# This makes sense. t /t 2 dt = 1. t t 2 + 1dt = 2 du = 1 3 u3/2 u=5

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1 1. (Line integrals Using parametrization. Two types and the flux integral) Formulas: ds = x (t) dt, d x = x (t)dt and d x = T ds since T = x (t)/ x (t). Another one is Nds = T ds ẑ = (dx, dy) ẑ = (dy, dx) in 2. (a). ompute the average of the polar angle on x 2 + y 2 = 4, y. The weighted average is fds/ ds In the case here, f = θ. We parametrize the curve with x(t) = (2 cos t, 2 sin t), t π. It s easy to see that tan θ = y(t)/x(t) = tan t and θ = t actually. Also ds = x dt = 2dt. Hence, the average is This makes sense. θ = π π t2dt π2 = 2dt 2π = π 2 (b). Let be y = ln x, 1 x 2. ompute x2 ds We parametrize the curve as x(t) = (t, ln t), 1 t 2. We can compute that ds = x dt = 1 + 1/t 2 dt. Hence, the integral is ˆ 2 1 t /t 2 dt = ˆ 2 1 t t 2 + 1dt = ˆ u 2 du = 1 3 u3/2 u=5 u=2 (c). Let F = ( 2y + 2x, 2x 2y) and is x(t) = (t, t 2 ), t 1. ompute the work done by F along the curve. The field is not conservative and the curve is not closed. evaluate directly. The work is F d x We have to Under the parametrization, F = ( 2t 2 + 2t, 2t 2t 2 ) and d x = (1, 2t)dt. Hence, the integral is (1 ( 2t 2 + 2t) + 2t(2t 2t 2 ))dt = (2t + 2t 2 4t 3 )dt = 2/3 (d). Let be the line segment from (1, 2) to ( 1, 2). Find the rate at which the amount of fluid flows across this curve where the velocity field is v = 4xyî y 2 ĵ. Parametrize the curve: x(t) = (1 2t, 2), t 1. The integral is v Nds. The curve is not closed and we have to compute directly. You may want to find N and ds respectively. If you do this, x = ( 2, ) and T = ( 1, ). Hence N = (, 1) and ds = x dt = 2dt. However, we can use the formula directly Nds = (dy, dx) = (dt, ( 2dt)) = (, 2)dt. Anyway, the integral is ( 4(1 2t) 2 4 ) ( 2 This means the flow is actually downward. ) dt = 8 1

2 2. (Line integrals of conservative fields the fundamental theorem) (a). Is the field F = (sin(y 2 ) + 4x 3 y, 2xy cos(y 2 ) + x 4 ) conservative? If yes, find the potential φ so that F = φ(note in our current textbook, there s a negative sign in the front of gradient.) We check that P y = 2y cos(y 2 ) + 4x 3 and Q x = 2y cos(y 2 ) + 4x 3. They are equal and the domain is simply connected. Then, yes. To find the potential, we find f first. ˆ f = P dx = x sin(y 2 ) + x 4 y + g(y) Using the fact f y = Q, we figure out g = and hence f = x sin(y 2 ) + x 4 y +. One possible φ = x sin(y 2 ) x 4 y (b). Let F = (2y + 2x, 2x 2y) and is x(t) = (t, t 2 ), t 1. ompute the work done by F along the curve. Notice that the field is conservative and we find f = x 2 + 2xy y 2 +. The starting point is (, ) while the end point is (1, 1). Hence ˆ f d x = f(1, 1) f(, ) = 2 (c). Let F = (2x + 2y, 2x + 2y, z). Is this field conservative? Let be the line segment from (1, 1, ) to (1, 2, 2). Find the line integral F T ds We check that F =. It s conservative. If you observe, you may guess out the function quickly f = x 2 + 2xy + y z2 +. If you can t, you may just integrate: ˆ f = (2x + 2y)dx = x 2 + 2xy + g(y, z) Then, f y = 2x + g y = 2x + 2y and hence g y = 2y. ˆ g = 2ydy = y 2 + h(z) Finally, f z = g z = h (z) = z. You solve h(z) = z 2 /2 + Noticing T ds = d x and using the fundamental theorem, the answer is f(1, 2, 2) f(1, 1, ) = ( ) ( ) = 7 3. (Line integrals on closed curve Green s theorem. Two versions) (a). Let be the boundary of the region y 1, y 2 x 1 with counterclockwise orientation. ompute y2 sin(x 2 )dx 2

3 Applying Green s, we have 2y sin(x 2 )da = 2 y 2 y sin(x 2 )dxdy hange the order of integration 2 ˆ x y sin(x 2 )dydx = x sin(x 2 )dx = 1 2 cos (b). Given F = (x 2 y, xy 2 ) and is the boundary of the unit circle oriented counterclockwisely. ompute the circulation of F on the curve. (irculation is F d x) The circulation equals x 2 ydx xy 2 dy = ( x 2 y 2 )da = ˆ 2π r 2 rdrdθ = π 2 (c). Let v = (2x 3, y 3 ) and be the circle x 2 + y 2 = 4 oriented counterclockwisely. ompute the outer flux of v on the curve. v Nds = vda = (6x 2 3y 2 )da = ˆ 2π ˆ 2 (6 r 2 cos 2 θ 3 r 2 sin 2 θ)rdrdθ We integrate on θ first using double angle formula and have π ˆ 2 (6 r 2 3 r 2 )rdr = π = 12π (d). ompute xdy (i). Along the boundary of ellipse x2 /4 + y 2 = 1 (ii). Along a shaped curve with the right loop oriented counterclockwisely, assuming the right loop encloses area 3 and the left loop encloses area 2. (i). We compute the line integral directly. x(t) = (2 cos t, sin t), t < 2π. Notice that t is no longer the polar angle opposed to the circle case. Then, the integral is 2π 2 cos t cos tdt = 2π (1+cos(2t))dt = 2π. Actually, by Green s, the integral is dxdy = Area. (ii). Applying Green s, we have da. However, the loop is counterclockwise on the right but clockwise on the left half. So the answer is Area(right) Area(left) = 3 2 = 1 3

4 (e*) Let be any closed in the plane that encloses the origin. Let v = ( y/(x 2 + y 2 ), x/(x 2 + y 2 )). ompute v d x. The idea is to isolate the singular point (, ) with a small circle with radius r because we can t apply Green s directly. Let 1 be the boundary of the small circle. Then, noticing P y = Q x for v, we have ˆ ˆ v d x = v d x 1 On this small circle, x = r cos t, y = r sin t or x(t) = (r cos t, r sin t), t < 2π. Using the parametrization, we can compute the line integral directly. The answer is 2π 4. (Surface Integrals Basic definition and parametrization) (a). Parametrize the surfaces and compute d S = NdA, da for (i) x, y 3, z = 1. (ii). z = xy, 1 x 2, 1 y 3. (iii). z = x2 + y 2, 1 z 2 (i). We let u = x, v = y and we see that z = 1 always. Hence, x(u, v) = uî + vĵ + ˆk = (u, v, 1) with u, v 3. NdA = xu x v dudv = ˆkdudv. Hence, da = NdA = ẑdudv = dudv (ii). Let x = u, y = v. Then, z = uv. Then, x(u, v) = (u, v, uv) with 1 u 2, 1 v 3. We compute that NdA = (1,, v) (, 1, u)dudv = ( v, u, 1)dudv. Then, da = ( v, u, 1) dudv = 1 + u 2 + v 2 dudv (iii). You can use artesian as well. Here, a better choice is to use cylindrical coordinates. Then, we see that x = r cos θ, y = r sin θ, z = r. Hence, we have x(r, θ) = (r cos θ, r sin θ, r) with 1 r 2, θ < 2π. We then, compute that NdA = (cos θ, sin θ, 1) ( r sin θ, r cos θ, )drdθ = ( r cos θ, r sin θ, r)drdθ Hence da = r 1 + 1drdθ = 2rdrdθ ompare this with flat plane in cylindrical coordinates, or polar coordinates. In polar, x = (r cos θ, r sin θ, z ) where z is a constant. You get da = rdrdθ, which is different from our da here. (b). Let S be the surface z = sin(xy) for x, y π/2. Set up the integral for (x + z)da S x(u, v) = (u, v, sin(uv)). da = x u x v dudv = v 2 cos 2 (uv) + u 2 cos 2 (uv) + 1dudv The integral is ˆ π/2 ˆ π/2 (u + sin(uv)) v 2 cos 2 (uv) + u 2 cos 2 (uv) + 1dudv 4

5 (c*). onsider the surface determined by F (x, y, z) = 1, 1 z 2 where F (x, y, z) = x 2 +y 2 z 2. Set up the integral S v NdA where v = (x, 1, ). We parametrize the surface as x(x, y) = (x, y, z(x, y)). differentiation, we have x x = (1,, x z ), x y = (, 1, y z ) Using implicit We have NdA = x x x y dxdy = ( x/z, y/z, 1)dxdy The integral is ( x2 z y z )dxdy Further, we solve z = x 2 + y 2 1 and we can plug in. The integration region is x 2 + y or 2 x 2 + y 2 5. Using polar coordinates, the integral is ˆ 2π ˆ 5 ( r2 cos 2 θ 2 r2 1 r sin θ r2 1 )rdrdθ 5. (hanging line integrals to flux integrals(surface version)-stokes Theorem) (a). Let be the intersection of x 2 + y 2 = 4 with x + y + z = 8. Evaluate (x+y)dx+(y+z)dy+(z+x)dz in two ways, where is counterclockwise when viewed above. This is the circulation F d x where F = (x + y, y + z, z + x). The first way is to evaluate this directly by parametrization: Let x = 2 cos t, y = 2 sin t. Then, z can be determined from the equation of the plane: z = 8 x y = 8 2 sin t 2 cos t. Hence, the curve can be parametrized as x(t) = (2 cos t, 2 sin t, 8 2 cos t 2 sin t). Also, d x = ( 2 sin t, 2 cos t, 2 sin t 2 cos t)dt. On this curve, the force is F = (2 cos t+ 2 sin t, 8 2 cos t, 8 2 sin t).the line integral is thus ˆ 2π (2 sin t+2 cos t)( 2 sin t)dt+(8 2 cos t)2 cos tdt+(8 2 sin t)(2 sin t 2 cos t)dt = 4π 4π 4π = 12π The second way is to use Stokes Theorem. We pick the surface to be the plane z = 8 x y inside the cylinder. The surface can thus be parametrized as x(x, y) = (x, y, 8 x y) with x 2 + y 2 4 5

6 We compute the curl F = ( 1, 1, 1). We then compute Hence, we have NdA = (1,, 1) (, 1, 1)dxdy = (1, 1, 1)dxdy ( 3)dxdy = ˆ 2π ˆ 2 ( 3)rdrdθ = 3 4π (b). Use Stokes Theorem to compute the circulation of F = (y 2 + z 2 )î + (x 2 + z 2 )ĵ + (x 2 + y 2 )ˆk along the boundary of the triangle cut from the plane x+y +z = 1 by the first octant, counterclockwise when viewed from above. F = (2y 2z, 2z 2x, 2x 2y). We use the surface as the triangle. Then, we parametrize the surface as x(x, y) = (x, y, 1 x y). We have NdA = (1, 1, 1)dxdy. Hence the integral is (2y 2z + 2z 2x + 2x 2y)dxdy = (c). Let S be the hemisphere x 2 + y 2 + z 2 = 9, z with normal pointing away from origin. Let v = ( y, x 3 + xy 2, xyz). ompute the flux: curl( v) NdA First of all, using Stokes, this is equal to v d x S where is the circle x 2 + y 2 = 9, z =. Here, one may parametrize the curve and get x = (3 cos t, 3 sin t, ), t < 2π and reduce the integral to ˆ 2π ( 3 sin t, 27 cos 3 t + 27 cos t sin 2 t, ) ( 3 sin t, 3 cos t, )dt Another smarter way is to notice dz = on the circle and thus the integral is actually ydx + (x 3 + xy 2 )dy in xy plane. Then, applying Green s Theorem, this is (1 + 3x 2 + y 2 )da 6

7 where is the disk. This can be evaluated using polar coordinates. (d). Let A(1,, ), B(, 2, ) and (,, 1). onsider the closed curve = AB + B + A. Let F = ( y, x, z). ompute the line integral F T ds. The integral is just F d x. The curve is counterclockwise when viewed above. Hence, we can apply Stokes Theorem and have: F d x = F NdA S, the flux of curl. Here S can be picked as the plane determined by the three points. It s easy to compute that F = (,, 2). Now we need to determine the plane: You can determine the normal n = AB A and use AP n = to figure out. However, according to the three points, it s easy to find the plane is x/1 + y/2 + z/( 1) = 1. In general, the plane that passes (a,, ), (, b, ), (,, c) is x/a + y/b + z/c = 1. Anyhow, the surface is parametrized as x(x, y) = (x, y, x + y/2 1), x + y/2 1, x, y NdA = ( 1, 1/2, 1)dxdy. Hence the integral is ˆ 2(1 x) 2dydx = 4(1 x)dx = 2 6. (Reducing flux integral on closed surfaces to volume integrals-ivergence Theorem) (a). Let R be the region inside the sphere x 2 + y 2 + z 2 = 4 and above xy plane. Let S be the boundary of this region which is thus closed. ompute the flux S v NdA where v = (xy 2, x 2 y + y 3 /3, x 2 z). Applying divergence theorem, we have vdv = (y 2 + (x 2 + y 2 ) + x 2 )dv This is good for spherical coordinates. 2(x 2 + y 2 ) = 2ρ 2 sin 2 φ, dv = ρ 2 sin φdρdφdθ and ρ 2, θ < 2π, φ π/2 ˆ 2π ˆ π/2 ˆ 2 2ρ 2 sin 2 φρ 2 sin φdρdφdθ 7

8 For sin 3 φ, just write it as (1 cos 2 φ) sin φ and do sub u = cos φ. (b). Let be the region x 2 + y 2 4, z 3. ompute the flux of v = ( y, x, z) through the boundary of. raw the field and explain intuitively why the flux is positive. Applying the divergence, ( + + 1)dV = dv The region is good for cylindrical coordinates. r 2, θ < 2π, z 3, dv = rdrdθdz. Answer is 12π (c). Let v = (y 2, xyz, xz 2 ). ompute the rate at which the fluid flows out of the cube x, y, z 1. We need to evaluate S v NdA. Applying divergence theorem, ( + xz + 2xz)dV = 3xzdxdydz = 3/4 (d). onsider S 1 : z = x 2 + y 2 and S 2 : z = x 2 + y 2. These two surfaces enclose a region R. The boundary of this region is S with outer normal N. ompute the flux integral S F NdA where F = (y, x, z 2 /2). The surface is closed. We just apply the ivergence theorem and have F dv = zdv R R The region is actually above z = x 2 + y 2 and below z = x 2 + y 2. The intersection of them is when x 2 +y 2 = 1 or x = y =. Anyway, this region can be written conveniently in cylindrical coordinate, which is r 1, θ < 2π, r 2 z r. Hence the integral is ˆ 2π ˆ r r 2 z[rdzdθdr] = 2π ( 1 2 r3 1 2 r5 )dr = π/12 7. (url, ivergence etc) Let v = (xy 2, x 2 y + y 3 /3, x 2 z). ompute the curl curl( v) = v and the divergence div( v) = v. ivergence is easy: v = (xy 2 ) x + (x 2 y + y 3 /3) y + (x 2 z) z = 2x 2 + 2y 2 8

9 url is i x xy 2 j y x 2 y + y 3 /3 k z x 2 z = i[(x 2 z) y (x 2 y +y 3 /3) z ] j[(x 2 z) x (xy 2 ) z ]+ k[(x 2 y +y 3 /3) x (xy 2 ) y ] = (, 2xz, ) 8. (Tangent planes, Implicit differentiation, chain rule) (a). onsider the implicit functions defined by xy yz+e xz = 3. ompute z/ x. If you forget the formula, use F (x, y, z(x, y)) = to recover. z derivative on x, you have F x + F z x =. The derivative is z x = F x F z = y + zexz y + xe xz Taking (b). Find the tangent plane of the surface defined in (a) at the point (, 1, z ). We first solve z. Plugging the values, +z +1 = 3 and hence z = 2. To compute the tangent plane of the level set, we compute the normal vector F = (y + ze xz, x z, y + xe zx ) = (1, 2, 1). The tangent plane is x y + 1 z 2 = (c). ompute the linear approximation of z = 2 x y at (1, 1), and the tangent line of the level set passing through (1, 1). Let f(x, y) = 2 x y. We have f(1, 1) = 1, f x = 2 x y ln 2 and f x (1, 1) = ln 2. Similarly, f y (1, 1) = ln 2. Hence the linear approximation is f(x, y) 1 + ln 2(x 1) ln 2(y 1) The graph of this function is the tangent plane of the graph. Also you should be able to compute the second order Taylor expansion. The tangent line to the level set is (ln 2)(x 1) (ln 2)(y 1) = 9

10 (d). If f(1, 2, 3) = ( 2, 3, 5). Let g(x, y) = f(xy, x + y, y 2 + 2x). ompute g x (1, 1) and g y (1, 1) Use the chain rule. We assume the function f(u, v, w). Then, we have g x (x, y) = f u u x + f v v x + f w w x = f u y + f v 1 + f w 2 Plugging in (1, 1), we have g x (1, 1) = f u (1, 2, 3) 1 + f v (1, 2, 3) 1 + f w (1, 2, 3) 2 = = 11 omputing g y (1, 1) is similar. The answer is (Volume integrals in spherical coordinates; cylindrical coordinates) (a). Find the moment of inertia about z-axis of the ball x 2 + y 2 + z 2 1 with density µ(x, y, z) = z 2 The moment of inertia about z axis is I z = (x 2 + y 2 )µ(x, y, z)dv This problem is good for spherical. ρ 1, θ < 2π, φ π and µ = ρ 2 cos 2 φ. x 2 + y 2 = ρ 2 sin 2 φ. Further, dv = ρ 2 sin φdρdθdφ Then, the integral is ˆ 2π ˆ π ρ 2 sin 2 φρ 2 cos 2 φρ 2 sin φdρdφdθ = 2π 7 ˆ π sin 3 φ cos 2 φdφ This integral can be evaluated by writing sin 3 φ = (1 cos 2 φ) sin φ and doing u = cos φ. You can finish it (b). Find the center of mass of the region inside x 2 + y 2 1, x 2 + y 2 4, bounded by z = x 2 + y 2 and xy plane with unit density. The domain is symmetric in x and µ(x, y, z) = 1 = µ( x, y, z). We know x = by symmetry. Similarly, ȳ =. We only have to compute z which is zµdv z = µdv The domain is good for cylindrical. In cylindrical, we have 1 r 2, θ < 2π, z r 2 and µ = 1. Further, dv = rdrdθdz. Hence, we have z = 2 2π r π 1 The center is (,, 63/45) zrdzdθdr r2 rdzdθdr = π 2 1 r5 dr 2π 2 1 r3 dr =

11 1. (2nd derivative test; Lagrange multiplier) (a). onsider f(x, y) = x 3 + y 3 3xy. Find all critical points on R 2 and classify them. f x = 3x 2 3y = and f y = 3y 2 3x =. Hence, x 4 x = and we have x =, x = 1. Using y = x 2, we have (, ), (1, 1). f xx = 6x, f xy = 3, f yy = 6y At (, ), f xx f yy f 2 xy <, the point is a saddle point; at (1, 1), the three numbers are 6, 3, 6 respectively and f xx f yy f 2 xy = 36 9 >, further f xx > and the point is a local min. (b). Find the minimum surface area of a rectangular box without bottom, provided the volume is V. Let the dimensions be x, y, z. f(x, y) = xy +2xz +2yz with the constraint g(x, y, z) = xyz = V. Then y + 2z = λyz x + 2z = λxz 2x + 2y = λxy Now, you have x(y + 2z) = y(x + 2z) = z(2x + 2y). Then, x = y, x = 2z. Hence x 2 x/2 = V or x = 3 2V. Hence y, z can be found. There must be a minimum value and we only have one candidate. This should be the point. Plugging the values and the area is... 11

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