Worked examples · Sheet 3

Areas of surfaces of revolution and portions of spheres and cylinders

Theory used: Area of a surface of revolution Area of a surface

Exercise 1

Compute the area of the lateral surface of the cone obtained by rotating about the $z$ -axis the segment with equations $x = t$ , $z = 2 t$ for $t \in [0, 3]$ .

g (t) = t, h (t) = 2 t, g^{'} (t)^{2} + h^{'} (t)^{2} = 1 + 4 = 5

Area = 2 π \int_{0}^{3} t 5 d t = 2 π 5 \cdot \frac{9}{2} = 95 π

The curve generating the surface is the segment $x = g (t) = t$ , $z = h (t) = 2 t$ , with $g \geq 0$ , so the formula for the area of a surface of revolution applies. The derivatives are $g^{'} = 1$ and $h^{'} = 2$ , therefore the factor $g^{'2} + h^{'2} = 5$ is constant. The integral reduces to $2 π 5 \int_{0}^{3} t d t$ , and since $\int_{0}^{3} t d t = \frac{9}{2}$ , the area equals $95 π$ . The same value is obtained from the elementary formula $π R L$ for the lateral surface of the cone, with base radius $R = 3$ and slant height $L = 9 + 36 = 35$ .

Exercise 2

Compute the area of the spherical cap

Σ = {(x, y, z) : x^{2} + y^{2} + z^{2} = 4, z \geq 1} .

r (φ, θ) = (2 sin φ cos θ, 2 sin φ sin θ, 2 cos φ), d σ = 4 sin φ d φ d θ

z \geq 1 ⟺ 2 cos φ \geq 1 ⟺ 0 \leq φ \leq \frac{π}{3}

Area = \int_{0}^{2 π} \int_{0}^{π /3} 4 sin φ d φ d θ = 8 π [- cos φ]_{0}^{π /3} = 8 π (1 - \frac{1}{2}) = 4 π

Cross-section with the plane

y = 0

: the cap is the part of the sphere with

z \geq 1

, that is, with colatitude

φ

between

0

and

\frac{π}{3}

The cap is a portion of the sphere of radius $2$ , so we use the spherical parametrization, whose area element is $d σ = R^{2} sin φ d φ d θ = 4 sin φ d φ d θ$ . The condition $z \geq 1$ is translated into the parameters: since $z = 2 cos φ$ , the inequality is equivalent to $cos φ \geq \frac{1}{2}$ , that is, $φ \in [0, \frac{π}{3}]$ , while $θ$ runs over the full turn. The integral in $θ$ gives $2 π$ and the one in $φ$ gives $1 - cos \frac{π}{3} = \frac{1}{2}$ , so the area is $4 π$ . The result is consistent with the formula $2 π R h$ for the area of a cap, with $R = 2$ and height $h = 2 - 1 = 1$ .

Exercise 3

Compute the area of the portion of the cylinder $x^{2} + y^{2} = 9$ lying between the plane $z = 0$ and the plane $z = x + 5$ .

r (θ, t) = (3 cos θ, 3 sin θ, t), d σ = 3 d θ d t

0 \leq t \leq 3 cos θ + 5, 3 cos θ + 5 > 0 for every θ

Area = \int_{0}^{2 π} \int_{0}^{3 c o s θ + 5} 3 d t d θ = 3 \int_{0}^{2 π} (3 cos θ + 5) d θ = 3 (0 + 10 π) = 30 π

The parameter domain of the portion of the cylinder:

0 \leq θ \leq 2 π

and

0 \leq t \leq 3 cos θ + 5

; the function

3 cos θ + 5

lies between

2

and

8

The cylinder of radius $3$ is parametrized by the angle $θ$ and the height $t$ , and the area element is $d σ = R d θ d t = 3 d θ d t$ , since for the cylinder the norm of the normal vector is the radius. On the cylinder $x = 3 cos θ$ , so the upper plane cuts the surface at height $t = 3 cos θ + 5$ ; since $3 cos θ + 5 \geq 2 > 0$ for every $θ$ , the upper height stays above the plane $z = 0$ and the parameter region is $0 \leq t \leq 3 cos θ + 5$ with $θ \in [0, 2 π]$ . The inner integral in $t$ gives the length $3 cos θ + 5$ of the generating line. In the integral in $θ$ the term $3 cos θ$ has zero integral over the period, therefore what remains is $5 \cdot 2 π = 10 π$ , and the area equals $30 π$ .