Homework Week 10 - Solutions

Math 311 - Fall 2012

October 30, 2012

p. 236, Problem 2

Find dS and dS , in terms of dθ and dφ for the surface with parametric equations

x = (1 + cos θ) cosφ, y = (1 + cos θ) sinφ, z = sin θ

Solution:R(θ, φ) = (1 + cos θ) cosφi(1 + cos θ) sinφj sin θk

∂R

∂θ= − sin θ cosφi− sin θ sinφj cos θk

∂R

∂φ= −(1 + cos θ) sinφi(1 + cos θ) cosφj

dS =

(∂R

∂θ× ∂R

∂φ

)dθdφ =

∣∣∣∣∣∣i j k

− sin θ cosφ − sin θ sinφ cos θ−(1 + cos θ) sinφ (1 + cos θ) cosφ 0

∣∣∣∣∣∣ dθdφdS = −(1 + cos θ)[cos θ cosφi+ cos θ sinφj+ sin θk]dθdφ

dS = (1 + cos θ)dθdφ

p. 236, Problem 4

Determine the element of surface area dS in the special case of the paraboloid of revolution z =x2 + y2. Solution:

R(x, y) = xi+ yj+ (x2 + y2)k

∂R

∂x= i+ 2xk,

∂R

∂y= j+ 2yk

dS =

(∂R

∂x× ∂R

∂y

)dxdy =

∣∣∣∣∣∣i j k1 0 2x0 1 2y

∣∣∣∣∣∣ dydx = [−2xi− 2yj+ k]dxdy

dS =√1 + 4(x2 + y2)dxdy

1

p. 236, Problem 5

Find the area of the section of the surface

x = u2, y = uv, z =1

2v2

bounded by the curves u = 0, u = 1, v = 0, and v = 3.Solution:

R(u, v) = u2i+ uvj+1

2v2k

∂R

∂u= 2ui+ vj,

∂R

∂v= 2ui+ vj

dS =

(∂R

∂u× ∂R

∂v

)dudv =

∣∣∣∣∣∣i j k2u v 00 u v

∣∣∣∣∣∣ dudv = [v2i− 2uvj+ 2u2k]dudv

dS · dS = (v4 + 4u2v2 + 4u2)du2dv2 = (v2 + 2u2)2du2dv2

Therefore, dS = (v2 + 2u2)dudv. The surface area is

S =

3∫0

1∫0

(v2 + 2u2) dudv =

3∫0

v2u+2

3u3

∣∣∣∣∣1

0

dv =

3∫0

v2 +2

3dv =

1

3v3 +

2

3v

∣∣∣∣∣3

0

= 11

p. 236, Problem 6

Consider the triangle with vertices (1, 0, 0), (0, 1, 0), and (0, 0, 1).

(a) Find a unit vector n normal to this triangle, pointing away from the origin.

(b) Determine cos γ for this vector.

(c) Supply the appropriate limits for the integral∫∫dxdy

| cos γ|

if it is to represent the area of this triangle.

(d) Evaluate the integral.

(e) Obtain the same answer by applying the area cosine principle to the projection of the triangleon the yz plane.

Solution: The plane that passes through the three given vertices is x + y + z = 1. The unit(outward) normal to this plane is n = 1√

3(i+ j+k). The angle between this vector and the vertical

unit vector is given by

cos γ = k · n =1√3

2

The limits of the integral are 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1− x.

S =

1∫0

1−x∫0

√3 dydx =

√3

2

We would get the same thing with

S =√3

1∫0

1−x∫0

dydx =√3

(1

2

)

3