TBIL-Calc Surface Areas of Revolution (AI4)

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Section 6.4 Surface Areas of Revolution (AI4)

Learning Outcomes

🔗
Compute surface areas of surfaces of revolution.

Subsection 6.4.1 Activities

Fact 6.4.1.

A frustum is the portion of a cone that lies between one or two parallel planes.

frustum — Figure 129. Plot of a frustum.

The surface area of the “side” of the frustum is:

2 π \frac{r + R}{2} \cdot l

where

r

and

R

are the radii of the bases, and

l

is the length of the side.

Note that if

r = R,

this reduces to the surface area of a “side” of a cylinder.

Activity 6.4.1.

Suppose we wanted to find the surface area of the the solid of revolution generated by rotating

y = \sqrt{x}, 0 \leq x \leq 4

about the

y

-axis.

Bounded region rotated about \(x\)-axis. — Figure 130. Plot of bounded region rotated about $x$ -axis.

(a)

Suppose we wanted to estimate the surface area with two frustums with

Δ x = 2 .

Bounded region rotated about \(x\)-axis. — Figure 131. Plot of bounded region rotated about $x$ -axis.

What is the surface area of the frustum formed by rotating the line segment from

(0, 0)

to

(2, \sqrt{2})

about the

x

-axis?

🔗
$2 π \frac{0 + \sqrt{2}}{2} \cdot 2$
🔗
$2 π \frac{0 + \sqrt{2}}{2} \cdot \sqrt{2^{2} + {\sqrt{2}}^{2}}$
🔗
$π {\sqrt{2}}^{2} \cdot 2$
🔗
$π {\sqrt{2}}^{2} \cdot \sqrt{2^{2} + {\sqrt{2}}^{2}}$

(b)

Bounded region rotated about \(x\)-axis. — Figure 132. Plot of bounded region rotated about the $x$ -axis.

What is the surface area of the frustum formed by rotating the line segment from

(2, \sqrt{2})

to

(4, 2)

about the

x

-axis?

🔗
$2 π \frac{4 + \sqrt{2}}{2} \cdot \sqrt{2}$
🔗
$2 π \frac{4 + \sqrt{2}}{2} \cdot \sqrt{6}$
🔗
$2 π \frac{4 + \sqrt{2}}{2} \cdot \sqrt{6 - 2 \sqrt{2}}$

(c)

Suppose we wanted to estimate the surface area with four frustums with

Δ x = 1 .

Bounded region rotated about \(x\)-axis. — Figure 133. Plot of bounded region rotated about $x$ -axis.

\begin{array}{cccccc} x_{i} & Δ x & r_{i} & R_{i} & l & Estimated Surface Area \\ x_{1} = 0 & 1 & 0 & 1 & \sqrt{1^{2} + 1^{2}} \\ x_{2} = 1 & 1 & 1 & \sqrt{2} & \sqrt{1^{2} + (\sqrt{2} - 1)^{2}} \\ x_{3} = 2 & 1 & \sqrt{2} & \sqrt{3} \\ x_{4} = 3 & 1 & 3 & 2 \end{array}

(d)

Suppose we wanted to estimate the surface area with

n

frustums.

Bounded region rotated about \(x\)-axis. — Figure 134. Plot of bounded region rotated about $x$ -axis.

Let

f (x) = \sqrt{x} .

Which of the following expressions represents the surface area generated bo rotating the line segment from

(x_{0}, f (x_{0}))

to

(Δ x, f (x_{0} + Δ x))

about the

x

-axis?

🔗
$π {(\frac{f (x_{0}) + f (x_{0} + Δ x)}{2})}^{2} \sqrt{(Δ x)^{2} + (f (x_{0} + Δ x) - f (x_{0}))^{2}} .$
🔗
$2 π \frac{f (x_{0}) + f (x_{0} + Δ x)}{2} \sqrt{(Δ x)^{2} + (f (x_{0} + Δ x) - f (x_{0}))^{2}} .$
🔗
$2 π \frac{f (x_{0}) + f (x_{0} + Δ x)}{2} Δ x .$

(e)

Which of the following Riemann sums best estimates the surface area of the solid generated by rotating

y = \sqrt{x}

over

[0, 4]

about the

x

-axis ? Let

f (x) = \sqrt{x} .

🔗
$\sum π {(\frac{f (x_{i}) + f (x_{i} + Δ x)}{2})}^{2} \sqrt{(Δ x)^{2} + (f (x_{i} + Δ x) - f (x_{i}))^{2}} .$
🔗
$\sum 2 π \frac{f (x_{i}) + f (x_{i} + Δ x)}{2} \sqrt{(Δ x)^{2} + (f (x_{i} + Δ x) - f (x_{i}))^{2}} .$
🔗
$\sum 2 π \frac{f (x_{i}) + f (x_{0} + Δ x)}{2} Δ x .$

Fact 6.4.2.

Recall from Fact 6.2.1 that

\begin{aligned} lim_{Δ x \to 0} \sqrt{(Δ x)^{2} + (f (x_{i} + Δ x) - f (x_{i}))^{2}} & = lim_{Δ x \to 0} \sqrt{(Δ x)^{2} (1 + {(\frac{f (x_{i} + Δ x) - f (x_{i})}{Δ x})}^{2})} \\ = lim_{Δ x \to 0} \sqrt{1 + {(\frac{f (x_{i} + Δ x) - f (x_{i})}{Δ x})}^{2}} Δ x \\ = \sqrt{1 + (f^{'} (x))^{2}} d x, \end{aligned}

and that

lim_{Δ x \to 0} \frac{f (x_{i}) + f (x_{i} + Δ x)}{2} = f (x_{i}) .

Thus given a function

f (x) \geq 0

over

[a, b],

the surface area of the solid generated by rotating this function about the

x

-axis is

S A = \int_{a}^{b} 2 π f (x) \sqrt{1 + (f^{'} (x))^{2}} d x .

Activity 6.4.2.

Consider again the solid generated by rotating

y = \sqrt{x}

over

[0, 4]

about the

x

-axis.

(a)

Find an integral which computes the surface area of this solid.

(b)

If we instead rotate

y = \sqrt{x}

over

[0, 4]

about the

y

-axis, what is an integral which computes the surface area for this solid?

Activity 6.4.3.

Consider again the function

f (x) = \ln (x) + 1

over

[1, 5] .

(a)

Find an integral which computes the surface area of the solid generated by rotating the above curve about the

x

-axis.

(b)

Find an integral which computes the surface area of the solid generated by rotating the above curve about the

y

-axis.

Subsection 6.4.2 Videos

Figure 135. Video: Compute surface areas of surfaces of revolution

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