********************************************************** 1. Evaluate the double or iterated integrals:

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1 Practice problems 1. (a). Let f = 3x 2 + 4y 2 + z 2 and g = 2x + 3y + z = 1. Use Lagrange multiplier to find the extrema of f on g = 1. Is this a max or a min? No max, but there is min. Hence, among the candidates, you can find the min using Lagrange multiplier. (b). Let f = 2x + 3y + z and g = 3x 2 + 4y 2 + z 2 = 1. Do the same questions as in (a). Both the max and min exist. Then, among the candidates, the largest one must be the max and the smallest must be the min. Use Lagrange multipliers. 2. Consider that we want to make a box with 6 faces. We want the volume to be 250 in 3. The cost for the top and bottom material is 4 c/in 2 and the cost for side material is 2 c/in 2. (a). Suppose the three dimensions are x, y, z(length, width, height). Write out the cost function f(x, y, z) and the constraint g(x, y, z). (b). Use Lagrangian multiplier to find the dimensions that minimize the cost. (c). Using the constraint, solve z out and then the cost f is reduced to a function of x, y only without any constraint. Find the critical points of this function and verify that it is a local minimum. The constraint is the volume g = xyz = 250. The cost is f = 4(2xy) + 2(2yz+2xz). To solve the equations, one trick we always use for xyz = C type of equations is to solve yz = 250/x. Then, 250λ =.... You can equate the right hand sides. Convince yourself that the candidate you find is a min instead of a max. 3. Find the points on the ellipse x 2 + xy + y 2 = 3 (since we have xy, this is not a circle!) that are closest and farthest from the origin. 4. Find the minimum value of f(x, y, z) = x+y +z with constraint xyz = 1, x > 0, y > 0, z > 0. Use your result to show that 3 3 abc a + b + c for any a > 0, b > 0, c > Consider that we want to find the optimal value of f with constraint g = C 1 and h = C 2. The constraint is the intersection of two surfaces which is usually a curve. Let s denote it by C. 1

2 (a). Suppose that P is a minimum of f along the constraint curve. Then, f is (A). Parallel with C. (B). Perpendicular with C. (C). Neither parallel nor perpendicular. (Choose one.) (b). Suppose that P is a minimum and g(p ) = 1, 2, 3, h(p ) = 2, 1, 2. Can 5, 4, 2 be f? If f = 5, 4, ζ, determine ζ. ********************************************************** 1. Evaluate the double or iterated integrals: x 3 + 1dA where = {(x, y) : 0 y 1, y x 1}. 1 1 y sin(x 2 + y 2 )dxdy First: change the order of integration; Second: polar. 2. Consider the ice-cream cone shaped lamina bounded by x 2 + y 2 = 1, y 0 and y = x 1, x 1. The density is δ = 1. Set up the integral for the total mass (which equals the area) in Cartesian coordinates using the order dxdy. (Hint: You need to break the region into two parts.) If I ask you to evaluate the integrals, will you change the coordinates for one of the regions? 3. Compute the volume of the solid bounded by y = 1, y = x + 1, y = 1 x,z = x 2 + y 2 and xy plane. Consider the three surfaces that have no z variable first: y = 1, y = x+1, y = 1 x. These three guys form a triangular region in xy plane. For the z, direction, it s clearly between z = 0 and z = x 2 + y 2. The double integral formulation is (x 2 + y 2 0)dA. The triple integral formulation is T dv = 1 y= 1 x=1 y x 2 +y 2 x=y 1 0 dzdxdy. 4. Set up the integral for the volume inside both x 2 +y 2 = 1 and x 2 +y 2 2y = 0, below z = 2y 2 + x 2 and above xy plane. ewrite it in polar coordinates. If this region has a mass density δ = x 2 + 2z 2, set up the integral for the moment of inertia about y axis (don t evaluate). 2

3 5. Evaluate the volume bounded by the paraboloid z = x 2 + y 2 and the plane z = y. This is the problem that involves the polar coordinates. integral it is (y x 2 y 2 )da In double where is the region bounded by x 2 + y 2 = y. Then, apply polar. This is one example in the lecture notes. This problem can also be evaluated using triple integral by cylindrical coordinates. It s essentially the same as the double integral one. 6. Write the following integral in the order dzdxdy and dxdydz 3/2 3/2 3/4 x 2 3/4 x 2 1 x 2 y x 2 y 2 f(x, y, z)dzdydx. 7. Let T be the region bounded by y = x 2, x = y 2, z = 0 and z = x + y. Find the triple integral T xydv. The key is to write out the region: 0 x 1, x 2 y x and 0 z x + y. 8. Let be the region bounded by 2x + y = 3, y 2x = 2, 2x + y = 1, 2x y = 4. Compute the double integral (16x 2 4y 2 )dxdy. We use change of variables u = 2x + y, v = y 2x. Then, we compute the Jacobian. 9. Compute the area bounded by xy = 1, y = 3/x, y = 2x 2, y = 4x 2. change of variables. u = xy, v = y/x Let be the region bounded by ye x 2 = 0, ye x 4 = 0, x 3 6y 1 = 0, x 3 6y 2 = 0 in the first quadrant. Evaluate the double integral I = (2e x y + x 2 e x )e x y da. 3

4 11. Consider that T is given by x 2 +y 2 /4+z 2 /9 1. Evaluate the average 1 value of f over T : V olume(t ) T f(x, y, z)dv, where f = x2. Do change of variables first u = x, v = y/2, w = z/3. Then, the region is transformed into a ball. 12. Find the volume under f(x, y) = x and above the region = {(x, y) : (x 1) 2 + y 2 1, x 2 + (y 1) 2 1}. convenient in cylindrical coordinates. 13. Consider the ball x 2 +y 2 +z 2 4. If the density inside x 2 +y 2 +z 2 = 1 is δ 1 = x 2 + y 2 + z 2 while the density outside it is δ = 1. What is the total mass of the ball? 14. Consider the the solid bounded by φ = π/6 and ρ = 2 cos φ. Suppose the density is given by δ = x 2 + y 2. Find the centroid and moment of inertia about z axis. Clearly, use spherical coordinates to set up the triple integral. Then, δ = ρ 2 sin 2 φ. 15. Evaluate T xydv where T is the region bounded by x2 +y 2 2x = 0 and x 2 + y 2 + z 2 = 4. What is the volume of this region? Cylindrical coordinates are convenient. The sphere becomes r 2 + z 2 = Set up the integral for the mass of the region contained in the sphere x 2 + y 2 + (z a) 2 = a 2 but below z = r with unit density. Spherical coordinates. Surface areas: Parametrize the surface y = f(x, z). Use this to compute the area of the plane y = 2x + 2z + 1 inside x 2 + z 2 = 1. Set up an integral for the surface area of the surface cut from x = y 2 + 3z 2 2z by z + 2y = 3, z = y 2. Consider the surface of revolution obtained by revolving x = f(z) about z axis. Parametrize this surface. Consider the fence S: x = 2 sin(t), y = 8 cos(3t), 0 t < 2π and 0 z 2. Set up the surface area integral S ds. 4

5 S is the surface z = θ, 0 θ π and 1 x 2 + y 2 4. Set up an iterated integral for the surface area. 5