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QUIZ 6, Version A : MATH 251, Section 505
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Write up your result, detail your calculations if necessary and BOX your final answer
Let S be the surface y 2 = x2 + z 2 between the planes y = 0 and y = 3 and a vector field F (x, y, z) =&lt;
−x, y, −z &gt;.
1. [16pts] Find a parametric equation of S.
2. [16pts] Find the flux of F across S (oriented upward).
˛
3. [18pts] Compute
F.dr by using Stokes’ Theorem where C is the boundary curve of S .
C
‹
~ where Ŝ is the closed surface S bounF.dS
4. [5pts] (Bonus) Use the divergence theorem to compute
Ŝ
ded by the plane y = 2.
1. A parametric equation of this part of the cone is r(u, v) =&lt; u cos v, u, u sin v &gt; where
(u, v) ∈ {(u, v) | 0 ≤ u ≤ 3, 0 ≤ v ≤ 2π} . We can also parametrize this seen as the graph of f (x, z) =
p
p
√
x2 + z 2 and r(x, z) =&lt; x, z 2 + x2 , z &gt; where (x, z) ∈ {(x, z) | x2 + z 2 ≤ 3} .
&acute;&acute;
2. The flux is
S F.N dS where N is a normal vector to S. From the previous question, N (u, v) =
&plusmn; &lt; u cos v, −u, u sin v &gt;. Since it’s oriented upward, we keep the sign such that the y-component is
positive on the domain of r, so N (u, v) =&lt; −u cos v, u, −u sin v &gt;.
ˆ ˆ
ˆ ˆ
ˆ
~=
F.dS
S
2π
ˆ
3
F.N dS =
F (r(u, v)).N (u, v)du dv
S
0
0
ˆ 2π ˆ 3
=
&lt; −u cos v, u, −u sin v &gt; . &lt; −u cos v, u, −u sin v &gt; du dv
0
0
ˆ 2π ˆ 3
=
u2 + u2 du dv = 32π
6 .
0
0
√
3. S has as boundary curve the circle C := { x2 + z 2 = 3, y = 3}. By Stoke’s Theorem,
ˆ
ˆ ˆ
~
F.dr =
CurlF.dS.
C
S
But CurlF =&lt; ∂x , ∂y , ∂z &gt; &times; &lt; −x, y, −z &gt;=&lt; 0, 0, 0 &gt; so
ˆ
ˆ ˆ
~= 0.
F.dr =
CurlF.dS
C
S
4. (Bonus) By divergence Theorem,
‹
ˆ ˆ ˆ
ˆ ˆ ˆ
~=
F.dS
divF dV =
−1dV,
Ŝ
E
E
where E is the solid region bounded by the cone S and the plane y = 2, described as follows
p
p
{(x, y, z) | x2 + z 2 ≤ y ≤ 2, x2 + z 2 ≤ 2},
and becomes in cylindrical coordinates {(r, θ, y) | 0 ≤ r ≤ 2, 0 ≤ θ ≤ 2π, r ≤ y ≤ 2}. So,
ˆ
ˆ ˆ ˆ
‹
~=−
F.dS
Ŝ
2π
ˆ
2ˆ 2
dV = −
ˆ
= −2π
0
E
2
rdy dr dθ,
0
0
r
2
r3
8π
2
r(2 − r)dr = −2π r −
= −
3 0
3
```