Math 212 Spring 2008: Solutions: HW #11
Instructor: S. Cautis
1. section 7.3 #2
As usual, denote the parametrization by
T (u, v) = (u2 − v 2 , u + v, u2 + 4v)
Then the tangent plane contains the two vectors
∂T
= (2u, 1, 2u)
∂u
and
∂T
= (−2v, 1, 4)
∂v
The point (−1/4, 1/2, 2) corresponds to the image of (u, v) = (0, 1/2)
under T (you can solve for u and v). Thus the tangent plane contains the
two vectors (0, 1, 0) and (−1, 1, 4).
The cross product to these vectors is (0, 1, 0) × (−1, 1, 4) = (4, 0, 1) so the
tangent plane is
(x + 1/4, y − 1/2, z − 2) · (4, 0, 1) = 0
or equivalently
4x + z = 1.
2. section 7.3, #6
We have
T (θ, φ) = (3 cos θ sin φ, 2 sin θ sin φ, cos φ)
so that
and
∂T
= (−3 sin θ sin φ, 2 cos θ sin φ, 0)
∂θ
∂T
= (3 cos θ cos φ, 2 sin θ cos φ, − sin φ)
∂φ
So a normal vector is
∂T
∂T
×
= (−2 cos θ sin2 φ, −3 sin θ sin2 φ, −6 sin φ cos φ)
∂θ
∂φ
where we used that sin2 θ + cos2 θ = 1 to get the z-coordinate.
The norm of this vector is the square root of
(−2 cos θ sin2 φ)2 +(−3 sin θ sin2 φ)2 +(−6 sin φ cos φ)2 = sin2 φ(4 cos2 θ sin2 φ+9 sin2 θ sin2 φ+36 cos2 φ)
One could try to simplify this further but there is no point because it will
not simplify to something very simple like in problem 5. The answer is
just
1
p
(2 cos θ sin φ, 3 sin θ sin φ, 6 cos φ)
2
2
4 cos θ sin φ + 9 sin2 θ sin2 φ + 36 cos2 φ
where we multiplied the normal vector vector by −1 and also cancelled a
factor of sin φ.
What should be clear is that the vector part looks similar to the original
coordinates of the surface. It is not clear to me how this tells you what the
surface is. However, going back to the original parametrization, we notice
that it is the same as the parametrization of the unit sphere except that
it has been dilated in the x and y directions. Thus we have a “squished”
sphere which also goes by the name of ellipsoid.
3. section 7.3, #12
Since
z > 0 we can write the surface as the graph of the function z =
p
2 − x3 − 3xy. Thus it has the natural parametrization
p
T (u, v) = (u, v, 2 − u3 − 3uv)
Now
−3u2 − 3v
∂T
= (1, 0, √
)
∂u
2 2 − u3 − 3uv
and
−3u
∂T
)
= (0, 1, √
∂v
2 2 − u3 − 3uv
The the normal vector is
∂T
∂T
3u2 + 3v
3u
, √
, 1)
×
=( √
3
∂u
∂v
2 2 − u − 3uv 2 2 − u3 − 3uv
Substituting u = 1 and v = 1/3 we get zero in the denominators
which is
√
bad. So let’s get rid by the denominators by multiplying by 2 2 − u3 − 3uv
to get
p
(3u2 + 3v, 3u, 2 2 − u3 − 3uv)
Now substituting we get
(4, 3, 0)
so the tangent plane is
(x − 1, y − 1/3, z) · (4, 3, 0) = 0
or equivalently
4x + 3y = 5
The reason we had to do this funny business of clearing the denominator is because in our parametrization we assumed z > 0 while the point
(1, 1/3, 0) is on the boundary of this domain (since it’s z-coordinate is
zero) – so we have to take some sort of limit.
The surface is also the zero level set of F (x, y, z) = x3 + 3xy + z 2 − 2.
Thus the normal vector is
(
∂F ∂F ∂F
,
,
) = (3x2 + 3y, 3x, 2z)
∂x ∂y ∂z
which equals (4, 3, 0) when you plug in x = 1, y = 1/3, z = 0. Thus we get
the same tangent plane as calculated above.
4. section 7.3, #14
a) If a sphere is parametrized in spherical coordinates then we calculated
in class that
Φθ × Φφ = R2 (− sin2 φ cos θ, − sin2 φ sin θ, − sin φ cos φ)
where the radius R in our case is 2.
√
The point (1, 1, 2) corresponds to θ = φ = π/4 so the normal vector
there is
1
1
1
4(− √ , − √ , − )
2 2 2 2 2
√
which dividing by − 2 gives
√
(1, 1, 2)
So the tangent plane is
(x − 1, y − 1, z −
√
√
2) · (1, 1, 2) = 0
b) The normal vector now is
∂f ∂f ∂f
,
,
) = (2x, 2y, 2z)
∂x ∂y ∂z
√
√
which at (1, 1, 2) equals (2, 2, 2 2). Dividing by 2 we get the same
normal vector as above (and hence the same tangent plane).
√
c) Here we parametrize as T (u, v) = (u, v, 4 − u2 − v 2 ). Then
(
Tu = (1, 0, √
and
−u
)
4 − u2 − z 2
−v
)
Tv = (0, 1, √
4 − u2 − v 2
Hence
u
v
Tu × Tv = ( √
,√
, 1)
2
2
4−u −v
4 − u2 − v 2
√
Plugging in (u, v) = (1, 1) (which corresponds to the point (1, 1, 2) under
the map T ) we get the normal vector
1
1
( √ , √ , 1)
2
2
which after multiplying by
√
2 we get the same normal vector as above.
Thus we see that all three calculations agree – i.e. they all give us the
same tangent planes.
5. section 7.4, #4
We have
∂(x, y)
−(R + cos φ) sin θ
= det
(R + cos φ) cos θ
∂(θ, φ)
(R − sin φ) cos θ
(R − sin φ) sin θ
= −(R+cos φ)(R−sin φ)
and
∂(y, z)
(R + cos φ) cos θ
= det
0
∂(θ, φ)
(R − sin φ) sin θ
cos φ
= (R+cos φ) cos θ cos φ
and
∂(x, z)
−(R + cos φ) sin θ
= det
0
∂(θ, φ)
(R − sin φ) cos θ
cos φ
= −(R+cos φ) sin θ cos φ
Plugging into formula 3 we get
Z Z q
Area(T ) =
(R + cos φ)2 (R − sin φ)2 + (R + cos φ)2 cos2 θ cos2 φ + (R + cos φ)2 sin2 θ cos2 φdθd
Z ZD p
(R + cos φ)2 (R − sin φ)2 + (R + cos φ)2 cos2 φdθdφ
=
=
Z
0
D
2π Z 2π
= 2π
Z
0
2π
0
p
(R + cos φ) R2 − 2R sin φ + 1dθdφ
p
(R + cos φ) R2 − 2R sin φ + 1dφ
R 2π
p
Now 0 cos φ R2 − 2R sin φ + 1dφ = 0 and using a table of integrals one
finds that
Z 2π p
R R2 − 2R sin φ + 1dφ = 2πR
0
so that the area is 2π(2πR) = (2π)2 R (as required).
On the other hand, the way the torus is parametrized it is the rotation
of the circle in the y − z plane of radius one and center (0, R, 0) around
the z-axis. Let’s
p consider justhe upper half of the circle given in the y − z
plane by z = 1 − (y − R)2 . Then by formulat 6 the area is
2π
Z
R+1
R−1
p
|y| 1 + f ′ (y)2 dy
p
where f (y) = 1 − (y − R)2 . Now f ′ (y) = √−(y−R)
1−(y−R)2
1 + f ′ (y)2 = 1 +
Thus the area is
Z
2π
R+1
R−1
so that
(y − R)2
1
=
2
1 − (y − R)
1 − (y − R)2
y
p
dy = 2π
1 − (y − R)2
Z
1
−1
y+R
p
dy
1 − y2
where we changed variables y 7→ y + R. Now
Z 1
p
y
dy = [ 1 − y 2 ]1−1 = 0
2
−1 1 − y
while
Z
1
−1
R
p
dy = Rπ
1 − y2
(using a table of integrals). Thus the area is 2π · Rπ and doubling that
(we just computed the area of half the torus) we get (2π)2 R as required.
6. section 7.4, #14
We have
Tu = (
∂x ∂y ∂z
,
,
)
∂u ∂u ∂u
and
∂x ∂y ∂z
,
,
)
∂u ∂u ∂u
Now taking the cross product and simplifying we arrive at the desired
expression.
Tv = (
7. section 7.4, #16
Write c(t) = (x(t), y(t)). Then rotating the surface about the y-axis generates a surface S which is parametrized by
T (t, θ) = (x(t) cos θ, y(t), x(t) sin θ)
Then
Tt = (x′ (t) cos θ, y ′ (t), x′ (t) sin θ)
while
Tθ = (−x(t) sin θ, 0, x(t) cos θ)
Then
Tt × Tθ = (x(t)y ′ (t) cos θ, −x(t)x′ (t), x(t)y ′ (t) sin θ)
Thus
||Tt ×Tθ || =
p
x(t)2 y ′ (t)2 + x(t)2 x′ (t)2 = |x(t)|||x′ (t), y ′ (t))|| = x(t)||c′ (t)||
where we removed the absolute value signs around x(t) since we are in the
right half of the plane so x(t) > 0. Thus the area is
Z
0
2π
Z
b
x(t)||c′ (t)||dθdt = 2π
Z
x(t)ds
c
a
Now the average value of the x coordinate is
Z
1
x̄ =
x(t)ds
l(c) c
so the area is
2π
Z
x(t)ds = 2πx̄l(c)
c
as required.
8. section 7.4 #22c
Using polar coordinates we have
Z Z q
A(S) =
fx2 + fy2 + 1dA
D
=
where f = xy 3 ex
2 2
2x3 y 4 ex y .
2 2
y
Z
1
0
Z
2π
0
so that fx = y 3 ex
q
fx2 + fy2 + 1rdθdr
2 2
y
+2x2 y 5 ex
2 2
y
and fy = 3xy 2 ex
2 2
y
+