APPM 2350 Spring 2006 Exam 3: Vector Calculus, Exams of Advanced Calculus

The instructions and problems for exam 3 of the appm 2350: vector calculus course held in spring 2006. The exam covers topics such as conservative vector fields, integrals over solid regions, and transformations of integrals. Students are required to solve various problems using vector calculus concepts.

Typology: Exams

2012/2013

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APPM 2350 Spring 2006 Exam 3 Apr 19, 2006 1
INSTRUCTIONS: Computers, calculators, books, notes, flailing monkeys, etc. are not
permitted. Some (possibly) useful formulae are attached. Write your name, your instruc-
tor’s name and your recitation section (and/or TA’s name) on the front of your bluebook.
Work all problems. Start each problem on a new page. Unless otherwise stated, show
all your work clearly and box your final answer; a correct answer with incorrect or no
supporting work will receive no credit, while an incorrect answer with relevant work may
receive partial credit.
1. (20 points)
(a) Find the constants a,b,cso that the vector field defined by F= (x+ 2y+azı+
(bx 3yz+ (4x+cy + 2z)ˆ
kis conservative.
(b) Is F=xyˆızˆ+x2ˆ
ka conservative vector field? If so, find a scalar function f
such that F=f.
(c) Evaluate RCF·drwhere Fis the vector field in (b) and Cis the curve x=t2,
y= 2t,z=t3from t= 0 to t= 1. You may leave your answer as a sum of
fractions.
2. (25 points) Geoff, the proprietor of Land O’ Lake Ice Cream Company, is testing a
new machine for pouring ice cream into the cartons for sale. Given a set of axes in
which the carton is the cuboid region {(x, y, z) : 0 x4,0y2,0z2},
the machine is pouring the ice cream in such a way that the distribution of Crunchy
Frog BitsTM in Geoff’s revolutionary frog-flavored ice cream is given by the equation
δ(x, y, z) = 2+ z2[bits/in3]. Megan, one of Geoff’s intrepid taste-testers, has scooped
out a lump of ice cream from the top of the carton in the shape of the hemisphere
{(x, y, z) : (x2)2+ (y1)2+ (z2)21, z 2}.
(a) Sketch the ice-cream carton with the hemisphere removed. Label your axes!
(b) Calculate the total number of frog bits in the whole carton (before the scoop is
removed).
(c) Show that the volume element dV is unchanged under translation of coordinates
i.e. u=xx0,v=yy0,w=zz0, where x0, y0, z0are constants.
(d) Use the result in (c), and another coordinate change, to set up the integral
necessary to calculate the total number of frog bits in Megan’s scoop using the
most natural set of coordinates. (You do NOT need to evaluate the integral.)
3. (20 points) Calculate the integral of f(x, y, z) = 2zover the wedge in the first octant
cut from the solid cylinder y2+z21 by the plane y=x.
4. (15 points) Convert Z1
1Z2y2
1
x
x2+y2dxdy into polar coordinates (but do not eval-
uate it).
pf3

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Download APPM 2350 Spring 2006 Exam 3: Vector Calculus and more Exams Advanced Calculus in PDF only on Docsity!

INSTRUCTIONS: Computers, calculators, books, notes, flailing monkeys, etc. are not

permitted. Some (possibly) useful formulae are attached. Write your name, your instruc-

tor’s name and your recitation section (and/or TA’s name) on the front of your bluebook.

Work all problems. Start each problem on a new page. Unless otherwise stated, show

all your work clearly and box your final answer; a correct answer with incorrect or no

supporting work will receive no credit, while an incorrect answer with relevant work may

receive partial credit.

  1. (20 points)

(a) Find the constants a, b, c so that the vector field defined by F = (x + 2y + az)ˆı +

(bx − 3 y − z)ˆ + (4x + cy + 2z)

k is conservative.

(b) Is F = xyˆı − zˆ + x

k a conservative vector field? If so, find a scalar function f

such that F = ∇f.

(c) Evaluate

C

F · dr where F is the vector field in (b) and C is the curve x = t

2 ,

y = 2t, z = t

3 from t = 0 to t = 1. You may leave your answer as a sum of

fractions.

  1. (25 points) Geoff, the proprietor of Land O’ Lake Ice Cream Company, is testing a

new machine for pouring ice cream into the cartons for sale. Given a set of axes in

which the carton is the cuboid region {(x, y, z) : 0 ≤ x ≤ 4 , 0 ≤ y ≤ 2 , 0 ≤ z ≤ 2 },

the machine is pouring the ice cream in such a way that the distribution of Crunchy

Frog Bits

TM in Geoff’s revolutionary frog-flavored ice cream is given by the equation

δ(x, y, z) = 2 + z

2 [bits/in

3 ]. Megan, one of Geoff’s intrepid taste-testers, has scooped

out a lump of ice cream from the top of the carton in the shape of the hemisphere

{(x, y, z) : (x − 2) 2

  • (y − 1) 2
  • (z − 2) 2 ≤ 1 , z ≤ 2 }.

(a) Sketch the ice-cream carton with the hemisphere removed. Label your axes!

(b) Calculate the total number of frog bits in the whole carton (before the scoop is

removed).

(c) Show that the volume element dV is unchanged under translation of coordinates

— i.e. u = x − x 0 , v = y − y 0 , w = z − z 0 , where x 0 , y 0 , z 0 are constants.

(d) Use the result in (c), and another coordinate change, to set up the integral

necessary to calculate the total number of frog bits in Megan’s scoop using the

most natural set of coordinates. (You do NOT need to evaluate the integral.)

  1. (20 points) Calculate the integral of f (x, y, z) = 2z over the wedge in the first octant

cut from the solid cylinder y

2

  • z

2 ≤ 1 by the plane y = x.

  1. (15 points) Convert

1

− 1

2 −y^2

1

x

x 2

  • y 2

dxdy into polar coordinates (but do not eval-

uate it).

  1. (20 points) Integrating over an ellipse R = {(x, y) :

x 2

a^2

y^2 b^2

≤ 1 } is difficult, even in

polar coordinates. However, the change of coordinates x = au cos(v), y = bu sin(v),

where u ≥ 0 and 0 ≤ v ≤ 2 π, can make the problem feasible.

(a) Write the integral

R

f (x, y) dA in the new coordinates (u, v).

(b) Use your result from (a) to find the area of R.

(c) Verify that your result in (b) makes sense when a = b.

— Useful and interesting formulae —

proj A

B =

A · B

A · A

A d =

P S × v|

|v|

d =

P S~ ·

n

|n|

s(t) =

t

t 0

|v(u)| du T =

dr

ds

v

|v|

N =

dT/ds

|dT/ds|

dT/dt

|dT/dt|

B = T × N

κ =

dT

ds

|v × a|

|v| 3

|f

′′ (x)|

[1 + (f ′ (x)) 2 ] 3 / 2

| x˙y¨ − y˙x¨|

[ ˙x 2

  • ˙y 2 ] 3 / 2

τ = −

dB

ds

· N

a = aT T + aN N where aT =

d

dt

|v|, aN = κ|v|

2

|a|

2 − a

2 T

df

ds

= Duf = (∇f ) · u

Discriminant:

2 f

∂x 2

2 f

∂y 2

2 f

∂x∂y

∇f = λ∇g g = 0

f (x, y) = f (0, 0) +

∂f

∂x

(0,0)

x +

∂f

∂y

(0,0)

y

2 f

∂x

2

(0,0)

x

2

  • 2

2 f

∂x∂y

(0,0)

xy +

∂f

∂y

2

(0,0)

y

2

n!

x

∂x

  • y

∂y

n

(0,0)

f +...

|E(x, y)| ≤

M

(|x − x 0 | + |y − y 0 |)

2 where |fxx|, |fxy|, |fyy| ≤ M