math 3132 test 1 solutions winter 2015

DATE: February 10, 2015

EXAMINATION: Engineering Mathematical Analysis 3COURSE: MATH 3132

UNIVERSITY OF MANITOBATERM TEST 1

PAGE: 1 of 6TIME: 1 hour

EXAMINER: Harland

1.[5] Calculate

ˆC

x2 dx+ y2 dy + z2 dz

where C consists of the line segment from (0, 0, 0) to (1, 2,−1) and from (1, 2,−1)to (3, 2, 0).

Solution:

We first note that

curlF = (0− 0)i− (0− 0)j + (0− 0)k = 0

which along with F begin continuous everywhere, means the vector field isindependent of path.

Method 1: Ignore the above and do it directly.

A parametrization of the first line (call the curve C1) is x = t, y = 2t, z = −t.and 0 ≤ t ≤ 1. Thus

ˆC1

x2 dx+ y2 dy + z2 dz =

ˆ 1

0

t2 dt+ 4t2(2 dt) + (−t)2(− dt)

=

ˆ 1

0

8t2 dt

=8

3t3∣∣∣∣10

=8

3

A parametrization of the second line (call the curve C2) is x = 1 + 2t, y =2, z = −1 + t. and 0 ≤ t ≤ 1. Thus

ˆC2


ˆ 1

0

(1 + 2t)2(2 dt) + 4(0 dt) + (−1 + t)2 dt

=

ˆ 1

0

(8t2 + 8t+ 2) + (t2 − 2t+ 1) dt

=

ˆ 1

0

9t2 + 6t+ 3 dt

= 3t3 + 3t2 + 3t∣∣10

= 9

Thus

ˆC

x2 dx+ y2 dy + z2 dz =8

3+ 9 =

35

3.

Method 2: Change the path

The easiest path is the straight line connecting (0, 0, 0) to (3, 2, 0) which canbe parameterized as x = 3t, y = 2t, z = 0. and 0 ≤ t ≤ 1. Thus





EXAMINER: Harland

ˆC


ˆ 1

0

9t2(3 dt) + 4t2(2 dt) + 0 dt

=

ˆ 1

0

35t2 dt

=35

3t3∣∣∣∣10

=35

3

Method 3: Find a potential function If ∇f = F = x2i + y2j + z2k then

fx = x2 ⇒ f(x, y, z) =x3

3+ C(y, z)

Hence fy is both y2 (from F) and∂C

∂yfrom f. Hence

∂C

∂y= y2 ⇒ C(y, z) =

y3

3+D(z)⇒ f(x, y, z) =

x3

3+y3

3+D(z)

Hence fz is both z2 (from F) anddD

dzfrom f. Hence

dD

dz= z2 ⇒ D(z) =

z3

3+K ⇒ f(x, y, z) =

x3

3+y3

3+z3

3Thus

ˆC

x2 dx+y2 dy+z2 dz = f(3, 2, 0)−f(0, 0, 0) =33

3+

23

3+

03

3−(

03

3+

03

3+

03

3

)=

35

3.

2. Let F(x, y, z) = (y3 − 6xz + z2)i + (3xy2)j + (2z − 3x2)k.

(a)[3] Determine with justification whether the integral

ˆC

F · dr is independent of

path.

Solution:

curlF = (0− 0)i− (−6x− (−6x+ 2z))j + (3y2 − 0)k = 2zj 6= 0.

Therefore the integral is not independent of path

(b)[5] Calculate

ˆC

F · dr where C is the line segment from (0, 0, 0) to (2, 1, 0).

Solution: The curve has parametrization x = 2t, y = t, z = 0 and thusdr = 2i + j

F · dr = (y3− 6xz + z2)(2 dt) + (3xy2) dt = 2(t3− 0− 0) + 6t3 dt = 8t3 dt

Thus ˆC

F · dr =

ˆ 1

0

8t3 dt dt = 2t4|10 = 2.





EXAMINER: Harland

3. (a)[2] State Green’s Theorem.

Solution: Let F = P i+Qj. If P and Q have continuous partial deriva-tives on a domain, then

‰C

F · dr =

¨R

(∂Q

∂x− ∂P

∂y

)dA

where R is the region bounded by C.

(b)[3] Use Green’s Theorem to calculate v where C is the triangle from (0, 0) to(2, 0) to (1, 1) back to (0, 0).

Solution:∂Q

∂x− ∂P

∂y= 1− 1 = 0.

Therefore

‰C

y dx+ x dy =

¨R

0 dA = 0.

(c)[6] Use Green’s Theorem to calculate

C

x2 dx − xy dy where C is the curve

bounded by y = x and y = x2 given below.

Solution:∂Q

∂x− ∂P

∂y= −y − 0 = −y.

Therefore

C

x2 dx− xy dy = −¨R

−y dA =

¨R

y dA

Now R is determined by x2 ≤ y ≤ x, 0 ≤ x ≤ 1. Thus





EXAMINER: Harland

I =

¨R

y dA

=

ˆ 1

0

ˆ x

x2y dy dx

=

ˆ 1

0

y2

2

∣∣∣∣xx2dx

=1

2

ˆ 1

0

x2 − x4 dx

=1

2

(x3

3− x5

5

∣∣∣∣10

=1

2

(1

3− 1

5

)=

1

15





EXAMINER: Harland

4. (a)[2] State Stokes’ Theorem.

Solution: Let F = P i+Qj+Rk. If P , Q and R have continuous partialderivatives on a domain, then

‰C

F · dr =

¨S

curl(F) · n dS

where S is a surface bounded by C.

(b)[6] Consider F(x, y, z) = z2i + 2xyj + 4y2k. Use Stokes’ Theorem to calculatethe work done (neglecting units) to move a particle along line segments from(0, 0, 0) to (3, 0, 0) to (3, 2, 1) to (0, 2, 1) and back to (0, 0, 0). (Hint: Thereis a plane containing those four points)

Solution: The four points form a parallelogram and thus there is a planeconnecting all of the points. The vectors 〈3, 0, 0〉 and 〈0, 2, 1〉 and thus

n = 〈3, 0, 0〉 × 〈0, 2, 1〉 = 〈0,−3, 6〉 ⇒ n =−j + 2k√

5

(Note that the curve is counterclockwise as viewed from above and thusn points upward.)

The equation of the plane itself is therefore z = 12y and thus dS =√

1 + 02 + (1/2)2dA =

√5

2dA.

Also curl(F) = (8y − 0)i− (0− 2z)j + (2y − 0)k = 8yi + 2zj + 2yk.

Hence

I =

¨Sxy

(8yi + 2zj + 2yk) · −j + 2k√5

(√5

2

)dA

=1

2

¨Sxy

−2z + 4y dA

=1

2

¨Sxy

−2

(1

2y

)+ 4y dA

=1

2

¨Sxy

3y dA

The projection is the rectangle 0 ≤ x ≤ 3, 0 ≤ y ≤ 2 and thus

I =1

2

¨Sxy

3y dA

=1

2

ˆ 3

0

ˆ 2

0

3y dy dx

=1

2

ˆ 3

0

3

2y2∣∣∣∣20

dx

=1

2

ˆ 3

0

6 dx

= 3x|30= 9





EXAMINER: Harland

5. (a)[2] State the Divergence Theorem.

Solution: Let F = P i+Qj+Rk. If P , Q and R have continuous partialderivatives on a domain, then

‹S

F · n dS =

˚V

div(F) dV

where V is a solid bounded by S.

(b)[6] Calculate

‹S

F · n dS if F(x, y, z) = 3xy2i + xezj + z3k and S is the surface

of the solid bounded by the cylinder y2 + z2 = 1 and the planes x = −1 andx = 2.

Solution: div(F = 3y2 + 0 + 3z2 = 3y2 + 3z2. Hence using polar (on yand z) yields

Thus

I =

˚V

(3y2 + 3z2) dV

=

ˆ 2

−1

¨Syz

(3y2 + 3z2) dAdx

=

ˆ 2

−1

ˆ 2π

0

ˆ 1

0

(3r2)(r)dr dθ dx

=

ˆ 2

−1

ˆ 2π

0

3

4r4∣∣∣∣10

dθ dx

=

ˆ 2

−1

ˆ 2π

0

3

4dθ dx

=

ˆ 2

−1

3

4θ

∣∣∣∣2π0

dx

=

ˆ 2

−1

3π

2dx

=3π x

2

∣∣∣∣2−1

=9π

2

math 3132 test 1 solutions winter 2015

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