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QUIZ 8, 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 z 2 = x2 + y 2 between z = 0 and z = 2.
1. [16pts] Find a parametric equation of S.
A parametric equation of this part of the cone is r(u, v) =< u cos v, u sin v, u > where
(u, v) ∈ {(u, v) | 0 ≤ u ≤ 2, 0 ≤ v ≤ 2π} . We can also parametrize this seen as the graph of f (x, y) =
p
p
p
x2 + y 2 and r(x, y) =< x, y, x2 + y 2 > where (x, y) ∈ {(x, y) | x2 + y 2 ≤ 2} .
2. [12pts] Compute the area of S.
A normal vector to S is
N (u, v) = ±(ru × rv ) = ±(< cos v, sin v, 1 > × < −u sin v, u cos v, 0 >) = ± < −u cos v, −u sin v, u > .
Z Z
A(S) =
2π
Z
Z
S
2π Z 2 √
Z
=
0
√
0
2
|N (u, v)|du dv,
dS =
0
0
2 2
√
u
2udu dv = 2 2π
,
2 0
= 4 2π .
3. [12pts] Find Rthe
R flux of F (x, y, z) =< −x, −y, z > across S and oriented upward.
The flux is
S F.N dS where N is a normal vector to S. From the previous question, N (u, v) =
± < −u cos v, −u sin v, u >. Since it’s oriented upward, we keep the sign such that the z-component
is positive on the domain of r, so N (u, v) =< −u cos v, −u sin v, u >.
Z Z
~=
F.dS
Z Z
Z
2π
Z
2
F.N dS =
F (r(u, v)).N (u, v)du dv
S
0
0
Z 2π Z 2
=
< −u cos v, −u sin v, u > . < −u cos v, −u sin v, u > du dv
S
0
Z
0
2π
Z
=
0
2
u2 + u2 du dv =
0
32π
.
3
I
4. [10pts] Compute
F.dr by using Stokes’ Theorem where C is the circle with center the origin and
C
radius 2.
p
S is a closed surface with the boundary curve the circle C := { x2 + y 2 = 2}. By Stoke’s Theorem,
Z
Z Z
~
F.dr =
CurlF.dS.
C
S
But CurlF =< ∂x , ∂y , ∂z > × < −x, −y, z >=< 0, 0, 0 > so
Z
Z Z
~= 0.
F.dr =
CurlF.dS
C
S