MATH 2423 #15:
QUESTION 1
#15: 1a) ANSWER: 67 pi / 2
SOLUTION: Cylindrical coordinates will be best for this
problem.
- DIV F = ( x 3 + z ) x + ( y 3 + z )
y + ( z 2 + xy ) z = 3x 2 + 3y 2 + 2z
= 3r 2 + 2z
- The lower z-limit is z = -1. The upper z-limit is z = 5 - x - y = 5 - rcos
0 +
rsin0
- The equation of the cylinder x 2 + y 2 = 1 becomes r = 1.
Therefore, r goes from 0 to 1 and
0 goes from 0 to 2pi.
- dV = r dz dr d
0
- The integral becomes
© 2000: Roxanne M. Byrne