Math 222

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Final Exam Solutions

Spring 2008

Suppose that over a certain region of space the electrical potential V is given by

V (x, y, z) = x2 + yz + xyz . (8 pts)

(a) Find the rate of change of the potential at P (1, 1, −1) in the direction of the vector v = (3, 4, 5) . Solution. We have ∇V = (2x + yz, z + xz, y + xy) , so ∇V (1, 1, −1) = (1, −2, 2) and the √ 1 (3, 4, 5) · (1, −2, 2) = 1/ directional derivative is | 1v| v · ∇V = 5√ 2. 2

(4 pts)

(b) In which direction does V change most rapidly at P ? Solution. The rate of change is greatest in the direction of the unit vector ∇V /|∇V | which , 2). in this case is (1, −2, 2)/|(1, −2, 2)| = ( 31 , −2 3 3

(4 pts)

(c) What is the maximum rate of change at P ? Solution. The maximum rate of change is |∇V (1, 1, −1)| = |(1, −2, 2)| = 3 .

(18 pts)

2

Determine all the local minima, local maxima, and saddles of the function f (x, y) =

x4 − x2 − 2xy + y 2 . Solution. The critical points satisfy ∇f = (4x3 − 2x − 2y, −2x + 2y) = (0, 0) . From the second coordinate we get x = y , and then the first coordinate equation becomes 4x3 − 4x = 0 , with solutions x = 0, ±1 . Thus the critical points are(0, 0) and ±(1, 1) .  12x2 − 2 −2 For the second derivative test the Hessian matrix is . At (0, 0) this is −2 2   −2 −2 with negative determinant −8 , so this critical point is a saddle. At ±(1, 1) the −2 2   10 −2 matrix is with positive determinant 16 . Since the upper left entry is positive −2 2 this means the critical points ±(1, 1) are local minima. (18 pts)

3

Using the method of Lagrange multipliers, find the points on the sphere x2 +y 2 +z 2 = 4

that are closest and farthest from the point (2, 1, −2) . Solution. We want to find max/min for the function f (x, y, z) = (x−2)2 +(y −1)2 +(z +2)2

with the constraint g(x, y, z) = x2 + y 2 + z 2 = 4 . The equation ∇f = λ∇g gives the three

equations 2(x − 2) = λ2x , 2(y − 1) = λ2y , and 2(z + 2) = λ2z . These equations are not satisfied if any of x, y, z is 0 , so we can divide by x , y , and z to solve the equations for λ 1

y−1 z+2 x−2 y−1 x−2 = = . The equation = gives xy − 2y = xy − x x y z x y z+2 y−1 = gives yz − z = yz + 2y so −z = 2y . Plugging so x = 2y . The equation y z x = 2y and z = −2y into the constraint equation gives 4y 2 + y 2 + 4y 2 = 4 , so y = ±2/3 . to get λ =

Thus we get the two solutions (x, y, z) = ±( 34 , 23 , − 43 ) . We compute f ( 34 , 23 , − 34 ) = 1 and

f (− 34 , − 32 , 43 ) = 5 , so ( 43 , 32 , − 34 ) is the closest point and (− 43 , − 23 , 34 ) is the farthest point.

(18 pts)

4

Use an appropriate change of variables to evaluate ZZZ

2

2

R

2

z cos x4 + y9 + 25 q y2 z2 x2 4 + 9 + 25

where R is the region defined by the inequalities

x2 4




+

dV

y2 9

+

z2 25

≤ 1 and

z2 25

≤

x2 4

+

y2 9

.

Solution. First we eliminate the various constants by the substitutions u = x/2 , v = y/3 , w = z/5 , hence x = 2u , dx = 2 du , y = 3v , dy = 3 dv , z = 5w , dz = 5 dw . Then dx dy dz = 2 · 3 · 5 du dv dw and the integral becomes 30

ZZZ

T

cos(u2 + v 2 + w2 ) √ du dv dw u2 + v 2 + w 2

where T is the region with u2 + v 2 + w2 ≤ 1 and w2 ≤ u2 + v 2 . In spherical coordinates these inequalities become ρ ≤ 1 and 30

Z

0

5

(9 pts)

2πZ 3π/4Z 1 π/4

0

π 4

≤φ≤

3π 4

. Thus we have the integral

1  3π/4  1 cos(ρ2 ) 2 2 ρ sin φ dρ dφ dθ = 30(2π) sin(ρ ) − cos φ ρ 2 0 π/4 √
= 30π sin(1) 2

Let R be the unit cube defined by 0 ≤ x ≤ 1 , 0 ≤ y ≤ 1 , 0 ≤ z ≤ 1 , and let

F(x, y, z) = (x + yz, y + xz, z + xy) . ZZ (a) Compute F · dS . ∂R

Solution. We can apply Gauss’s theorem to get

ZZ

ZZZ

F · dS = div(F) dV . We have R Z∂R ZZ div(F) = 1 + 1 + 1 = 3 , so the integral becomes 3 dv which is 3 times the volume

of the cube R , which is 1 , so the answer is 3 . 2

R

(9 pts)

(b) Determine whether the line integral

Z

C

F · ds is path-independent. Compute this line

integral for C the straight line segment from (0, 0, 0) to (1, 1, 1) . Solution. By inspection F is the gradient of the function f (x, y, z) =

x2 +y 2 +z 2 2

+xyz . This

implies that the line integral is path-independent, and its value is f (1, 1, 1) − f (0, 0, 0) = 5/2 . (18 pts)

6

Compute

Z

2xy dx + (x2 + 2x) dy

C

where C consists of the two closed curves shown in the figure at the right.

Solution. Since C consists of closed curves we can apply Green’s theorem. The curl of the 2 vector field Z Z (2xy, x + 2x) is 2x + 2 − 2x = 2 . Then Green’s theorem says the line integral equals 2 dA where R is the region inside the two curves. Thus the line integral equals R

twice the area inside the two curves. Counting squares, this area is 23 so the value of the line integral is 46 . 7 (6 pts)

Let F(x, y) be the vector field

(a) Compute curl(F) .

 x−y x+y  , . x2 + y 2 x2 + y 2

Solution. x2 + y 2 − (x + y)2x − [(x2 + y 2 )(−1) − (x − y)2y] curl(F) = (x2 + y 2 )2 = (6 pts)

x2 + y 2 − 2x2 − 2xy + x2 + y 2 + 2xy − 2y 2 =0 (x2 + y 2 )2

(b) Compute the line integral of F around the circle x2 + y 2 = a2 . Solution. If we could apply Green’s theorem the answer would be 0 since curl(F) = 0 . However, Green’s theorem doesn’t apply because F is not defined at the origin. Instead we compute the line integral directly, using the parametrization x = a cos t , y = a sin t , so dx = −a sin t dt and dy = a cos t dt . The line integral is then Z 2π h Z 2π (a cos t − a sin t)(−a sin t) (a cos t + a sin t)(a cos t) i + dt = dt = 2π a2 a2 0 0 3

(6 pts)

(c) In which of the following regions is F the gradient field of some function f (x, y) defined throughout the region? Give reasons for your answers. (The middle region is the points (x, y) with x > 0 .)

Solution. For the first region, if F were a gradient field then the line integral around any closed curve in the region would be zero, but we saw in part (b) that this is not true, so F can’t be a gradient field in this region. The second region is simply connected, so the fact that curl(F) = 0 implies that F is a gradient field in this region. The third region is a subregion of the second region, so since F is the gradient of some function f in the second region, F is also the gradient of the same function f when restricted to the smaller third region. (6 pts)

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(a) Compute

Z

xy 2 dx+x dy+z 2 dz where S is the portion of the cylinder x2 +y 2 = 1 ∂S

lying between the graphs z = f (x, y) and z = g(x, y) for arbitrary smooth functions f and g such that f (x, y) < g(x, y) for all (x, y) . (Use the orientation for ∂S induced from the outward normal orientation for S .) Solution. There are two ways to do this using Stokes’ theorem. In the first way we compute the surface integral of curl(F) over the cylindrical surface S . Here F = (xy 2 , x, z 2 ) , and one computes that curl(F) = (1 − 2xy) k. The normal vectors n to S are horizontal since the tangent planes to S are vertical. Thus the normal vectors have the form (a, b, 0) so we have RR RR curl(F)·n = 0 and hence S curl(F)·dS = S curl(F)·n dS = 0 . By Stokes’ theorem this implies that the given line integral R F · ds is zero. This is the first method. ∂S 4

The second method is to apply Stokes’ theorem twice, once to the surface S1 formed by the part of the graph z = f (x, y) inside the cylinder, and once to the surface S2 which is the part of the graph z = g(x, y) inside the cylinder. For the surface S1 , since this is part of the graph z = f (x, y) , we have ZZ

S1

curl(F) · dS =

ZZ

D

2

(0, 0, 1 − 2xy) · (−fx , −fy , 1) dx dy =

2

where D is the unit disk x +y ≤ 1 in the xy -plane. Notice that

ZZ

(1 − 2xy) dx dy

ZZ

(1−2xy) dx dy does

D

D

not depend on the function f , so we would get the same answer for the surface S2 with g in place of f . Thus by Stokes’ theorem the line integrals of F around ∂S1 and ∂S2 are the same. This is using the orientations of ∂S1 and ∂S2 that are counterclockwise when viewed from above. On ∂S1 this is the same orientation as in ∂S , but for ∂S2 it is the

(6 pts)

opposite orientation. Since changing the orientation changes the sign of the line integral, R this implies that ∂S F · ds = 0 , which finishes the proof using the second method. Z (b) Using your answer in part (a), show that the line integral xy 2 dx + x dy + z 2 dz C

takes the same value for all closed curves C formed by the intersection of the cylinder with a graph z = f (x, y) . (Use the orientation of C that is counterclockwise, viewed from above.) Solution. If you used the second method for doing part (a) then you have already done this.

If you used the first method, all you have to observe is that by switching the orientation of the upper curve, the sum of the integrals over the two curves oriented as in the figure becomes the difference between the two integrals when they are oriented as in part (b). Since this sum or difference is zero by part (a), the result in (b) follows. (6 pts)

(c) Compute the value of the line integral in (b) using a convenient choice of the curve C . Solution. To compute the integral, the simplest thing is to take C to be the unit circle R in the xy -plane. Thus z = 0 and the line integral is C xy 2 dx + x dy . This can be computed either directly, via the parametrization x = cos t , y = sin t , or by applying 5

Green’s theorem (or Stokes’ theorem) to get ZZ

D

(1 − 2xy) dx dy =

Z

0

2πZ 1

(1 − 2r 2 cos θ sin θ) r dr dθ

0 2π  2

1  2r 3 r − cos θ sin θ dθ = 2 3 0 0 Z 2π   θ 1  2π 1 2 = − cos θ sin θ dθ = − sin2 θ = π 2 3 2 3 0 0 Z

[This calculation could be simplified by noting that the term −2xy contributes 0 to the value of the integral by a symmetry argument: Replacing x by −x changes the sign of RR −2xy . After deleting the −2xy term we are left with D 1 dx dy , the area of the disk,

which is π .] (18 pts)

9

Find the surface area of the boundary surface ∂R of the solid region R consisting of

all points (x, y, z) satisfying x2 + z 2 ≤ 1 and y 2 + z 2 ≤ 1 . Solution. The challenge here is to see what the intersection of the two solid cylinders looks like. For a start, the figure on the left below shows the solid cylinder along the y -axis with a hole drilled out of it by the cylinder along the x -axis. This hole is the region R .

The region R has a lot of symmetry. The parts in each octant have the exact same shape, so we can compute the surface area of the part of ∂R in the first octant and then multiply by 8 . The part in the first octant is shown in the figure at the right. This has a further symmetry given by reflection across the vertical plane x = y , dividing it into two symmetric halves, indicated by the dashed and undashed lines. For the part with undashed lines the curved upper surface is part of the graph of the function g(x, y) = (1 − y 2 )1/2 , RR q so this has surface area T gx2 + gy2 + 1 dx dy where T is the triangular shadow of this 6

curved surface in the xy -plane. We have q

gx2 + gy2 + 1 =

s

y2 1 − y2

+1 =

The surface area of this piece is then Z 1Z 0

0

y

1 p

1 − y2

dx dy =

Z

1 0

s

1 y2 + 1 − y2 =p 2 1−y 1 − y2

y p

1 − y2

1 dy = −(1 − y 2 )1/2 = 1 0

Since this is one-sixteenth of the total area of ∂R , the total area is 16 .

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