[16 .6] Parametric Surfaces and Their Areas
19. The plane through the origin that contains the vectors i− j and j−k The vector equation of the plane can be written as
γ(u, v) = u(i − j) + v(j − k)
= ui + (v − u)j + (−v)k Thus, x= u , y = v − u , z = −v
23. In spherical coordinates, parametric equations are
x= 2 sin ϕ cos θ, y = 2 sin ϕ sin θ, z = 2 cos ϕ The intersection of the sphere with the cone z = √
x2+ y2 corresponds to 2 cosϕ = 2| sin ϕ| ⇒ ϕ = π
Thus, x4 = 2 sin ϕ cos θ, y = 2 sin ϕ sin θ, z = 2 cos ϕ where 0 ≤ θ ≤ 2π, 0 ≤ ϕ ≤ π4
24. In spherical coordinates, parametric equations are
x= 4 sin ϕ cos θ, y = 4 sin ϕ sin θ, z = 4 cos ϕ
The intersection of the sphere with the plane z= 2 corresponds to z = 4 cos ϕ = 2 ⇒ ϕ = π The intersection of the sphere with the plane z= −2 corresponds to 3
z= 4 cos ϕ = −2 ⇒ ϕ = 2π
Thus, we can parametrize this surface as x3 = 4 sin ϕ cos θ, y = 4 sin ϕ sin θ, z = 4 cos ϕ where 0 ≤ θ ≤ 2π, π3 ≤ ϕ ≤ 23π
34. γ(u, v) = (u2+ 1) i + (v3+ 1) j + (u + v) k γu= 2u i + 0 j + 1 k
γv = 0 i + 3v2j+ 1 k
⇒ γu× γv = −3v2i− 2u j + 6uv2 k
The point (5,2,3) corresponds to u = 2, v = 1 in this surface, so a normal vector to the surface at this point is (−3, −4, 12).
Thus, the tangent plane at (5,2,3) is
−3(x − 5) − 4(y − 2) + 12(z − 3) = 0 ⇒ 3x + 4y − 12z = −13 35. γ(u, v) = u cos v i + u sin v j + v k
γu= cos v i + sin v j + 0 k γv = −u sin v i + u cos v j + 1 k
⇒ γu× γv = sin v i − cos v j + u k At u = 1, v = π
3, γ(1,π 3)= (1
2,
√3 2 ,π
3), so a normal vector at this point is (
√3 2 , −1
2, 1).
Thus, the tangent plane at u = 1, v = π 3 is
√3 2 x− 1
2y+ z = π 3 41. z= f (x, y) = 1
3 − 1 3x− 2
3y and D is the disk x2+ y2 ≤ 3 So the area of the surface is
A(S )=
"
D
√
1+ (∂z
∂x)2+ (∂z
∂y)2dA=
"
D
√
1+ (−1
3)2+ (−2 3)2dA
=
√14 3
"
D
dA=
√14
3 A(D)=
√14 3 π(√
3)2 = √ 14π
1
44. z = f (x, y) = 1 + 3x + 2y2and D is the triangle with vertices (0,0), (0,1), (2,1) i.e. 0 ≤ x ≤ 2y, 0≤ y ≤ 1
So the area of the surface is
A(S ) =
"
D
√
1+ (∂z
∂x)2+ (∂z
∂y)2dA=
"
D
√1+ (3)2+ (4y)2dA
=
∫ 1 0
∫ 2y 0
√10+ 16y2dxdy=
∫ 1 0
2y√
10+ 16y2dy
= 1
16 · 2 3
(10+ 16y2)32 10= 1
24(2632 − 1032)
45. z= f (x, y) = xy, and D is the disk x2+ y2≤ 1 So the area of the surface is
A(S ) =
"
D
√
1+ (∂z
∂x)2+ (∂z
∂y)2dA=
"
D
√1+ y2+ x2dA
Let x = r cos θ, y = r sin θ with 0 ≤ r ≤ 1, 0 ≤ θ ≤ 2π Then A(S ) =
∫ 2π
0
∫ 1
0
√1+ r2rdrdθ =
∫ 2π
0
dθ
∫ 1
0
r√
1+ r2dr
= 2π · 1 2 · 2
3(1+ r2)32 10 = 2π 3 (2√
2− 1)
61. Solve the equations x2+ y2+ z2= 4z and z = x2+ y2⇒ z2− 3z = 0 ⇒ z = 0 or z = 3 Thus, the surfaces intersect on z= 0 and z = 3
But x2+ y2+ z2 = 4z ⇒ x2+ y2+ (z − 2)2 = 4, so z = 3 intersects the upper hemisphere.
∴ z = 2 + √
4− x2− y2and D={
(x, y)|x2+ y2 ≤ 3} So the area of the surface is
A(S ) =
"
D
√
1+ (∂z
∂x)2+ (∂z
∂y)2dA=
"
D
√
1+ ( −x
√4− x2− y2)2+ ( −y
√4− x2− y2)2dA
=
"
D
√ 4
4− x2− y2dA=
∫ 2π 0
∫ √3 0
√ 2
4− r2rdrdθ = 2π[
−2√
4− r2]√3 0 = 4π
62. Let S be the face of the surface that intersects the positive y-axis.
It can be parametrized asγ(x, θ) = (x, cos θ, sin θ) for −π2 ≤ θ ≤ π2 and x2+z2 ≤ 1 ⇒ −√
1− z2≤ x≤ √
1− z2⇒ − cos θ ≤ x ≤ cos θ
Thenγx = (1, 0, 0), γy = (0, − sin θ, cos θ),and γx× γy = (0, − cos θ, − sin θ) ⇒ |γx× γy| = 1
∴ A(S ) =
∫ π2
−π2
∫ cosθ
− cos θ1dxdθ =
∫ π2
−π2 2 cosθdθ = [2 sin θ]−π2π
2 = 4
Since the complete surface consists of four congruent faces, the total surface area is 4× 4 = 16
2
