Practice problems **********************************************************

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1 Practice problems I will not test spherical and cylindrical coordinates explicitly but these two coordinates can be used in the problems when you evaluate triple integrals. 1. Set up the integral without evaluation. The volume inside (x 1) 2 + y 2 + z 2 = 1, below z = 3r but above z = r. This problem is very tricky in cylindrical or Cartesian since we must divide the region into several blocks if we use these two. It is convenient to use spherical. π/6 φ 3π/4 by z = 3r and z = r. The sphere is ρ = 2 sin φ cos θ. The range for θ is from π/2 to π/2. This is because the sphere is on the right of yz plane. At the starting point ρ = and cos θ =. This means it starts from π/2. Lastly, ρ 2 sin φ cos θ 2. Set up the integral for the volume under the cone with vertex at (,, H), base in z = H/2 with radius R, inside x 2 + y 2 = R 2 but above xy plane. 3. Set up the integral for the volume outside r = 1 inside ρ = 2 in both cylindrical and spherical coordinates. The tricky part is the spherical. In spherical, the two intersects at 2 sin φ = 1 or π = π/6, 5π/6. Hence, in spherical, θ < 2π, π/6 φ 5π/6. For the ρ part, r = 1 is ρ sin φ = 1. Hence, 1/ sin φ ρ 2 ********************************************************** 1. Green s theorem (a) Circulation and flux Compute the line integral of x 2 y + xe x3, xy sin 2 (e y ) over the curve consists of the segments from (, ) (2, ) (2, 4) and y = x 2 oriented counterclockwise. Compute the line integral C P dx + Qdy where P = xy, Q = x 2 and C is the loop of the curve in the first quadrant whose polar equation is r = sin(2θ). We have Q x P y = 2x x = x. Hence, D xda. In polar, θ π/2, r sin(2θ) π/2 sin(2θ) 1 r cos θrdrdθ

2 (b) The integral π/2 sin 3 (θ) cos 4 (θ)dθ can be done by doing substitution u = cos θ Let F = xy 2 i + x 2 yj. Let C be the union of x-axis( x 2) and y = 4 x 2, oriented counterclockwise. Compute the flux C F nds in two ways. The first way is Green s theorem F nds = D F da. Then, you have D (x2 + y 2 )da which can be finished using polar. The second way is to parametrize directly. nds = dy, dx. Then, P dy Qdx. We parametrize the boundary with two pieces. R is the region inside (x 1) 2 + y 2 = 1 but outside x 2 + y 2 = 1. Let C be the boundary of R oriented counterclockwise. Let F = e 2x2 3x + y, y. Let C 1 be part of C which is on x 2 +y 2 = 1 and C 2 be the part of C which is on (x 1) 2 +y 2 = 1. Compute C 2 F T ds by computing the circulation on C using Green s and C 1 F dr (Hint: C 1 e 2x2 3x dx = by the Fundamental theorem). Find the area of the region enclosed by r(t) = sin(2t), sin(t) above x-axis. For the loop, t π. Then, A = ydx = π sin(t)2 cos(2t)dt. u = cos t and cos(2t) = 2 cos 2 t 1 = 2u 2 1. Answer is 4/3 2. Surface integrals (2 types) (a) Parametrization, and the surface integral of a function (b) Flux Consider the surface of revolution obtained by revolving x = f(z) about z axis. Parametrize this surface. If a z b, set up the integrals for the surface area. Use z and θ as the parameters. x = f(z) cos θ, y = f(z) sin θ, z = z Consider the fence S: x = 2 sin(t), y = 8 cos(3t), t < 2π and z 2. Set up the surface integral S 2yzdS. Use t, z for the parameters. r = 2 sin t, 8 cos(3t), z. Compute the flux of F = 2, 2, 3 across the surface S: r(u, v) = u + v, u v, uv, u, v 1 We compute directly that nds = r u r v dudv =... 2

3 Compute the flux of G = 2x, x y, y+z through the surface S, which is the portion of the plane 2x 3y + 5z = inside x 2 + y 2 = 1 oriented upward. The surface has a boundary and the field is not a curl(the divergence is nonzero). We parametrize the surface r = x, y, 1 5 (3y 2x), x2 + y 2 1. Then, the flux is Φ = G (r x r y )dxdy D Note that r x r y always points up for a function graph and we are fine. D is the disk. For this problem, use polar. Don t forget to write z in terms of x, y Parametrize the upper hemi-sphere with radius 1. Then compute the flux of F = x 2,, across it. r = sin φ cos θ, sin φ sin θ, cos φ. φ π/2, θ < 2π, ρ 1 Transform the integrals into double integrals without evaluating: S (z2 y 2 )ds and S (z2 y 2 )dxdy. S is the part of x = y 2 + z 2 on the left of x = 9. S F nds. F = x, y, z and S is the cone z = r between z = 2 and z = 5, with the normal pointing downward. 3. Stokes Theorem Let F = yz, xz, z 2 and S is 4x 2 + y 2 = e z between z = and z = 2 ln 2, with the normal point upward. Compute F nds S Let F = zi+xj yk. Let C be the intersection between x 2 +y 2 = 1 and z + y = 3 oriented counterclockwise if viewed from above. Compute C F T ds. If the curve is the intersection between x 2 + y 2 + (z 3) 2 = 1 and z + y = 3, do the problem again. Let F = 3y, 2x, x 2 y 2 z 2. Let S be the surface x 2 + y 2 + (z 1) 2 = 2 above xy plane, oriented upward. Compute ( F ) nds S 3

4 Suppose F = xy 2 z 2 + y, x 2 yz 2 + z, x 2 y 2 z + x. C is the hexagon with vertices (2,, 1), (1,, 2), (, 1, 2), (, 2, 1), (1, 2, ) and (2, 1, ), which are all in the plane x + y + z = 3. Compute the circulation of the field over this curve. (Hint: The best way is to split a conservative field first. Anyway, using Stokes theorem directly will give you the same answer.) Apply Stokes. F = y, z, x because the extra field is (x 2 y 2 z 2 /2) which has zero curl. If you don t notice this, it is fine since it will be canceled as well. Then, the order is clockwise. D 1, 1, 1 1, 1, 1 dxdy = 3Area(D) = 9 Compute the line integral of G = x 2 2y, 2e y z, z 3 + 3x along the curve r(t) = cos t, sin t, cos t sin t where t : 2π. Note that the curve is closed since r() = r(2π). Then, we can throw away a conservative field and have 2y, z, 3x. At this point, you can either use direct computation or use Stokes s theorem. If we apply Stokes s, noticing that the curve is on z = xy. 4. Divergence theorem (a) Computation and applications Let a = 2, 3, 4, b = 1, 1, 2 and c = 2, 3, 5. Let T be the parallelepiped determined by 2a + b, b, 3b 2c. S is the boundary of T with outer normal n. Compute F nds, T where F = x 2 1, y 2xy + e 8z, 3z + ln(1 + x 2 ). Compute the flux S F nds where F = y3 + z 2, xy xz 2, xe y. S is the boundary of the solid x + y + z 1, x, y, z, with n being the outer normal. Compute P dydz + Q dzdx + R dxdy, S where P, Q, R = x + e y8 z 9, y + 3y 2 + ln(x 8 + 1), z + cos(xy). S is the surface of the upper hemi-ball with radius 1, oriented outward. By divergence theorem, T (1+1+6y +1)dV. The integral of y is zero. Hence, 3V ol(t ) = π13 = 2π 4

5 (b) Surface independence Let F = y 2 z z 2, 4 xy, 3 + xz. Let S consists of the cylinder x 2 + y 2 = 1, z 1 and z = r, z 1. The normal of the cone is upward while the normal of the cylinder is outward. Compute S F n ds Compute S F nds where F = y 2 z z 2, 4 xy, 3 + xz and S is the surface of revolution by revolving z = x 2 + 3x 2, z about z axis and the normal is downward. Consider the surface z = (x 2 + y 2 1)(x 4 + y 4 + 1) for z, oriented upward. Compute S F nds where F = xy3 + y 2 z, x 2 z 2 x 2 y, zx 2 zy 3. Divergence free. We use surface independence. The surface we pick then is the one in z = plane. Hence, it is the region inside x 2 + y 2 1 =. We find that F k =. The final answer is therefore zero. let S be r(t, z) = (1 z) 8 3 cos t, (1 z) 8 2 sin t, z and t < 2π, z 1. The normal is upward. Compute the flux of F = y 2 z z 2, 4 xy, 3 + xz through this surface. ***************************************************** For 2nd derivative test, it is simple enough. I will test in Midterm 3 but no practice problems here... The following are all about Lagrange multiplier. 1. (a). Let f = 3x 2 + 4y 2 + z 2 and g = 2x + 3y + z = 1. Use Lagrange multiplier to find the candidates for extrema of f on g = 1. Identify if there is max among them, if yes, which one? How about min? No max, but there is min. Hence, among the candidates, you can find the min by evaluating function values. (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. 2. Consider that we want to make a box with 6 faces. We want the volume to be 25 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). 5

6 (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 = 25. 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 = 25/x. Then, 25λ =.... 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 >, y >, z >. Use your result to show that 3 3 abc a + b + c for any a >, b >, c >. 6