MATH 2423 #15: QUESTION
3
#15: 3a) ANSWER: Yes, 67 pi / 2.
SOLUTION:
- The surface is made up of 3 segments, two planes and a cylinder.
- Top segment: x + y + z = 5 inside the cylinder x 2 + y 2 = 1.
- the projection of the surface is the circle x 2 + y 2 = 1.
- GRAD ( x + y + z ) = < ( x + y + z ) x , ( x + y + z )
y , ( x + y + z ) z > = <1,1,1>
- ( x + y + z ) z = 1
- For the upward normal: dS = ( <1,1,1>/1) dA xy
- F • dS = < x 3 + z , y 3 + z , z
2 + xy > • ( <1,1,1>/1) dA xy =
( x 3 + y 3 + xy + z 2 + 2z ) dA xy =
( x 3 + y 3 + xy + ( 5 - x - y ) 2 + 2 ( 5 - x - y ) )
dA xy
- The best coordinate system is the polar coordinate system:
- F • dS == ( r 3 cos 3
0 - r 3 sin 3 0 + r 2
cos0 sin0 + ( 5 -
rcos0 - rsin0 ) 2 + 2 ( 5 - rcos0 -
rsin0 ) ) r dr d0
- The integral becomes
- Bottom segment: z = -1 inside the cylinder x 2 + y 2 = 1.
- GRAD ( z ) = < 0 , 0 , 1 >
- ( z ) z = 1
- For the downward normal, dS = ( < 0 , 0 , 1 >/(-1 ) dA xy
- F • dS = < x 3 + z , y 3 + z , z
2 + xy > • <0,0,-1> dA xy = - ( z 2 + xy )
dA xy =
- ( 1 + xy ) dA xy
- Polar coordinates is again the best coordinate system.
- F • dS = - ( 1 + r 2 cos
0
sin0 ) r dr d0
- The cylinder cannot be projected down to the xy-plane since it is perpendicular to the
xy-plane.
Instead, project it to the xz-plane. It must be broken into tow parts, the right and left half.
- The right half: x 2 + y 2 = 1, y > 0
- To get the projected region, you must eliminate y from the equations.
- The intersection between x 2 + y 2 = 1 and z = -1 is z = -1
- The intersection between x 2 + y 2 = 1 and y = 0 is x =
+ 1
- The intersection between x 2 + y 2 = 1 and x + y + z = 5 is
x 2 + ( 5 - x - z ) 2 = 1
The bottom loop of the ellipse is for the right half, the top loop is for the right half.
For
this part z = 5 - x - ( 1 - x 2 ) ½. ( Notice that ( 1 - x 2 ) ½ = 5 - x - z = y
> 0. )- GRAD ( x 2 + y 2 ) = < 2x , 2y , 0 > = 2 < x ,
y, 0 >
- ( x 2 + y 2 ) y = 2y
- For the right normal, dS = ( 2 < x , y, 0 >/(2y) )dA xz
- F • dS = < x 3 + z , y 3 + z , z
2 + xy > • ( < x , y, 0 >/y )dA xz =
( x 4 + xz + y 4 + yz )/y dA xz =
[( x 4 + xz + + ( 1 - x 2 ) 2 + ( 1 - x 2
)1/2 z )/( 1 - x 2 ) ½ ] dz dx
- The integral becomes
- The left half has most of the same values except
- z = 5 - x + ( 1 - x 2 ) ½ because y = 5 - x - z = - ( 1 -
x 2 ) ½ < 0.
- dS = ( 2 < x , y, 0 >/(-2y) )dA xz because the normal must
have a negative y-component
and y = - ( 1 - x 2 ) ½ .
- F • dS = - [( x 4 + xz + + ( 1 - x 2 )
2 - ( 1 - x 2 )1/2 z )/( 1 - x 2 )
½ ] dz dx
- Adding these four integrals gives 67 pi / 2 which agrees with problem 1.
© 2000: Roxanne M. Byrne