Docsity
Docsity

Prepare for your exams
Prepare for your exams

Study with the several resources on Docsity


Earn points to download
Earn points to download

Earn points by helping other students or get them with a premium plan


Guidelines and tips
Guidelines and tips

Calculus III Exam 3 Solutions: Polar, Cylindrical, Vector Calculus & Green's Theorem - Pro, Exams of Advanced Calculus

Solutions to exam 3 of math 2433 calculus iii at the university of houston. Topics covered include evaluating integrals using polar and cylindrical coordinates, calculating jacobian determinants, triple integrals, finding mass of a metal region, verifying vector fields, and applying green's theorem. Solutions also include the area of an ellipse.

Typology: Exams

2011/2012

Uploaded on 05/10/2012

littleprincess2021
littleprincess2021 🇺🇸

1 document

1 / 5

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
Math 2433 Calculus III —– Exam 3
University of Houston
May 1, 2012
Name (printed):
Last four digits of your PeopleSoft ID:
Your recitation class (please circle one):
Zhiheng Liu (9:00-10:00 am, SEC 206)
Nidhi Jaiswal (9:00-10:00 am, SEC 205)
Ricky Ng (8:00-9:00 am, SEC 102)
Ilija Jegdic (9:00-10:00 am, SEC 204)
Closed books and notes. NO CALCULATORS. Show all your work for all credits. An-
swers, even correct obtained through guesswork, incorrect procedure or no procedure, are
worthless.
1 2 3 4 5 6 7 8
1
pf3
pf4
pf5

Partial preview of the text

Download Calculus III Exam 3 Solutions: Polar, Cylindrical, Vector Calculus & Green's Theorem - Pro and more Exams Advanced Calculus in PDF only on Docsity!

Math 2433 Calculus III —– Exam 3

University of Houston

May 1, 2012

Name (printed):

Last four digits of your PeopleSoft ID:

Your recitation class (please circle one):

Zhiheng Liu (9:00-10:00 am, SEC 206)

Nidhi Jaiswal (9:00-10:00 am, SEC 205)

Ricky Ng (8:00-9:00 am, SEC 102)

Ilija Jegdic (9:00-10:00 am, SEC 204)

Closed books and notes. NO CALCULATORS. Show all your work for all credits. An-

swers, even correct obtained through guesswork, incorrect procedure or no procedure, are

worthless.

  1. (13 points) Using polar coordinates, evaluate the integral

D

sin(x

2

  • y

2

) dx dy

where D is the region 1 ≤ x

2

  • y

2 ≤ 4.

Solution: Using the polar coordinates, write

(r, θ)

0 ≤ θ ≤ 2 π

f ix θ, 1 ≤ r ≤ 2

Then

Ω

sin(x

2

  • y

2

) dx dy =

2 π

0

2

1

sin(r

2

)r dr dθ = ...

  1. (12 points) Let f (x, y) = (x + y, 2 xy) be a vector-valued function. Calculate its

Jacobian determinant det D(f ).

Solution: f = (f 1

, f 2

) where f 1

= x + y and f 2

= 2xy. Then

det D(f ) = det

∂f 1

∂x

∂f 2

∂x

∂f 1

∂y

∂f 2

∂y

= det

1 2 y

1 2 x

= 2x − 2 y.

  1. (13 points) Let C be the curve given by

~r(t) = (1 + sin t)

i + (1 + sin

2

t)

j, 0 ≤ t ≤

π

and

F is the vector field

F (x, y) = 2x ~i + 2y ~j.

Verify that

F = ▽ f where f (x, y) = x

2

  • y

2 , and then compute the work

C

F (~r) · d~r done

by

F on a particle moving along C.

Solution: ▽ f = ▽(x

2

  • y

2 ) = 2x~i + 2y~j =

F. Then by the fundamental theorem of

calculus, the work

W =

C

F (~r) · d~r = f (~r(

2 i

)) − f (~r(0)) = f (1 + 1, 1 + 1) − f (1, 1) = ...

  1. (12 points) Let ~v = xy~i − 2 yz~j. Calculate the divergence div ~v and the curl curl ~v.

Solution: v 1

= xy, v 2

= − 2 yz, v 3

= 0. Then

div ~v =

∂v 1

∂x

∂v 2

∂y

∂v 3

∂z

= y − 2 z,

and

curl ~v = det

i

j

k

∂x

∂y

∂z

v 1

v 2

v 3

= det

i

j

k

∂x

∂y

∂z

xy − 2 yz 0

= det

∂y

∂z

− 2 yz 0

i − det

∂x

∂z

xy 0

j + det

∂x

∂y

xy − 2 yz

k

= (0 + 2y)

i − (0 − 0)

j + (0 − x)

k = 2y~i − x

k.

  1. (13 points) By applying Green’s theorem to evaluate the line integral

C

x

2

y dx +

x

3

dy

where C is the square Ω with vertices (0, 0), (2, 0), (2, 2) and (0, 2).

Solution: Let P = x

2 y and Q =

1

3

x

3

. Applying Green theorem, we have

C

x

2 y dx+

1

3

x

3 dy =

∫ ∫

Ω

∂Q

∂x

∂P

∂y

)dxdy =

Ω

(x

2 − x

2 )dxdy = 0.

  1. (12 points) Find the area of the ellipse

x

2

4

y

2

9

= 1 by using the formula area =

1

2

C

x dy − y dx.

Solution: The area is 6π (see an example in the lecture notes, page 205).