alculus III Math 33 pring 7 Final exam May 3rd. uggested solutions This exam contains twenty problems numbered 1 through. All problems are multiple choice problems, and each counts 5% of your total score. Problem 1 Find the center of the sphere x + y + z 6x + y + 6 =. A) ( 6,, ) B) ( 3, 1, ) ) (,, ) ) (1, 1, 1) F) (3, 1, 3) G) (6,, ) H) (6,, 6) E) (3, 1, ) ompleting squares gives (x 3) + (y + 1) + z = 6 + 9 + 1 = 4. This is a sphere with center (3, 1, ) and radius. Problem ompute curl 5x + 3yz, 7y + xz, 3z + 3xy. A) 3z, x, 3y B) z,, x ) 1x, 14y, 6z ) 3xy xz, 3yz 3xy, xz 3yz E) x,, z F) 5x, 7y, 3z G) 3x, 3y, z H) 3yz, xz, 3xy Let F (x, y, z) = P, Q, Z = 5x + 3yz, 7y + xz, 3z + 3xy. Then the curl of F is curl F = F R = y Q z, P z R x, Q x P y = 3x x, 3y 3y, z 3z = x,, z. 1
Problem 3 alculate the divergence of 5x + 3yz, 7y + xz, 3z + 3xy. A) x 3y 3z B) x z ) 1x + 14y + 6z ) E) 1 F) 5x + 7y + 3z G) 3x + 3y + z H) 3xy + xz + 3yz We calculate div 5x + 3yz, 7y + xz, 3z + 3xy = (5x +3yz) x + (7y +xz) y = 1x + 14y + 6z. + (3z +3xy) z Problem 4 Let be the part of the paraboloid z = 3 x y above the xy-plane, with upward orientation. alculate the surface integral (flux integral) of F (x, y, z) = x, y, z over. A) π B) 3π ) 5π ) 7π H) 15π E) 9π F) 11π G) 13π Because is the graph of a function, z = g(x, y), we can regard x and y as parameters and calculate the surface integral using the formula F d ( = P z Q z + R) da. x y This gives F d = [ x( x) y( y) (3 x y ) ] da = 3 ( x + y 1 ) da. The region is the projection of to the xy-plane, which is the disk centered at the origin with radius 3. witching to polar coordinates, F d = π 3 3 ( r 1 ) r dr dθ = π 3 [ 1 4 r4 1 r] 3 dθ = 9 π 4 dθ = 9π.
Problem 5 alculate F d r, where F (x, y) = xy, x + y and is the curve in the figure below. y 3 1 1 1 3 4 5 6 1 x 3 A) B) 1 ) ) 3 E) 4 F) 5 G) 6 H) 7 The function f(x, y) = x y + 1 y satisfies f = F. The Fundamental Theorem for Line Integrals says that F d r = f(, 3) f(1, 1) = 9 3 = 3. 3
Problem 6 Let a =, 1, and b = 1,,. alculate a, b, a b, a b. What is the largest value? A) B) ) 5 ) 6 E) 3 F) 4 G) 5 H) 6 We calculate a =, 1, = + 1 + ( ) = 5, b = 1,, = ( 1) + + ( ) = 3, a b =, 1, 1,, = + + 4 = 6, a b =, 1, 1,, =,, 1 = 3. Problem 7 Find the arc length of the curve given by r(t) = t t, cos(3t), sin(3t), t 3. A) B) 6 ) 1 ) 14 E) 18 F) G) 6 H) 3 ince r (t) = 3 t, 3 sin(3t), 3 cos(3t), we get r (t) = 9t + 9 sin 3t + 9 cos 3t = 9t + 9 = 3 t + 1. Therefore, the arc length of this curve is 3 r (t) dt = 3 3 t + 1 dt = [ 3 3 (t + 1) t + 1 ] 3 = 8 4 1 = 14. 4
Problem 8 What kind of surface is described by the parametrization r(u, v) = u cos v, u sin v, u, u 1, v π? A) Ellipsoid B) Paraboloid ) Plane ) one E) Hyperboloid of one sheet F) Hyperboloid of two sheets G) Hemisphere H) ylinder One way to solve this is to observe that z = x + y, where x = u cos v, y = u sin v, and z = u. The equation z = x + y is that of a paraboloid. Alternatively, we can look at the grid lines obtained keeping either u or v constant. If u is constant we get circles, while if v is constant the grid lines are parabolas. 5
Problem 9 Evaluate 4 by reversing the order of integration. A) 5 B) 3 ) 9 ) 469 9 9 5 5 H) 6 5 y 1 + x3 dx dy E) 5 F) 47 G) 9 9 9 5 In the picture below, the lines y = 4 and x = and the parabola y = x have been graphed. The integral is taken over the region bounded by the parabola, the line x = and the x-axis. 5 4 3 1 1 3 4 5 Integrating first with respect to y, we can rewrite the integral as follows: x [ 1 + x3 dy dx = ] y 1 + x 3 y=x dx = x 1 + x 3 dx. y= The substitution u = 1 + x 3 gives du = 3x dx. Hence x 1 + x 3 dx = 9 1 1 3 u du = 1 3[ u u] 9 3 1 = (7 1) = 5. 9 9 6
Problem 1 Let F be the vector field F (x, y, z) = x, e sin(z), (5 + 3y ) 7. Let E be the solid bounded by the cylinder x + y = 1 and the planes z = and z = 4, while is the (positively oriented) boundary surface of E. Use the ivergence Theorem to evaluate F d. A) π B) π ) 3π ) 3π E) 4π F) 3 π G) 5π H) 6π By the ivergence Theorem, F d = div F dv, E where E is the solid described above. Here div F = 1 + + = 1, so F d = 1 dv = volume(e) E ince E is a cylinder with radius 1 and height 6, we know that volume(e) = π 1 6 = 6π. 7
Problem 11 Which of the following statements are true? I) The vectors 1, 3, 4 and 1, 3, 4 are orthogonal. II) The vector 1,, 5 is longer than the vector 3, 3, 1. III) The cross product of 1,, 1 and, 3, 3 is parallel to 3, 1, 3. A) None of them B) Only I) ) Only II) ) Only III) E) I) and II) F) I) and III) G) II) and III) H) All of them I) The dot product of the two vectors is nonzero, which implies that they are not orthogonal. II) 1,, 5 = 6 and 3, 3, 1 = 19, so this statement is correct. III) 1,, 1, 3, 3 = 3, 1, 3, so they are indeed parallel. 8
Problem 1 Find the curvature of the curve r(t) = ( sin t ) i + ( cos t ) j, π t π, at the unique point where the tangent to the curve is parallel to the x-axis. A) B) 1 ) ) 3 E) 4 F) 5 G) 6 H) 7 The short answer is that the curve is a circle of radius 1, and therefore has constant curvature 1. If we do not make that observation, we can carry out the following calculations. We first find the tangent vector r and the unit tangent vector T to the curve, r (t) = cos t, sin t, r (t) = (cos t) + ( sin t) = 1, T (t) = r (t) r (t) = r (t) = cos t, sin t. The tangent r (t) is parallel to the x-axis when its y-component is zero. That is, when sin t =, which implies t =. To find the curvature at t =, we also calculate T (t) = sin t, cos t, T (t) = ( sin t) + ( cos t) = 1. Thus, the curvature is κ() = T () r () = 1. 9
Problem 13 Let be the region in the xy-plane bounded by the upper semicircle x + y = 1 and the x-axis, and let be the positively oriented boundary of. Use Green s Theorem to evaluate xe cos x dx + ( x 4 y + 4x y 3) dy. A) B) 4 ) 8 ) 1 E) 16 F) G) 4 H) 8 Green s Theorem says that xe cos x dx + ( x 4 y + 4x y 3) dy = = ( (x 4 y + 4x y 3 ) x 8x 3 y + 8xy 3 da. ince is a circular region, we convert to polar coordinates: 8x 3 y + 8xy 3 da = = = π 1 π 8xy ( x + y ) da = π 1 4r 5 ( sin θ cos θ) dr dθ = [ 3 r6] r=1 r= sin θ dθ = π ( ) xe cos x ) da y 8(r cos θ)(r sin θ)(r ) r dr dθ π 1 4r 5 (sin θ) dr dθ sin θ dθ = [ 1 cos θ] π =. 3 3 1
Problem 14 The space curve is parametrized by x = t 3, y = t, z = t 4, t 1. Evaluate ( 3x + 8yz ) ds. A) 1 (6 6 1) B) 1 (17 17 1) 34 4 1) ) 1 (6 6 1) ) 1 18 (3 3 13 E) 1 15 (17 17 1) F) 1 16 (8 8 1) G) 1 19 (15 15 1) H) 1 13 (9 9 1) We first calculate ds = ( dx dt Using the parametrization, ) ( + dy ) ( dt + dz ) (3t ) dt dt = ( ) ( ) + 1 + 4t 3 dt = 1 + 9t4 + 16t 6 dt. ( ) 1 ( 3x + 8yz ds = 3t 3 + 8t 5) 1 + 9t 4 + 16t 6 dt. To integrate this we use the substitution u = 1 + 9t 4 + 16t 6, for which du = dt 36t3 + 96t 5 = 1 ( 3t 3 + 8t 5). Therefore, ( 3x + 8yz ) ds = 6 1 [ 1 1 u du = 1 u u ] 6 = 1 (6 6 1). 1 3 1 18 11
Problem 15 Find a constant M such that where R is the rectangle R xe y da = M area(r), R = { (x, y): x 4, y 1 }. A) ( e 1 ) B) e G) 1 ( e 1 ) H) 1e ) 3 ( e 1 ) ) 3e E) 6 ( e 1 ) F) 6e Note that M will correspond to the average value of the function xe y over the rectangle R. The area of the rectangle R is 1 =. Let us compute the integral, R xe y da = 4 1 This implies that M = 3(e 1). xe y dy dx = 4 [ ] xe y y=1 dx = (e 1)[ 1 x] 4 = 6(e 1). y= 1
Problem 16 Use tokes Theorem to evaluate F d r, where F (x, y, z) =,, x and is the boundary of the part of the plane 3x + y + z = 6 in the first octant. ( is oriented counterclockwise when viewed from above.) A) B) 4 ) 8 ) 1 E) 16 F) G) 4 H) 8 We use tokes Theorem to write F d r = curl F d, where is the part of the plane 3x + y + z = 6 bounded by (oriented upward). The curl of F is x j, so using the formula F d ( = P z Q z + R) da, x y we calculate curl F d = x( ) da = 4x da. Here is the projection of to the xy-plane. This will be the triangle bounded by the x- and y-axes, and the line 3x + y = 6 y = 3 3 x. Thus, 4x da = 4 3 3 x x dy dx = 4 3x 3 x dx = 4 [ 3 x 1 x3] = 8. 13
Problem 17 Find the volume of the solid bounded by the cylinder x +y = 6, and the two planes y + z = 7 and y + z = 14. A) 6π B) 7π ) 1π ) 36π E) 4π F) 63π G) 84π H) 5π Using cylindrical coordinates, we need to evaluate the following integral, π 6 14 y 7 y 1 r dz dr dθ = = 7 π 6 π ( (14 y) (7 y) ) r dr dθ = 7 π 6 [1 π r] r= 6 dθ = 7 3 dθ = 1 π = 4π. r= r dr dθ Alternatively, we could look at it from a geometrical point of view: If we cut off the bottom (skew) part and place it on top, then we get a circular cylinder with radius 6 and height 7. The volume of such a cylinder is πr h = π 6 7 = 4π. 14
Problem 18 Which of the vector fields are conservative? x, y, y e x, y e x, yz, xz, xy A) None of them B) Only x, y ) Only y e x, y e x ) Only yz, xz, xy E) x, y and y e x, y e x F) x, y and yz, xz, xy G) y e x, y e x and yz, xz, xy H) All of them = P, while a three- y A two-dimensional vector field F = P i + Q j is conservative if Q x dimensional vector field F = P i + Q j is conservative if curl F =. For x, y, we find that so it is conservative. For y e x, y e x, while Therefore, it is not conservative. Finally, ( ) ( ) x = = y, y x ( y e x) = ye x, y ( y e x) = y e x. x curl yz, xz, xy = x x, y y, z z =,, =, such that yz, xz, xy is conservative. 15
Problem 19 Given the function f(x, y) = 5x y + 5 3 y3 5x 5y + 7. How many local maxima, local minima, and saddle points does f have? A) 1 local minimum, and 1 saddle point B) 1 local maximum, and 1 saddle point ) 1 local maximum, and 1 local minimum ) 1 local maximum, 1 local minimum, and saddle points E) 1 local maximum, local minima, and 1 saddle point F) local maxima, 1 local minimum, and 1 saddle point G) local minima, and saddle points H) local maxima, and saddle points We first find the partial derivatives of f, f f = 1xy 1x, = x y 5x +5y 1y. The critical points are then where f = f =. The critical points are (, ), (, ), (1, 1), ( 1, 1). x y Next we calculate the second derivatives, f = 1y 1, x f = 1y 1, y f x y = 1x. The discriminant is = (1y 1) 1x. At the point (, ): > and f < local maximum, x At the point (, ): > and f > local minimum, x At the point (1, 1): < saddle point, At the point ( 1, 1): < saddle point. This gives us 1 local maximum, 1 local minimum and saddle points. 16
Problem Evaluate ( x + y ) d, where is the surface parametrized by x = u + e v, y = u, z = v + 3, u 3, v 1 ln(). A) 1 1 B) ) 1 1 ) 3 3 1 1 E) 3 3 F) 4 4 G) 4 4 3 3 H) 5 5 3 3 Let r be the vector function r(u, v) = u + e v, u, v + 3, u 3, v 1 ln(), which describes. Then ru r v =, 1, e v,, =, 4, e v = + e v. Using the parametrization, (x + y) = ( u + e v + ( u) ) = e v, so ( x + y ) d = 1 ln 3 e v 1 + e v ln du dv = 3 e v + e v dv. To integrate this, we use the substitution w = + e v. Then dw dv = ev, and 1 ln 3 e v + e v dv = 3 1 1 [ w dw = 3 w w ] = 1 1. 3 1 17