Prepare for your exams
Get points
Guidelines and tips
Sell on Docsity
Docsity AI

Prepare for your exams

Study with the several resources on Docsity

Earn points to download

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

Guidelines and tips

Sell on Docsity

Docsity AI

Log in Sign up

Prepare for your exams

Study with the several resources on Docsity

Find documents

Prepare for your exams with the study notes shared by other students like you on Docsity

Search for your university

Find the specific documents for your university's exams

Docsity AINEW

Summarize your documents, ask them questions, convert them into quizzes and concept maps

Explore questions

Clear up your doubts by reading the answers to questions asked by your fellow students

Earn points to download

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

Share documents

20 Points

For each uploaded document

Answer questions

5 Points

For each given answer (max 1 per day)

All the ways to get free points

Get points immediately

Choose a premium plan with all the points you need

Study Opportunities

Choose your next study program

Get in touch with the best universities in the world. Search through thousands of universities and official partners

Community

Ask the community

Ask the community for help and clear up your study doubts

Free resources

Our save-the-student-ebooks!

Download our free guides on studying techniques, anxiety management strategies, and thesis advice from Docsity tutors

Solutions for Homework Problems - Multivariable Calculus | MATH 2210, Assignments of Mathematics

Utah State University (USU)Mathematics

Prof. Bryan J. Bornholdt

Material Type: Assignment; Professor: Bornholdt; Class: MULTIVARI CALCULUS (QI)(H); Subject: Mathematics; University: Utah State University; Term: Unknown 1989;

Typology: Assignments

Pre 2010

Uploaded on 07/30/2009

koofers-user-y4l 🇺🇸

9 documents

1 / 5

This page cannot be seen from the preview

Don't miss anything!

MATH 2210 12.6 Surface Area

Solutions to selected homework problems

Note: For graphs of the form

),( yxfz 

, we may view the surface parametrically

as

kjir ),(),( yxfyxyx 

. Thus, we have

kir ),(),( yxfyx

xx



and

kjr ),(),( yxfyx

yx



. Consequently, the cross product is

1),(),(  jirr yxfyxf

yxyx

so the surface area of a graph

),( yxfz 

over a region D is given by



































D

dA

y

z

x

z

SA

2

1)(

. (1)

Problem 2: We want to find the area of the surface defined by

1052  zyx

that lies

inside the cylinder

9

22

 yx

. Solving for z, we get

yxz 5210 

. Using equation

(1) above, we have

   

units sq. 309

2

1

302

30

521)(

3

0

2

0

3

0

22



























r

ddrr

dASA

D

Problem 5: We have

22

xyz 

so

kir x

x

2

and

kjr y

y

2

so

kjirr  yx

yx

22

which gives

 

units sq. 551717

6

14

3

2

8

1

2

14

144)(

4

1

2

3

2

1

2

0

2

0

2

1

2

22





















































r

rdrrd

drdrr

dAyxSA

D

Discover Assignments of Mathematics Utah State University (USU)

Partial preview of the text

Download Solutions for Homework Problems - Multivariable Calculus | MATH 2210 and more Assignments Mathematics in PDF only on Docsity!

MATH 2210 12.6 Surface Area

Solutions to selected homework problems

Note: For graphs of the form

z  f ( x , y )

, we may view the surface parametrically

as r^ (^ x ,^ y )^ x i  y j  f ( x , y ) k. Thus, we have r^ x (^ x ,^ y ) i^  fx ( x , y ) k and

r ( x , y ) j f ( x , y ) k x y

 

. Consequently, the cross product is

r  r  f ( x , y ) i  f ( x , y ) j  1 x y x y

so the surface area of a graph

z  f ( x , y )

over a region D is given by















^ 













 

D

dA

y

z

x

z

A S

2 2

Problem 2: We want to find the area of the surface defined by

2 x  5 y  z  10

that lies

inside the cylinder 9

2 2

x  y . Solving for z , we get z^ ^10 ^2 x ^5 y. Using equation

(1) above, we have

9 30 sq. units

2

1

2 30

30

( ) 1 2 5

3

0

2

0

3

0

2 2















 



    

r

rdr d

AS dA

D

Problem 5: We have

2 2

z  y  x so r i x k

x

  2 and r^ j y k

y

  2

so

r  r  x i  y j  k x y 2 2

which gives

 17 17 5 5  sq.units

4

1

2

3

2

1

2

0

2

0

2

1

2

2 2



r

d r rdr

r rdr d

AS x y dA

D

Problem 10:

Find the area of the surface

2 2

x  y  z inside the cylinder 9

2 2

y  z .

Note that we will view x as a function of y and z. Therefore, we can represent the surface

as the vector function r^ (^ y^ , z )^ ( y  z ) i  y j  z k

2 2

. Thus, the surface area is given by

 37 37 1  sq.units

1 4 Let 1 4

37

1

2

3

37

1

2

0

2

0

3

0

2

2 2

(^22)



u

d u du

r rdrd u r

y z dxdy

dA

z

x

y

x

AS dA

D

y z

r r

Observations: Notice that if the surface is flat (a plane as in #2), the quantity

u v

r  r is a constant that rescales the area of the underlying region of integration.

We can think of the region of integration as the projection of the surface onto the

given plane (say, xy plane) and that u v

r  r

rescales the area accordingly.

In the case of a curved surface as in the other problems, the region of integration is

still the projection of the surface onto the xy plane but the variable form accounts

for the curvature of the surface.

Problem 24: There are several ways to do this problem. The simplest may be finding

the area of the surface that intersects the positive x axis and taking the region of

integration in the yz plane to be 1

2 2

y  z . As such, the surface ‘above’ this region is

2

x  f ( y , z ) 1  z so it has vector representation

r   z i  y j  z k

2

As a result,

2 2

2

z z

z

y z

r  r  

so we may obtain the area of this surface by

(By symmetry)

1

0

1

0

2

1

2

2 2

 





 

dy dz

z

AS dA

z

y z

y z r r

However, this integral is improper since it is undefined at z = 1. Hence, we have

lim 4

( ) lim 4

1

0

1

0 0

1 2

0

1

0

1 2

2















 

t

dz

z

y

dy dz

z

A S

t

z t

t

t z

t

However, this is only ONE of four such faces of intersection. Consequently, the total

surface area is 16 square units.

Alternate Solution

Using polar coordinates in the yz plane, we have the vector representation

r  x i cos  j sin  k

for

   and   1 

2 2 x z 1 sin  1 sin  cos  cos 

2 2

   x     x . So

we have

r  i

x and^

r  j  k 

 sin cos

so

r  r cos  j sin  k  1

x .

This gives

16 sq. units

4 ( 4 )

4 2 cos

4 1

( )

2

cos

1

2 2



 



 





 

 





 



d

dx d

AS dA

y z

x

r r

Solutions for Homework Problems - Multivariable Calculus | MATH 2210, Assignments of Mathematics

Related documents

Partial preview of the text

Download Solutions for Homework Problems - Multivariable Calculus | MATH 2210 and more Assignments Mathematics in PDF only on Docsity!

MATH 2210 12.6 Surface Area

Solutions to selected homework problems

Note: For graphs of the form

, we may view the surface parametrically

as r^ (^ x ,^ y )^ x i  y j  f ( x , y ) k. Thus, we have r^ x (^ x ,^ y ) i^  fx ( x , y ) k and

. Consequently, the cross product is

so the surface area of a graph

over a region D is given by

Problem 2: We want to find the area of the surface defined by

that lies

inside the cylinder 9

x  y . Solving for z , we get z^ ^10 ^2 x ^5 y. Using equation

(1) above, we have

Problem 5: We have

z  y  x so r i x k

  2 and r^ j y k

so

which gives

 17 17 5 5  sq.units

Problem 10:

Find the area of the surface

x  y  z inside the cylinder 9

y  z .

Note that we will view x as a function of y and z. Therefore, we can represent the surface

as the vector function r^ (^ y^ , z )^ ( y  z ) i  y j  z k

. Thus, the surface area is given by

 37 37 1  sq.units

Observations: Notice that if the surface is flat (a plane as in #2), the quantity

r  r is a constant that rescales the area of the underlying region of integration.

We can think of the region of integration as the projection of the surface onto the

given plane (say, xy plane) and that u v

rescales the area accordingly.

In the case of a curved surface as in the other problems, the region of integration is

still the projection of the surface onto the xy plane but the variable form accounts

for the curvature of the surface.

Problem 24: There are several ways to do this problem. The simplest may be finding

the area of the surface that intersects the positive x axis and taking the region of

integration in the yz plane to be 1

y  z . As such, the surface ‘above’ this region is

x  f ( y , z ) 1  z so it has vector representation

As a result,

so we may obtain the area of this surface by

However, this integral is improper since it is undefined at z = 1. Hence, we have

A S

However, this is only ONE of four such faces of intersection. Consequently, the total

surface area is 16 square units.

Alternate Solution

Using polar coordinates in the yz plane, we have the vector representation

for

   and   1 

   x     x . So

we have

x and^

so

x .

This gives