z k sin(φ)(x ı + y j + z k)da = R 1 3 cos3 (φ) π 2π dθ = div(z k)dv = E curl(e x ı + e x j + e z k) d S = S

Transcription

1 Mathematis H Name (please print) Final xamination May 7, 28 Instrutions: Give brief, lear answers. Use theorems whenever possible. I. Verify the Divergene Theorem for the vetor field F(x,y,z) z k on the sphere x 2 + y 2 + z 2 with the outward normal. Using the formulas F d D F ( r u r v ) dd and r φ r θ a sin(φ)(x ı + y j + z k), we have 2π π z k d while for the solid unit ball, sin(φ)os 2 (φ)dφdθ z k sin(φ)(x ı + y j + z k)da 2π 3 os3 (φ) π 2π dθ sin(φ)z 2 da 2 3 dθ 4 3 π, div(z k)dv dv vol() 4 3 π. II. Use tokes Theorem to alulate surfae x + y + z that lies in the first otant. (e x ı + e x j + e z k) d r, where is the boundary of the portion of the One alulates that url(e x ı + e x j + e z k) e x k. Then, letting be the portion of the plane in the first otant and using tokes Theorem, we have (e x ı + e x j + e z k) d r url(e x ı + e x j + e z k) d e x k d. Now is the graph of the funtion z x y over the domain in the xy-plane given by x, y x. Using the formula for a surfae integral on a surfae that is the graph of a funtion, we find e x k d e x d y e x dxdy e y dy ( e + e ) e 2. III. Let f be a salar funtion of three variables and let F be a vetor field on a 3-dimensional domain. Writing F as P ı + Q j + k, verify that div(f F) f div( F) + f F. div(ff) div(fp ı + fq j + f k) x (fp) + y (fq) + z (f) ) + f x P + f P x + f y Q + f y + f z + f ( P z f x + f div( F) + ( f x f ı + y j + f z k ( f x P + f y Q + f z ) y + z ) (P ı + Q j + k) f div( F) + f F.

2 Page 2 IV. The piture to the right shows a parameterization of the one z x 2 + y 2. It is parameterized by letting θ be the polar angle in the xy-plane, and h be the z-oordinate. The parameterization is x hos(θ) y hsin(θ) z h, where the parameter domain in the θh-plane onsists of θ 2π, h. x z y. alulate r h and r θ. r h os(θ) ı + sin(θ) j + k and r θ hsin(θ) ı + hos(θ) j. h 2. For the line h, θ π in, draw the orresponding points 4 on. Do the same for the line h 2, θ π. At the intersetion 2 point of these two urves on, draw the vetors r θ and r h. z 2π θ r h r θ x y 3. Is the upward normal (i. e. the one with positive k-omponent) equal to r θ r h or to r h r θ? It is r h r θ.

3 Page 3 Name (please print) V. (9) Let be the surfae given by the parametri equations x u 2, y usin(v), z uos(v), with the parameter domain given by u 3, v π/2.. alulate d in terms of d. We have r u 2u ı + sin(v) j + os(v) k and r v uos(v) j usin(v) k. so ı j k r u r v 2u sin(v) os(v) u ı + 2u 2 sin(v) j + 2u 2 os(v) k, u os(v) u sin(v) d r u r v d 2. Find a normal vetor to at the point (4,, 3). u 2 + 4u 4 sin 2 (v) + 4u 4 os 2 (v) d u 2 + 4u 4 d. At this point u 2 4 so u is 2 (beause u 3). From the seond oordinate, 2sin(v), so sin(v) 2 and therefore v π 6. valuating r u r v u ı + 2u 2 sin(v) j + 2u 2 os(v) k at (2,π/6), we have the normal vetor 2 ı + 4 j k. 3. Use the parameterization to alulate (x ı + z j) d. Using the formula F d D F ( r u r v ) dd, we have (x ı + z j) d (x ı + z j) ( u ı + 2u 2 sin(v) j + 2u 2 os(v) k)d 3 xu + z2u 2 os(v)d u 3π ( 2 + 2u 3sin2 (v) 2 π/2 3 π/2 ) du 3 u 3 + 2u 3 sin(v)os(v)dv du ( 2 π 2 )u 3 du 34 ( 2 π ). 4 2 VI. Apply the Divergene Theorem to show that if is the boundary of the solid, then: (a) The volume of is (x ı + y j + z 3 k) d. + y j + z 3 (x ı k) d div(x ı + y j + z 3 k)dv 3dV dv vol(). 3 (b) If n is the unit outward normal on, then D n f d f dv, where n is the unit normal to the surfaes and f is the Laplaian f xx + f yy + f zz. The Divergene Theorem gives D n f d f n d f d div( f)dv. ine div( f) div(f x ı + f y j + f z k) fxx + f yy + f zz f, the last integral equals f dv.

4 Page 4 VII. (9) Five positive numbers x, y, z, u, and v are multiplied together. That is, the produt funtion is P xy z uv.. Assume that eah of x, y, and z is inreasing at.2 units per seond, and eah of u and v is dereasing at. units per seond. Find the rate of hange of the produt at a moment when all of the numbers exept v equal, and v 2. The hain ule gives d(xyzuv) dt x (xyzuv)dx dt + y (xyzuv)dy dt + z (xyzuv)dz dt + u (xyzuv)du dt + v (xyzuv)dv dt yzuv dx dt + xzuvdy dt + xyuvdz dt + xyzvdu dt + xyzudv dt peializing to the moment with the given information, we obtain the rate of hange 2. alulate the differential of P d(xyzuv) yzuv dx + xzuv dy + xyuv dz + xyzv du + xyzudv 3. This time, assume that eah of x, y, and z is less than or equal to, and eah of u and v is less than or equal to 5. Use the differential dp to estimate the maximum possible error in the omputed produt P that might result from rounding eah number off to the nearest whole number. ounding off to the nearest integer allows any of dx, et., to be as large as.5. In eah of the four-term produts, x, y, and z are at most and u and v are at most 5, so the maximum error is /2. VIII. For the funtion sin(2x + y + z), alulate eah of the following. (a) The diretional derivative (2,,2) in the diretion toward the origin. We have (sin(2x+y +z)) 2os(2x+y +z) ı+os(2x+y +z) j+os(2x+y +z) k, so (sin(2x+y + z))(2,,2) 2os(7) ı+os(7) j +os(7) k. The vetor from (2,,2) to the origin is 2 ı jj 2 k, whih has length 3, so a unit vetor in that diretion is 2 3 ı 3 j 2 3 k. To find the diretional derivative, we alulate (sin(2x + y + z))(2,,2) ( 2 3 ı 3 j 2 3 k) 7 3 os(7). (b) The maximum rate of hange at (2,,2), and the diretion in whih it ours. The maximum rate of hange is (sin(2x + y + z))(2,,2) 6os(7), and ours in the diretion of (sin(2x + y + z))(2,,2) 2os(7) ı + os(7) j + os(7) k.

5 Page 5 Name (please print) IX. (a) For eah of the following multiple integrals, rewrite the integral to hange the order of integration as requested x 2 f(x,y)dy dx, write to integrate first with respet to x. The domain of integration is the bottom half of the disk of radius 3, so hanging the order of integration 3 y 2 gives f(x,y)dxdy. 3 3 y 2 (b) y respet to z. f(x,y,z)dz dxdy, write to integrate first with respet to y, then with respet to x, then with The region of integration is the spae above the square [,] [,] in the xy-plane and below the plane z y. For fixed z and x, the range of y is z y, and the possible xz-values are the square [,] [,]. o we have y f(x,y,z)dz dxdy z f(x,y,z)dy dxdz. X.. 2. Let P(x,y) and Q(x,y) be funtions with ontinuous partial derivatives, defined on the retangle [a,b] [,d] in the xy-plane. Verify the following fats. (Do not try to use Green s Theorem; it is not the best way to verify these fats, and that approah would not be appropriate anyway, sine these fats are some of the key steps in the proof of Green s Theorem). D x da d Q(b,y)dy d d Q(a,y)dy D x da d b Q(b,y) Q(a,y)dy a x dxdy d d Q(b,y)dy x xb dy xa d where we used the Fundamental Theorem of alulus to integrate x. P dx + Qdy d Q(a,y)dy Q(b,y)dy, where is the straight line segment from (b,) to (b,d). Parameterizing as x b, y t for t d, we have dx dt and dy dt, so P dx + Qdy d P(b,t) dt + Q(b,t)dt d Q(b,t)dt d Q(b,y)dy.

6 Page 6 XI. alulate (y ı x j) d r, where T is the equilateral triangle that has one side the straight line from (, ) T to (2,) and lies in quadrants I and IV. ( x) Using Green s Theorem, (y ı x j) d r T x (y) y da 2dA, whih is 2 times T the area of T. ine T has base of length 2, a little geometry shows that the area of T is ( 3/2)(2) 2 /2, 3, so the answer is 2, 3. XII. A ertain onservative vetor field F is of the form (y 3 + ) 2 os(x) ı + g(x,y) j, for some funtion g(x,y). Let be the upper half of the unit irle, oriented lokwise. alulate F d r. The vetor field is defined on the entire plane, and sine it is onservative, the line integral is pathindependent. o instead of using, we may use the straight path whih goes from (,) to (,) along the x-axis. That is, F d r F d r ((y 3 + ) 2 os(x) ı + g(x,y) j ) d r os(x) ı + g(x,y) j ) d r, sine y on. We also have os(x) ı + g(x,y) j ) d r os(x)dx + g(x,y)dy os(x)dx, sine on, dy dt for any parameterization. Taking the parameterization x t, we have os(x)dx os(t)dt 2 sin().