Lesson 10 6 Surface Integrals.pdf


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G   g x ,  g y ,1  2 x, 2 y, 1
So, F G  x2  2xy  y 2  1 , and we have the following integral:
2
2
 F GdA    x  2 xy  y  1  dA 

S1





S1

2 1

0 0  r

2

0

3



 2r 3 cos sin   r drd 

2

0

2 1

0 0  r

2



 2r 2 cos sin   1 rdrd

1

 r 4 2r 3
r2 

cos

sin



 d
4
3
2

0
2

1 2
1
1
1 2
1 
2 2 3


  cos sin    d     sin     
2
3
2 0
4
2
2
4 3
4

Therefore, our surface integral is equal to

3
2

.

Internet Activity III
The following two videos provide step-by-step solutions to some examples of surface
integrals of a vector field over a surface. Each video is approximately 6-7 minutes in length.
http://www.youtube.com/watch?v=y-gsqWf3Gms
http://www.youtube.com/watch?v=9H4Q-FEKwGw

View each one, and follow along on your own paper.