Practice problems 1. Consider a right circular cone of uniform density. The height is H. Let s say the distance of the centroid to the base is d. What is the value d/h? We can create a coordinate system such that the base is in xy plane and the z-axis is the axis. Assume the radius of the base is R. Then, the region can be written in cylindrical coordinates as r R, θ < 2π, z H R (R r). The centroid can be computed by d = z = 1 m zδdv. In our case, we can cancel δ and have z = 1 V zdv 2. 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 3. Set up the integral for the moment of inertia about z axis inside both ρ = 2 and r = 2 cos θ, outside r = 1 and above xy plane. The density is δ = x 2 + y 2 + z 2. This problem is good for cylindrical coordinates. π/3 θ π/3 and 1 r 2 cos θ, z 4 r 2 4. Find the centroid of the ice-cream cone enclosed by x 2 + y 2 + z 2 = 1, z and z = r 1, z. Suppose the density is δ = 1 (Hint: break the region into two parts. One for cylindrical and one for spherical.) x = ȳ =. z = 1 V T zdv. The region can be written as the union of two φ π/2, θ < 2π, ρ 1 and r 1, θ < 2π, r 1 z 5. Find the total mass of the ice-cream cone inside x 2 + y 2 + (z 1) 2 = 1 and above z = 3r, assuming the density is δ = 1. This is the example in the book. ρ = 2 cos φ. φ π/6, θ < 2π, ρ 2 cos φ 1
6. 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 7. (The practice problems for 14.8 will be together with the one for surface integrals.) ********************************************************** 1. Line integrals (a) Parametrization Parametrize the curve y = x 2 r = t, t 2 Parametrize x 2 + 4y 2 = 1 r(t) = cos t, 1 2 sin t, t < 2π Parametrize the boundary of the region bounded by x-axis, y = x and x = 1. C = C 1 +C 2 +C 3. C 1 : r(t) = t,, t 1. C 2 : r(t) = 1, t, t 1. C 3 : r = t, t, t : 1 Parametrize the ellipse formed by the intersection of x 2 +y 2 = 1 and x + z =. r(t) = cos t, sin t, cos t (b) Usual line integrals (2 types) Consider the curve x 2 /4 + y 2 = 1 with x, y 1/2. If the density (per unit length) is δ = y/x, compute the moment of inertia I y. I y = C x2 δds = C xyds. r = 2 cos t, sin t. t π/6 Compute the line integral of F = 3y, 2x over the curve y = x 2 for y 1 oriented from right to left. Let r = t 3, t 2, t, t 1. Compute C F T ds where F = e yz,, ye yz (c) Conservative field. Compute the line integral C (x3 + y)dx + xdy where C is the curve jointing (, ) and (1, 1). Justify your answer. The field is conservative and is (x 4 /4 + xy) Let C be r(t) = ln(1 + t 9 ), t 3 + 1, t 1, t 1. Compute C xdy + ydx + dz The field is (xy+z). Or you can notice that it is d(xy+z) 2
Show that F = (3y 3 1xz 2 )i+9xy 2 j 1x 2 zk is irrotational and thus conservative. Find a potential function φ. C F T ds where F = zexz + e x, 2yz, xe xz + y 2. r = e t2, e t3, t 4. t [, 1] It is irrotational and thus conservative. φ = e xz + y 2 z Let C be r(t) = cos 4 t, sin 4 t, 7, t < 2π and F = x 3 2. Green s theorem z, y 3, y + z 3. Compute the line integral C F d r. (Hint: Split out a conservative field. The integral of the one you split will be zero since it is on a closed curve.) The field F = x 3, y 3, z 3 is conservative. Since the curve is closed, we only need to compute C z,, y dr = C 7dx+ ydz = 7 C dx = because 1,, is again conservative. (a) Circulation and flux Compute the line integral of x 2 y + xe x3, xy sin 2 (e y ) over the rectangle with vertices (, ), (2, ), (, 3) and (2, 3) oriented counterclockwise. C F dr = P dx + Qdy. P = x 2 y + xe x3, Q = xy sin 2 (e y ). Then, Q x P y = y x 2. 3 2 (y x 2 )dxdy 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) r cos θrdrdθ 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. 3
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. If v = y 2, xy, and C is the ellipse x 2 /9+y 2 /4 = 1, compute the circulation F dr in two ways. x y (b) Suppose v =,. Suppose C is a simple closed x 2 +y 2 x 2 +y 2 curve in the plane, oriented counterclockwise. What values can C v nds be? (Hint: This field is divergence free.) Since it is divergence free except at the origin, we discuss two cases. If C doesn t enclose the singular point, the answer is zero by Green s. If the curve encloses the origin, then we use a small circle to isolate the origin. The flux then equals the flux on the small circle. The answer is 2π. Then, the answer could be or 2π Suppose F = y x,. Suppose C is a simple closed x 2 +y 2 x 2 +y 2 curve in the plane, oriented counterclockwise. What values can C F T ds be? (Hint: This field is curl free or P y = Q x.) (c) Compute the area bounded by y = x 3, y-axis and y = 1 using both line integral and double integral. 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 Find the area of the region enclosed by one arch of the cycloid x(t) = a(t sin(t)), y(t) = a(1 cos(t)) and x-axis. Example in class 3. Surface integrals (2 types) (a) Parametrization, surface area element and the surface integral of a function 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. Use x, z. r = x, f(x, z), z = x, 2x + 2z + 1, z, x 2 + z 2 1. ds = r z r x dzdx 4
(b) Flux Consider the surface of revolution obtained by revolving x = f(z) about z axis. Parametrize this surface. 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 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. Can we use Stokes s theorem here? Why? 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 S F nds where F = y, x, z and S is the surface z = θ, θ π and 1 x 2 + y 2 4. Parametrize the surface. r = r cos θ, r sin θ, θ, 1 r 2, θ π. Then, nds = r r r θ drdθ 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 =... 4. Stokes Theorem 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 in two ways. The first way is direct computation by parametrization. r(t) = cos t, sin t, 3 sin t. 5
For the second way, we can apply Stokes s theorem. S F nds. Then, we have to parametrize the surface r = x, y, 3 y. If the vector field is F = z + x 3 + sin(x), x + e y2 + e y, y cos(8z) z 1, can you reduce this to the problem here? Why? Yes. The extra field G = x 3 + sin(x), e y2 + e y, cos(8z) z 1 is conservative. The integral on the closed curve is zero. 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 Apply Stokes and reduce it to the circulation over the curve x 2 + y 2 + ( 1) 2 = 2. On this curve, the field is F = 3y, 2x,. The curve is r = cos t, sin t,. The computation is easy. 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. Hence, the surface is r = x, y, xy, x 2 + y 2 1. We have D (2 + 3x y)dxdy. However, x2 + y 2 1 is the disk. The integral of x, y are zero. Then, 2 Area(D) = 2π 5. Divergence theorem (a) Computation and applications 6
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. By divergence, we have T ( + x + )dv. This volume is x 1, y 1 y, z 1 x y. Set up the volume integral and integrate. Let F = x + e y8 z 9, y + 3y 2 + ln(x 8 + 1), z + cos(xy). Let T be the upper hemi-ball with radius 1. Compute the flux of this field out of T. By divergence theorem, T (1 + 1 + 6y + 1)dV. The integral of y is zero. Hence, 3V ol(t ) = 3 2 3 π13 = 2π Suppose v is the gravitational field generated by a cloud of mass. The physical law tells us that S v nds = 4πGm where m is the total mass inside S. Suppose the density of the mass is δ. Then, the total mass is m = T δdv. Using the divergence theorem, show that δ = 1 4πG v. If the gravitational field is given by v = 3e x + 4y 2, 2y 2 + e yz, z 2 + xyz. Compute δ(,, 1)/δ(,, ) This is a direct application of the divergence theorem. It is easy (b) Surface independence Let F = GM r be the gravitational field generated by the r 3 Earth. S is any surface that does not go across the Earth. Find the flux S F nds Two cases, the first case is when the surface doesn t enclose the origin and the answer is zero. The second case is when the surface encloses the origin. We use a small sphere to isolate the origin. Since the field is divergence free, we can reduce the flux to be on the small sphere. 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. 7
The surface is a surface with boundary r(t, z = ) = 3 cos t, 2 sin t,. The field is divergence free. Hence, we can apply the surface independence technique and have the flux to be equal to the flux through the part inside r(t, z = ) in xy plane. Hence, D F kda = D 3dA. The area can be computed using whatever way you want. You can either use the line integral or use double integrals by change of variable. If you notice it is an ellipse, you can get the answer quickly as well. 8