Test 2 2013 Solutions

ID # : March 9, 2013

Lab # : Instructors: Dr. A. Contreras

Dr. J.-P. Gabardo Dr. A. French Test duration: 75 min.

Instructions: You must use permanent ink. Tests submitted in pencil will not be con-sidered later for remarking. This exam consists of 9 problems on 14 pages (make sure you have all 14 pages). The last two pages are for scratch or overflow work. The total number of points is 50. Do not add or remove pages from your test. No books, notes, or “cheat sheets” allowed. No calculator or other electronic devices are allowed. There is a formula that might be useful on the last page. GOOD LUCK!

Part I: Enter your answer in the appropriate box. Provide all details and fully justify your answer in order to receive credit.

1. (a) (4 pts.) Find the equation of the plane tangent to the surface with equation

x2+y2−z2 = 0 at the point P = (3,4,5).

Solution. The surface is the level surface F(x, y, z) = 0 of the function F(x, y, z) = x2+

y2−z2. We have

∇F(x, y, z) =

∂F

∂x, ∂F

∂y, ∂F

∂z

=h2x,2y,−2zi and

∇F(3,4,5) = h6,8,−10i The equation of the tangent plane at P = (3,4,5) is thus

6 (x−3) + 8 (y−4)−10 (z−5) = 0 or 3x+ 4y−5z = 0.

(b) (4 pts.) Prove that any plane tangent to the surface x2+y2−z2 = 0 passes through the origin.

Solution. If (x0, y0, z0) is a point on the surface, we have x02 + y20 − z02 = 0. Since ∇F(x0, y0, z0) = h2x0,2y0,−2z0i and the tangent plane to the surface at that point has equation

x0(x−x0) +y0(y−y0)−z0(z−z0) = 0.

For the point (0,0,0) to satisfy the equation of the tangent plane, we need −x2₀−y₀2+z₀2 = 0,

2. (6 pts.) Evaluate the integral

Z 1

0

Z 1

√

y

√

x3_{+ 1dx dy}

by changing the order of integration. Solution. We have

I :=

Z 1

0

Z 1

√

y

√

x3_{+ 1dx dy =}

Z Z

D

√

x3_{+ 1dA}

where D is the region of type II

D={(x, y), 0≤y≤1, √y≤x≤1}.

Expressing D as a region of type I, we have

D={(x, y), 0≤x≤1, 0≤y≤x2}.

Therefore,

I =

Z 1

0

Z x2

0 √

x3_{+ 1dy dx=}

Z 1

0

x2√x3_{+ 1dx}

=

2 9(x

3_{+ 1)}3/2

1

0 = 2

9(2

3. (a) (4 pts.) Find the values of A and B for which the vector field F(x, y, z) = hB y+y z, x+z3+x z, A y z2+x yi is conservative.

Solution. We have

∇ ×F(x, y, z) =





i j k

∂ ∂x

∂ ∂y

∂ ∂z

B y+y z x+z3+x z A y z2+x y 



= (A z2+x−3z2−x)i+ (y−y)j+ (1 +z−B−z)k = (A−3)z2i+ (1−B)k=0

if and only if A= 3 andB = 1.

(b) (4 pts.) Find a potential function for the conservative vector fieldFfound in part (a). Solution. We have

∂f

∂x =y+y z, ∂f

∂y =x+z

3

+x z, ∂f

∂z = 3y z

2

+x y. (1)

Integrating both sides of the first equation yields f(x, y, z) = x y+x y z+C(y, z). Differen-tiating with respect to y, we obtain

∂f

∂y =x+x z+ ∂C

∂y =x+z

3 +x z,

showing that ∂C_∂y =z3 _{and C(y, z) = y z}3_{+D(z). We have thus}

f(x, y, z) =x y+x y z+y z3+D(z) and

∂f

∂z =x y+ 3y z

2_+D0

(z) = 3y z2+x y.

ThusD(z) = D, a constant. TakingD= 0, we find thus the potential function

(c) (3 pts.) Let F be the conservative vector field found in part (a). Compute the line integral

Z

C

F·dr where C is any path starting at (0,0,0) and ending at (1,1,1). Solution. We have

Z

C

4. LetDbe the region in the first quadrant of thex, y plane bounded by the curvesy=x2

and x=y2 and letC be the positively oriented boundary of D. (a) (5 pts.) Compute the line integral

I

C

(x−y)dx+ (x+y)dy

directly after parametrizing both parts of C appropriately.

Solution. The 2 curves intersect at (0,0) and (1,1). The curve C consists of two parts: • C1: the part of the parabolay =x2 from (0,0) to (1,1),

• C2: the part of the parabolax=y2 from (1,1) back to (0,0).

We can parametrize C1 by letting x = t, y = t2, where 0 ≤ t ≤ 1. We have x0(t) = 1,

y0(t) = 2t and

Z

C1

(x−y)dx+ (x+y)dy=

Z 1

0

(t−t2)(1) + (t+t2) (2t)dt

=

Z 1

0

2t3+t2+t dt=

t4 2 + t3 3 + t2 2 1 0 = 4 3.

The curve traversingC2in the opposite direction (i.e. from (0,0) to (1,1) can be parametrized by letting y=t, x=t2, where 0≤t ≤1. We have x0(t) = 2t,y0(t) = 1 and

Z

C2

(x−y)dx+ (x+y)dy =−

Z 1

0

(t2−t)(2t) + (t2+t) (1)dt

=−

Z 1

0

2t3−t2 +t dt=

Z 1

0

−2t3+t2−t dt=

−t 4 2 + t3 3 − t2 2 1 0 =−2

3. Thus,

I

C

(x−y)dx+(x+y)dy=

Z

C1

(x−y)dx+(x+y)dy+

Z

C2

(x−y)dx+(x+y)dy= 4 3−

2 3 =

(b) (5 pts.) Compute the line integral in part (a) using Green’s theorem. Solution. The region D bounded by C can be expressed as the type I region

D={(x, y),0≤x≤1, x2 ≤y≤√x}.

Using Green’s theorem,

I

C

(x−y)dx+ (x+y)dy =

Z Z

D

∂(x+y)

∂x −

∂(x−y)

∂y dA

=

Z 1

0

Z

√

x

x2

2dy dx= 2

Z 1

0 √

x−x2dx= 2

2 3x

3/2₋ 1 3x

3

1

0

PART II: Multiple choice. Indicate your choice very clearly. There is only one correct answer in each multiple-choice problem. Circle the letter (A,B,C,D or E) corresponding to your choice. Ambiguous answers will be marked as wrong. You don’t need to justify your answers and no negative marks are given for a wrong answer.

5. (3 pts.) Letg = div (F+ curl(F)) where

F(x, y, z) =−x yi+x z2 ln(y2+ 1)j+ exyzk.

The value ofa for which g(a,2,0) = 0 is (A) −2

(B) −1 (C) 0

→ (D) 1 (E) 2

Solution. Since div (curl(F)) = 0, we have

g(x, y, z) = div (F+ curl(F)) = div (F) = ∂(−x y)

∂x +

∂(x z2 ln(y2+ 1))

∂y +

∂(exyz)

∂z

=−y+ 2x y z 2

1 +y2 +x y e

xyz

.

6. (3 pts.) Let

F(x, y, z) = eyi+x eyj+ (z+ 1)ezk and suppose C is path parametrized by

r(t) =t sin3(πt/2)i+t2et−1j+t100k, 0≤t≤1.

Which of the following is equal toR_CF·dr? (Hint. Is the vector field conservative?)

(A) 0 →(B) 2e

(C) 4e

(D) −2e

(E) −4e

Solution. By inspection, a potential function for F is given by

f(x, y, z) = x ey+

Z

(z+ 1)ezdz =x ey+z ez

Since r(0) =h0,0,0i and r(1) =h1,1,1i, we have

Z

C

7. (3 pts.) Evaluate the line integral

I =

Z

C

x ds,

where C is the arc of parabola y=x2 _{fron (0,0) to (1,1).}

Solution. (A) I = 1 (B) I = 2

√ 2−1

6 (C) I = 0

(D) I = √

5 2 → (E) I = 5

√ 5−1 12

Solution. We can parametrize C using r(t) = ht, t2i for 0≤t≤1. We have r0(t) =h1,2ti and kr0(t)k=√1 + 4t2_.

Thus,

I =

Z 1

0

t√1 + 4t2_{dt =}

1

12(1 + 4t 2₎3/2

1

0 = 5

8. (3 pts.) Let R be the solid region inside the cylinder x2 _+y2 _{= x, above the x, y} plane and below the paraboloid z = 2 +x2_+y2_{. Then the volume of R can be obtained by} computing the double integral:

Solution. (A)

Z π₂

−π

2 Z 1₂

0

2r+r3dr dθ

→(B)

Z π₂

−π

2 Z cosθ

0

2r+r3drdθ

(C) Z 1 0 Z √ 1 4−(x−

1 2)2

0

2 +x2+y2dy dx

(D)

Z π₂

0

Z sinθ

0

2 +r2dr dθ

(E)

Z 1₂

0

Z

√

1 4−x2

−√1 4−x2

2 +x2+y2dy dx

Solution. R is the region above the regionD in thex, y plane inside the circle x2+y2 =x

and below the graph of z = 2 +x2_+y2_{. Thus,} Vol(R) =

Z Z

D

2 +x2+y2dA

The circle x2+y2 =x can be expressed as the polar curve r = cosθ, −π

2 ≤ θ ≤

π

2 and the region D can be expressed as the region

D∗ ={(r, θ), −π

2 ≤θ ≤

π

9. (3 pts.) Compute the integral 1 2

I

C

x dy−y dx whereC is the polar curver= 3 + cosθ, 0≤θ≤2π.

(Hint. Use Green’s theorem.) →(A) 19π

2 (B) 15π

2 (C) 0

(D) −7π 2 (E) −3π

2

Solution. By Green’s theorem, we have 1

2

I

C

x dy−y dx=

Z Z

D

1dA

whereD is the region inside the curve C. This region can be expressed in polar coordinates as

D∗ ={(r, θ), 0≤θ≤2π,0≤r ≤3 + cosθ}.

Therefore, 1

2

I

C

x dy−y dx=

Z 2π

0

Z 3+cosθ

0

r dr dθ = 1 2

Z 2π

0

(3 + cosθ)2dθ

= 1 2

Z 2π

0

9 + 6 cosθ+ cos2θ dθ = 1 2

Z 2π

0

9 + 6 cosθ+1 + cos(2θ)

2 dθ

=

19

4 θ+ 3 sinθ+

sin(2θ) 8

2π

0

SCRATCH

Some formulas you may use:

cos2t = 1 2 +

1

2cos(2t), sin 2_t= 1

2 − 1

2cos(2t).

2 cosαcosβ = cos(α−β) + cos(α+β), 2 sinαsinβ = cos(α−β)−cos(α+β),

2 sinαcosβ = sin(α+β) + sin(α−β) cosht= e

t_+e−t

2 sinht =

et_−e−t

2

I

C

P dx+Q dy=

Z Z

D

∂Q

∂x −

∂P