Calculus III 2004 Summer Practice Final 8/3/2004

Transcription

1 .. Calculus III 4 ummer Practice Final 8/3/4. Compute the following limits if they exist: (a) lim (x,y) (,) e xy x+. cos x (b) lim x. (x,y) (,) x 4 +y 4 (a) ince lim (x,y) (,) exy and lim x + 6 in a (x,y) (,) neighbourhood of (, ), we use quotient property of limits to get lim (x,y) (,) e xy x +. (b) If we approaches (, ) along y, we have, by l hospital rule, cos x x sin x x cos x lim lim lim. (x,) (,) x 4 x 4x 3 x x This last limit tends to as x. Thus, the lim (x,y) (,) does not exist.. Find the absolute maximum and minimum for the function f (x, y, z) x + y + z on the ball B (x, y, z) x + y + z ª. Write B U B where U {(x, y, z) x + y + z < } and B {(x, y, z) x + y + z }. ince f (,, ) 6 (,, ), wehavenocriticalpointson U. cos x x x 4 +y 4

2 To find the critical points on B, letf (x, y, z) x + y + z and g (x, y, z) x + y + z. By the method of Lagrange multiplier, we have f λ g, that is, (,, ) λ (x, y, z). o, we have the following system of equations: λx λy λz x + y + z. By λx,weknowλ 6. We have x y z. By λ the last equation, we have 3.Thisimpliesthatλ ± 3 4λ ³. Moreover, we have two critical points (x, y, z) 3, 3, 3 or (x, y, z) ³ 3, 3, 3. ince we have only two candidates of absolute maximum and minimum, one must be the absolute maximum and another ³ must be the absolute minimum. It is easy to see that f 3, 3, is the absolute maximum and f ³ 3, 3, is the absolute minimum. 3. Evaluate x + y + z dxdydz over the sphere {(x, y, z) x + y + z }. Using spherical coordinates, let x ρ sin φ cos θ, y ρ sin φ sin θ and z ρ cos φ where ρ, φ π and θ π.

3 3 o, x + y + z ρ and the Jacobian is ρ sin φ. Therefore, x + y + z dxdydz π π π π µ µ π ρ 4 dρ Ã! ρ 5 5 ρ ρ ρ sin φdθdφdρ ρ4 sin φ dθdφdρ µ π sin φdφ dθ ³ cos φ π φ θ π θ ( ( ) ( )) π 5 4π etermine whether the integral xy dxdy exists where [, ] [, ]. If it exists, compute its value.

4 4 This integral is improper whenever xy.let δ, [δ, ] [ε, ]. Wehave δ,ε as δ and ε and xy is welldefined and integrable in δ,ε. Therefore, lim δ, δ lim xy dxdy lim δ δ lim δ δ lim dx δ δ x h 4 lim x δ h lim 4 4 i δ δ 4. xy dydx " µ # xy dx x y x x lim x dx x xδ i ln δ ince lim δ ln δ lim RR δ δ δ x+y dxdy ln. x +xy+y lim δ δ δ lim δ δ,wehave 5. Evaluate y e x dxdy.

5 By drawing the graph of the region, we change the order of integration into 5 y e x dxdy x ex xe x dx x e e. e x dydx 6. Evaluate z d where is the portion of the cylinder x + y 4between the planes z and z x +3. Use cylindrical coordinates for the cylinder. Let x cosθ, y sinθ and z z where z x +3cosθ +3 and θ π. o, the parametrization Φ of is Φ (θ, z) (cosθ, sinθ, z) where θ π and z cosθ +3. We calculate o, we have T θ T z ( sinθi +cosθj) (zk) cosθi +sinθj. kt θ T z k.

6 6 Therefore, z d π cosθ+3 π π 3 6π. z () dzdθ 3 ( cos θ +3)3 dθ 8cos 3 θ +36cos θ +54cosθ +7 dθ 7. Auniformfluid that flows vertically downward (like, heavy rain) is described by the vector field F (x, y, z) (,, ). Find the total flux through the cone z p x + y,x +y. Parametrize the cone by letting x r cos θ, y r sin θ and z p q x + y (r cos θ) +(rsin θ) r where r and θ π. o, the parametrization Φ of is Φ (r, θ) (r cos θ, r sin θ, r) where r and θ π. We calculate (since our fliud flows downward, we calculate T θ T r instead of T r T θ.) T θ T r ( r sin θi + r cos θj) (cos θi +sinθj + k) r cos θi + r sin θj rk. The flux is the surface integral of F over, thatis, RR F d. Therefore, we have F d F (T θ T r ) dθdr π. Φ π π (,, ) (r cos θ, r sin θ, r) drdθ rdrdθ

7 8. Evaluate C x ydx + ydy where C is the bournary of the region between the curve y x and y x 3 where x. By Green s theorem, (y) x ydx + ydy C x (x y) dxdy y where C is the boundry of with counterclockwise orientation. We can describe as x and x 3 y x. Hence, (y) x (x y) dxdy y x. x dydx x h 3 x y x i dx yx 3 x 3 + x 5 dx 9. Let F yzi +( x +3y +)j + x + z k. Evaluate ( F) d where is the cylinder x + y a, z (without the top and bottom). We have i j k F x y z yz x +3y + x + z (x y) j +( z) k. We can parametrize as x a cos θ, y a sin θ and z z where θ π and z. Thus, the parametrization Φ 7

8 8 of is Φ (θ, z) (a cos θ, a sin θ, z). where θ π and z. We calculate T θ T z ( a sin θi + a cos θj) (zk) a cos θi + a sin θj. Therefore, ( F) d ( F) (T θ T z ) dθdz π π (, (a cos θ)+a sin θ, z) (a cos θ, a sin θ, ) dzdθ a sin θ cos θ +a sin θ dθ π a sin θ sin θ cos θ dθ a π. By tokes theorem, ( F) d F ds where is the boundary of the surface. Therefore, we have two parts of. One is x + y a and z. Another is x + y a and z. We parametrize x + y a by letting x a cos θ and y a sin θ where θ π. Hence, dx a sin θdθ and dy a cos θdθ. Moreover, in both cases, dz.

9 Thus, for the first one, we have F ds (yz) dx +( x +3y +)dy + x + z dz π π ( a cos θ +3a sin θ +)(a cos θdθ) a cos θ +3a sin θ cos θ +a cos θ dθ a π. For the second one, we have F ds (yz) dx +( x +3y +)dy + x + z dz π π π ( a sin θ ) ( a sin θdθ)+( a cos θ +3a sin θ +)(a cos θdθ) a sin θ a cos θ +3a sin θ cos θ +a cos θ dθ a sin θ a +3a sin θ cos θ +a cos θ dθ 3a π. Therefore, ( F) d F ds a π 3a π 4a π. Why do we get a different answer? oes this mean that the tokes theorem is not valid in this case? If so, why? Or, do we need to use different orientations on two parts of boundary of?. (a) how that F 6xy (cos z) i +3x (cos z) j 3x y sin z k is conservative. 9

10 (b) Find f such that F f. (c) Evaluate the integral of F along the curve x cos 3 θ, y sin 3 θ, z where θ π. (a) ince F i j k x y z 6xy (cos z) 3x (cos z) (3x y sin z), F is conservative. (b) Assume F is the gradient of f (x, y, z). Then, f must satisfy f 6xy cos z x f 3x cos z y f 3x y sin z. z Integrate the firstonewithrespecttox,wehavef (x, y, z) 3x y cos z + h (y, z). Hence, we have 3x cos z f y 3x cos z+ h h. Therefore, we have,thatis,h (y, z) y y g (z). Thus,f (x, y, z) 3x y cos z+g (z). o, 3x y sin z f z 3x y cos z + g (z). We conclude that g (z),that is, g (z) C where C is a constant. This implies that f (x, y, z) 3x y cos z + C. (c) ince F is a gradient, we have F ds c f ds c ³ ³ π f 3.. Let c ³ cos 3 π f (c ()) ³ sin 3 π cos 3 cos 3 sin 3 cos F x yi + z 8 j xyzk. Evaluate the integral of F over the surface of the unit cube.

11 Let be the surface of the unit cube W. By Gauss divergence theorem, we have F d ( F) dv. We calculate W F xy ++( xy). Thus, we have F d () dv. W