Midterm 4 Solutions
1. Find the flux of the field
F(x, y, z) = xi + zj − yk
p
over the surface x = y 2 + z 2 between x = 0 and x = 1, oriented
outwards.
Parametrizing the surface in cylindrical coordinates, we get
r(r, θ) =< r, r cos θ, r sin θ >
rr =< 1, cos θ, sin θ >
rθ =< 0, −r sin θ, r cos θ >
rr × rθ =< r, −r cos θ, −r sin θ >
An outward orientation for this cone will point in the negative xdirection, so we’ll have to use
rθ × rr =< −r, r cos θ, r sin θ > .
ZZ
ZZ
F · dS =
< x, z, −y > ·(rθ × rr ) dA
S
Z
2π
Z
D
1
< r, r sin θ, −r cos θ > · < −r, r cos θ, r sin θ > dr dθ
=
0
0
2π
Z
Z
=
0
1
(−r2 + r2 sin θ cos θ − r2 sin θ cos θ) dr dθ
0
2π
Z
Z
=
0
Z
=
0
2. Evaluate
RR
S
2π
1
(−r2 ) dr dθ
0
1
2π
(− ) dθ = −
3
3
curl F · dS, where
F(x, y, z) =< z sin(x2 y) − y, x + y 3 z 2 , z
p
√
x2 + 2y cos 2x + 1 >
and S is the half of the sphere x2 + y 2 + z 2 = 9 where z ≥ 0, oriented
away from the origin.
I will use Stokes’ Theorem.
ZZ
Z
curl F · dS =
S
∂S
1
F · dr
The curve ∂S is the circle of radius 3 in the plane z = 0. The orientation of this curve is counterclockwise in the xy-plane. I will
parametrize this by
r(t) =< 3 cos t, 3 sin t, 0 >, 0 ≤ t ≤ 2π.
Z
Z 2π
(9 sin2 t + 9 cos2 t) dt
F · dr =
0
∂S
Z
2π
9 dt = 18π
0
3. Find a parametrization for each surface.
(a) z = 2x2 + 2y 2 in polar coordinates
r(r, θ) =< r cos θ, r sin θ, 2r2 > with 0 ≤ θ ≤ 2π and r ≥ 0
(b) the part of the sphere x2 +y 2 +z 2 = 4 above the cone z =
p
x2 + y 2
r(θ, φ) =< 2 sin φ cos θ, 2 sin φ sin θ, 2 cos φ > with 0 ≤ θ ≤ 2π and
0 ≤ φ ≤ π/4
4. Consider the surface parametrized by x = r, y = r cos θ, and z =
r sin θ. Find the tangent plane to the surface at the point (x, y, z) =
(3, √32 , √32 ). Write the tangent plane in the form x + Ay + Bz = C.
This point corresponds to (r, θ) = (3, π4 ).
rr =< 1, cos θ, sin θ >=< 1, √12 , √12 >
rθ =< 0, −r sin θ, r cos θ >=< 0, − √32 , √32 >
rr × rθ =< 3, − √32 , − √32 >
n · (r − r0 ) = 0
3
3
3
3
< 3, − √ , − √ > · < x − 3, y − √ , z − √ ) = 0
2
2
2
2
y
z
x− √ − √ =0
2
2
5. Find the area of the surface y = x2 − z 2 inside the circle x2 + z 2 = 4.
Parametrize the surface by r(x, z) =< x, x2 − z 2 , z > with x2 + z 2 ≤ 4.
rx =< 1, 2x, 0 >
rz =< 0, −2z, 1 >
√
|rx × rz | = | < 2x, −1, −2z > | = 4x2 + 1 + 4z 2
2
A=
ZZ √
Z
2π
Z
2
1 + 4x2 + 4z 2 dx dz =
0
D
√
1 + 4r2 r dr dθ
0
2
Use the u-substitution u = 1 + 4r .
2π 2
π
(1 + 4r2 )3/2 |20 = (173/2 − 1)
8 3
6
RR
6. Find the integral S (yz + 2y 2 − 2y) dS, where the surface S is the
part of the plane 2x + 2y + z = 2 in the first octant.
Parametrize the plane by r(x, y) =< x, y, 2−2x−2y > with 0 ≤ x ≤ 1
and 0 ≤ y ≤ 1 − x.
rx =< 1, 0, −2 >
ry =< 0, 1, −2 >
√
|rx × ry | = | < 2, 2, 1 > | = 4 + 4 + 1 = 3
ZZ
1
Z
2
1−x
Z
[y(2 − 2x − 2y) + 2y 2 − 2y]3 dy dx
(yz + 2y ) dS =
0
S
0
1
Z
1−x
Z
=
(−6xy) dy dx
0
0
Z
=
1
−3x(1 − x)2 dx
0
Z
=
1
(−3x + 6x2 − 3x3 ) dx
0
3
3
1
= − x2 + 2x3 − x4 |10 = −
2
4
4
3