fx, y, z dy dx dz. fx, y, z dx dz dy + y /4 4 4z fx, y, z dx dz dy. Express, as an iterated triple integral, the volume of the solid given by x + y z x + 4y + 4. The surfaces intersect when x + y = x + 4y + 4, i.e. x + y = 9, so the projection in the xyplane is a circle of radius centered at, x x+4y+4 V = dz dy dx 9 x x +y. ircle whether the following quantities are vectors, scalars, or are nonsensical that is, the statement is not defined or does not make sense: Vector Scalar Nonsense f Vector Scalar Nonsense F Vector Scalar Nonsense F Vector Scalar Nonsense F Vector Scalar Nonsense f. Sketch the domain of integration and express as an iterated integral in the opposite order. Do not evaluate the integral. e e x fx, y dy dx Since y = e x implies ln y = x, our integral is: e ln y fx, y dx dy 4. Evaluate y + x da, where D is the region bounded by y = x, x = and the x axis. D
5 M7Q Multivariable alculus Review Problems for Exam  Page 5 of D is shown above. So our integral is + x, we get 4 x u du = 4 ln 4 ln = ln. 4 y + x dy dx = x dx, which when letting u = + x 5. Evaluate D e y da, where D is the region bounded by x y, x 4. If we choose to integrate in y first, e y has no elementary antiderivative. Thus we should integrate in y x first. We get e y dx dy = ye y dy = e u du = e Evaluate the volume of the solid under fx, y = x + over the triangle with vertices,, 5,,, 5, pictured below. This is neither a horizontally nor vertically simple region, so we must use a sum of two integrals regardless of whether we select to integrate in x or y first. Let s choose to integrate in y.
6 M7Q Multivariable alculus Review Problems for Exam  Page 6 of The line between, and 5, is, finding the equation from two points: y = x +. The line between, and, 5 is y = x. The line between, 5 and 5, is y = 8 x. So we will get x 5 8 x two integrals: x + dy dx + x + dy dx. x+ For our first integral, x x+ 7 9 =. For our second integral, = 5 5 x+ x + dy dx = 8 x x+ 5 x x + dx = x + dy dx = Thus the volume of this region is + 4 = 4 5 x x x + dx = 5 x + x + 5 dx = 4. 8 x x x + dx x dx = 7. alculate the line integrals, a x + y ds where is the straight line segment from, to 4, 6. xt = + t, yt = + 4t, t 4t + 4 dt = + 7t5 dt = 65 x + y ds = + t + + b c x dy y dx where is the semicircle x + y = 4, y, oriented counterclockwise. xt = cos t, yt = sin t, t π π 4 cos t + 4 sin tdt = π 4dt = 4π. π F dr, where Fx, y =< x + y, x y > and is as pictured. cos t cos t sin t sintdt =
7 M7Q Multivariable alculus Review Problems for Exam  Page 7 of Evaluate Let P x, y = x+y and Qx, y = x y. Then P y = = Q, and F is conservative x with fx, y = x +xy y a potential function for F. Thus, F dr = f, f, = 4+ = integral represent? ds where the curve is parametrized by rt = 4t, t, t for t 5. What does this This integral represents the length of the curve. r t = 4,,. So r t = =. ds = dt = 5 = alculate the total mass of a metal tube in the helical shape rt = cost, sint, t distance in centimeters for t π if the mass density is δx, y, z = z g/cm. To find the total mass M, we must calculate δds where is parametrized by rt. r t = sin t, cos t, t, so r t = sin t + cos t + 4t = π + 4t. So M = δds = t + 4t dt. Let u = + 4t, du = 8t dt, so we have 8 +6π u / du = [ ] + 6π / = g.. Find the work done by the force field F = x, y, z on a particle moving along the piecewise linear path from,, to,,, and from,, to,,. Work is the line integral over the vector field, so we will split this into two line integrals over F, one for each line segment. The line segment from,, to,, has parametrization r t = t,, = t, t, t, with t. r t =,,, Fr t = t, 4t, t. So our line integral over the first line segment is t, 4t, t,, dt = t + 8t t dt = 8. Our second line segment has parametrization r t =,, + t,, = + t,, + t, with t. r t =,,, Fr t = + t, 4, t. Thus our line integral over the second line segment is + 8t + 8t + t t + dt = + t, 4, t,, dt = 9t + 6t + dt = + + = t + 4t, 4, t t +,, dt =
8 M7Q Multivariable alculus Review Problems for Exam  Page 8 of. Let F = x sinx, e y + e z, e y e z. a Show that F is not conservative. Notice that e y + e z = e z and e y e z = e y. Since these partials are z y not equal, F can not be conservative. b ompute F. F = x x sinx+ e y + e z + e y e z = sinx+x cosx+e y +e z y z c ompute F. F y F z, F z F x, F y F = e y + e z,,. x. Let d = x + y + z and F = x, y, z. d a Show that fx, y, z = lnd is a potential for F. Notice that fx, y, z = ln x + y + z = ln x + y + z, so f = x x + y + z, y x + y + z, z x + y + z = x, y, z. d b Find the divergence of F. Notice that x x d = x + y + z xx x + y + z = d x d 4. Similarly for the other partials we get F = d x d 4 + d y d 4 + d z d 4 = d d d 4 = d = x + y + z. c Verify that the curlf =. We calculate each component of the curl on a line below. z y x + y + z y zy z x + y + z = x + y + z yz x + y + z =, x z x + y + z z xz x x + y + z = x + y + z zx x + y + z =, y x x + y + z x yx y x + y + z = x + y + z xy x + y + z =. Since each component of the curl is, curl F =,,.. Let Fx, y, z be a vector field and rt, a t b be a parameterization of a curve. ircle the correct completion of each sentence. Vector Line Integral Scalar Line Integral Nonsense F dr Vector Line Integral Scalar Line Integral Nonsense F dr Vector Line Integral Scalar Line Integral Nonsense F ds Vector Line Integral Scalar Line Integral Nonsense F dr
9 M7Q Multivariable alculus Review Problems for Exam  Page 9 of 4. a ompute y, x dr where is the three quarters of the circle of radius oriented clockwise from, to,. We notice that y, x = f with fx, y = xy, so y, x dr = f, f, = =. b ompute y, x dr from, to,. where is the three quarters of the circle of radius oriented clockwise Since y, x is not conservative, we must parameterize the curve. One option is to notice that has the usual parameterization rt = cost, sint, π 4 t 7π 4, so y, x dr = 7π 4 π 4 sint, cost sint, cost dt = 7π 4 π 4 dt = π. 5. Let fx, y = e x +y. ompute f dr where is the line segment from, to,. f dr = f, f, = e 8 e 6. Let Fx, y = y + xy, x + xy a Find a potential function for F.. fx, y = arctanxy b alculate F dr where is any curve from the origin to a, a and a is any nonzero constant. F dr = arctanaa arctan = π Let Fx, y, z = x + y, y, z. a Show that F is not conservative. Notice that x y = and x + y =. Since these partial derivatives are not y equal, F cannot be conservative. b alculate the work done by the vector field Fx, y, z = x+y, y, z along the curve parameterized by rt = t, t, t, t.
10 M7Q Multivariable alculus Review Problems for Exam  Page of W = F dr = = t + t, t, t, t, t dt t + t + 4t + 6t 5 dt = t + t + t 4 + t 6 = 8. The following statement is false. If F is a gradient vector field, then the line integral of F along every curve is zero. Which single word must be added to make it true? LOSED path 9. Find a potential function for the vector field Fx, y, z =< z sec x, z, y + tan x >. fx, y, z = zy + z tanx.. Let fx, y, z = x yz, and let be a path with parameterization ct = Evaluate f dr. t, sin πt 4, t, t [, ]. et Alternative formulation: integrate over a piecewise defined curve. f dr = fc fc = f4,, f,, = 6. For what values of b and c will Fx, y, z = y + czx, ybx + cz, y + cx be a conservative gradient vector field? We want curlf =< cy,, b y >=, so b = c =.. a Is Fx, y, z =< x cos y, cos y x sin y, z > conservative? Justify your answer. F =,, x sin y + x sin y =<,, >, and all partial derivatives of component functions are continuous, so yes, F is conservative. b alculate F dr, where is the curve rt =< te t, tπ, + t >, t. If fx, y, z = x cos y + sin y + z, then f = F. So, F dr = fr fr.. Use Green s Theorem to calculate wedge shaped region pictured. x y dx + x ydy, where is the closed curve bounding the
11 M7Q Multivariable alculus Review Problems for Exam  Page of π/4 x y dx + x ydy = r r dr dθ = π/4 4 r4 R dθ = π. x y x y da = x y R x + y da = 4. Verify that the vector field Fx, y, z =< xy + z, x +, x + z > is conservative and calculate the work done by F in moving an object from,, to,,. F =,, z x =, so F is conservative. Let fx, y, z = x y + xz + z. Then f = F and the work done is W = F dr = f,, f,, = = alculate y dx + xy dy where is the counterclockwise oriented simple closed curve consisting of the piece of the parabola y = x between, and, together with the piece of the x axis between, and,. Use Green s theorem. y dx + xy dy = D x = y dx = x + x 4 dx = 8 5. y + y da = x y dy dx
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