QUIZ 10 : MATH 251, Section 516
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Write up your result, detail your calculations if necessary and BOX your final answer
I
1. [50pts] Use Stokes’ Theorem to evaluate
F.dr where C, oriented counterclockwise is the part of
C
the plane 3x + y + z = 3 in the first octant and F = hxz, 2xy, 3xyi.
Z Z
2. [50pts] Use Divergence Theorem to calculate
F.dS for F = h3y 2 z 3 , 9x2 yz 2 , −4xy 2 i and S the
S
surface of the cube with vertices (±1, ±1, ±1).
Correction.
1. The plane 3x + y + z = 3 with the xy-plane forms a surface S in R3 which we may describe as the
graph z = f (x, y) = 3 − 3x − y, with D = {(x, y) | 0 ≤ y ≤ 3 − 3x, 0 ≤ x ≤ 1}. F is a vector field
with its differentiable component functions in R3 . By Stokes’ Theorem,
I
Z Z
F.dr =
curlF.dS,
C
S
where
i
j
k
∂z CurlF = ∂x ∂y
xz 2xy 3xy = h3x, −3y + x, 2yi
Since S is the graph of f , a normal vector to S is ~n = ∇F , where F (x, y) = z − f (x, y) since the
normal vector is upward (it means z ≥ 0).
~n = h3, 1, 1i.
So,
Z Z
Z Z
~n
|~n|dA,
|~n|
D
Z Z
Z 1Z
=
curlF.~ndA =
curlF.dS =
S
curlF.
D
Z
0
3−3x
h3x, −3y + x, 2yii • h3, 1, 1idy dx,
0
1 Z 3−3x
Z
9x − 3y + x + 2ydy dx =
=
0
0
0
1
y2
10xy −
2
3−3x
dx,
0
Z 1
9 2
9
69x2 9
=
30x − 30x − + 9x − x dx =
39x −
− dx,
2
2
2
2
0
0
1
2
3
39x
69x
9x
69
21
7
=
−
−
= 15 −
=
= .
2
6
2 0
6
6
2
Z
1
2
Z Z
Z Z Z
2. Thanks to Divergence Theorem,
F.dS =
S
divF dV ,
R
where R is the cube {(x, y, z) | −1 ≤ x ≤ 1, −1 ≤ y ≤ 1, −1 ≤ z ≤ 1}.
divF = ∂x (3y 2 z 3 ) + ∂y (9x2 yz 2 ) + ∂z (−4xy 2 ) = 0 + 9x2 z 2 + 0 = 9x2 z 2 .
So,
Z
Z Z Z
1
Z
1
Z
1
9x2 z 2 dx dy dz,
divF dV =
−1
Z 1
R
=
−1
−1
Z
2
9x dx.
−1
=
1
3x3 −1 .
1
2
Z
1
z dz.
−1
3 1
z
3
dy, by Fubini’s Theorem
−1
2
.2 = 6. .2 = 8 .
3
−1