Multiple Choice. Compute the Jacobian, (u, v), of the coordinate transformation x = u2 v 4, y = uv. (a) 2u 2 + 4v 4 (b) xu yv (c) 3u 2 + 7v 6

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1 .(5pts) y = uv. ompute the Jacobian, Multiple hoice (x, y) (u, v), of the coordinate transformation x = u v 4, (a) u + 4v 4 (b) xu yv (c) u + 7v 6 (d) u (e) u v uv 4 Solution. u v 4v u = u + 4v 4..(5pts) Find sin(x + y ) da where D is the disk centered at the origin of radius a. D (a) π ( cos(a ) ) (b) π cos(a ) (c) π ( sin(a ) cos(a ) ) (d) π arcsin(a) (e) π arccos(a) Solution. onvert to iterated polar integral: sin(x +y ) da = D π cos(r a ) π dθ = cos(a ) dθ = π ( cos(a ) ) a sin(r )r dr dθ =

2 .(5pts) Let be a thin wire lying as the straight line x+y = in the first quadrant. Suppose the charge density on the wire is given by q(x, y) = x y. Then the total charge on the wire is q ds. Which number below is the total charge? (a) 4 (b) (c) (d) 6 (e) Solution. To parametrize the wire we can use r(t) = t, t, t. charge = q ds. Furthermore ds =, dt = dt so ( ) charge = ( t) t dt = ( 4t) dt = (t t ) = ( 4 ) = 4 4.(5pts) The density function of the quarter of the disk of radius in the first quadrant is given by the formula ρ(r, θ) = cos(θ). Which iterated integral below gives the moment about the x-axis of the quarter of this disk in the first quadrant. (a) r sin(θ) cos(θ) dr dθ (b) r sin(θ) cos(θ) dr dθ(c) 4 r sin(θ) cos(φ) dr dθ dφ (d) r sin(θ) cos (θ) dr(e) dθ r sin(θ) cos (θ) dr dθ Solution. D yρ da = The answer, if you care, is r sin(θ) cos(θ) dθ = r sin(θ) cos(θ)r dr dθ 9 sin(θ) cos(θ) dθ = 9 sin (θ) π = 9

3 5.(5pts) Let A be a thin plate in the xy plane. The precise shape is irrelevant. Suppose the mass of A is 5 and that the center of mass of A is at (, ). What is the moment of A about the x-axis? (a) (b) 5 (c) 5 (d) 5 (e) 6 5 Solution. ȳ = M x mass so = M x 5. 6.(5pts) Let F (x, y) = y, 4x. alculate r(t) = t, t, t. F d r where is the curve (a) (b) 4 (c) (d) 4 (e) 4 Solution. (t + 4t ) dt = dr = t, dt and F (t) = t, 4t and 6t dt = t =. F dr = t, 4t t, dt =

4 7.(5pts) Let E be the tetrahedron bounded by the planes x+y +z = 6, x y =, x =, and z =. Write a -fold iterated integral with respect to dy dx, which computes the volume of E. (a) 6 x x (6 x y) dy dx (b) 6 6 x (6 x y) dy dx (c) (e) 6 6 x x 6 x (6 x y) dy dx (d) (6 x y) dy dx 6 6 x x (6 x z) dy dx Solution. The region of integration for the double integral is the triangle determined by the lines obtained by intersecting x + y + z = 6, x = y =, and x = with the xy-plane. These lines are y = x and y = 6 x. The region of integration is y=6 x y=x The point of intersection of the two lines is (, ). The tetrahedron is the solid under the plane z = 6 x y and above z =. So the iterated integral is 6 x x 6 x y dy dx 8.(5pts) Evaluate R 4x + 9y da, where R is the region bounded by the ellipse 4x + 9y = 6. (a) 4π (b) 6 π (c) (d) 6π (e) 8

5 Solution. If we write u = x and w = y, the region becomes the interior of u + w = 6, in other words, the disk centered at the origin of radius 6, say D. Since det = 6, R(4x + 9y ) da = u + w 6 da = 6 (r)r dθ dr = 6 πr dr = D 6 6 π r 6 = π 6 = π = 4π

6 Partial redit 9.(pts) Find the mass of a thin wire in the shape of a parabola x = y, y, if the density is given by ρ(x, y) = 6y. Solution. The mass is ρ ds where the curve can be parametrized by r(y) = y, y, y. ds = r (y) dy = y, dy = + 4y dy. Hence the mass is 6y y dy u=+4y = u du = u = 6 (7 ) = =.(pts) Rewrite the following integral x 4 x y x +y (a) (6pts) Using cylindrical coordinates. y x + y + z dz dy dx. z Solution: Notice the domain of integration is a quarter of a cone in the first octant and the octant with negative x and positive y and z. Because we are in these octants θ π and r. Now for cylindrical coordinates z remains unchanged. Therefore x 4 x y x +y 4 r = r y x + y + z dz dy dx z r + z r sin(θ) r dz dr dθ. z (b) (6pts) Using spherical coordinates. Solution: Now z = 4 x y is a sphere of radius so ρ. z = at an angle of π from the positive z-axis. So x 4 x y y x + y + z dz dy dx z = = x +y π ρ sin(θ) sin(φ) ρ sin(φ) dρ dθ dφ ρ cos(φ) ρ sin(θ) sin (φ) cos(φ) dρ dθ dφ x +y is a cone

7 x y.(pts) Evaluate da, where R is the parallelogram in the xy-plane bounded R 4x 5y by x y =, x y = 5, 4x 5y =, 4x 5y = 8. Solution. The form of the region and the integrand suggests trying u = x y and w = 4x 5y. The boundary in uw space becomes u =, u = 5, w =, w = 8. Let W be this region. so The Jacobian calculation is (u, w) (x, y) = R x y 4x 5y da = W det 4 5 u w da = ln(8) u 5 = and (x, y) (u, w) = 5 8 = 5 4 ln(8) u w dw du = 5 u ln(w) 8 du =.(pts) The density of the semi-disk y of radius at (x, y) is ρ(x, y) = x + y. Find the center of mass given that the mass is 4π. Solution. By symmetry, the density and the region are symmetric with respect to the y-axis so x =. Next mass = (x + y ) da and M x = y(x + y ) da. Polar seems a good way to go so M x = Then y = D mass = r sin(θ)r r dr dθ = π = 64 π = 6 5π x D r r dr dθ = π r4 = 4π as given. 4 r 5 dθ = sin(θ) dθ = ( cos(θ) π ) = 64 5

8 .(pts) What ( is the work done by the force field F (x, y) = x ı + y j to move a particle from π ) (, ) to, along the curve r(t) = t, sin t? HINT: Work done = Force Displacement. Solution. F d r = t, sin(t), cos(t) dt = t + sin (t) π = π 8 +. t + sin(t) cos(t) dt =