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PAGE 1

6.5 + 6.6 Length and Surface Area

How long is the length of \( f(x) \) from \( x = a \) to \( x = b \)?

Graph of y=f(x) on x-y axes showing a curve starting at x=a and ending at x=b.

Same idea as last time: find the length of one small piece then accumulate using integration.

A curve with a circular dashed line highlighting a small segment for closer inspection.
A right triangle with legs delta x and delta y, where the hypotenuse approximates a curved segment.

is approximately the length of the green piece

the length \( L \) of the curve (green)

\[ L \approx \sqrt{(\Delta x)^2 + (\Delta y)^2} \]

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\[ \sqrt{(\Delta x)^2 + (\Delta y)^2} = \sqrt{(\Delta x)^2 \left[ 1 + \frac{(\Delta y)^2}{(\Delta x)^2} \right]} = \sqrt{1 + \left( \frac{\Delta y}{\Delta x} \right)^2} (\Delta x) \]

now shrink the interval: \( \Delta x \to dx \)

\[ \frac{\Delta y}{\Delta x} \to \frac{dy}{dx} \]

so, the exact length of the small piece is \( \sqrt{1 + \left( \frac{dy}{dx} \right)^2} dx \)

now accumulate ALL from \( x = a \) to \( x = b \)

so, the exact length of \( y = f(x) \) from \( x = a \) to \( x = b \) is

\[ \int_{a}^{b} \sqrt{1 + \left( \frac{dy}{dx} \right)^2} dx \]

or

\[ \int_{a}^{b} \sqrt{1 + (y')^2} dx \]

\( y = f(x) \) must be continuous and differentiable on \( a \le x \le b \)

\( f'(x) \) must exist on \( a \le x \le b \)

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Arc Length Example

Example: y = \(\frac{2}{3}(x^2 - 1)^{3/2}\) from \(x = 1\) to \(x = \sqrt{10}\)

The arc length formula is given by:

\[L = \int_{a}^{b} \sqrt{1 + (y')^2} \, dx\]
Coordinate graph showing a curve starting at (1,0) and increasing to x equals square root of 10.

Step 1: Find the Derivative

First, we compute the derivative \(y'\):

\[y' = \frac{2}{3} \cdot \frac{3}{2} (x^2 - 1)^{1/2} \cdot 2x = 2x(x^2 - 1)^{1/2}\]

Step 2: Set up and Evaluate the Integral

Substitute \(y'\) into the arc length formula:

\[\int_{1}^{\sqrt{10}} \sqrt{1 + [2x(x^2 - 1)^{1/2}]^2} \, dx = \int_{1}^{\sqrt{10}} \sqrt{1 + 4x^2(x^2 - 1)} \, dx\]

Simplify the expression inside the square root:

\[= \int_{1}^{\sqrt{10}} \sqrt{4x^4 - 4x^2 + 1} \, dx\]

Note: \(4x^4 - 4x^2 + 1 = (2x^2 - 1)^2\)

\[= \int_{1}^{\sqrt{10}} (2x^2 - 1) \, dx = \dots = \left[ \frac{17}{3}\sqrt{10} + \frac{1}{3} \right]\]
Result: \(\frac{17}{3}\sqrt{10} + \frac{1}{3}\)
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Arc Length with Respect to y

Back to \(\sqrt{(\Delta x)^2 + (\Delta y)^2}\)

Now factor out \((\Delta y)^2\):

\[\sqrt{(\Delta y)^2 \left( \frac{(\Delta x)^2}{(\Delta y)^2} + 1 \right)} = \sqrt{1 + \left( \frac{\Delta x}{\Delta y} \right)^2} (\Delta y)\]

Shrink interval: \(\Delta y \to dy\), \(\frac{\Delta x}{\Delta y} \to \frac{dx}{dy}\)

The equivalent formula for length of \(x = g(y)\) from \(y = c\) to \(y = d\) is:

\[\int_{c}^{d} \sqrt{1 + \left( \frac{dx}{dy} \right)^2} \, dy\]
Graph of x equals g of y showing a curve segment between y equals c and y equals d on the y-axis.
PAGE 5

Switching Variables for Arc Length

Sometimes the switch in variable is for convenience.

Sometimes we have to switch, for example, when \( f'(x) \) does not exist somewhere on \( a \le x \le b \) in:

\[ \int_{a}^{b} \sqrt{1 + [f'(x)]^2} \, dx \]

Example

\( y = x^{2/3} \) from \( x = 0 \) to \( x = 1 \)

\[ y' = \frac{2}{3} x^{-1/3} = \frac{2}{3 x^{1/3}} \]
\[ \int_{0}^{1} \sqrt{1 + \left( \frac{2}{3 x^{1/3}} \right)^2} \, dx \]

DNE at \( x = 0 \)

Graph of y = x^(2/3) from x=0 to x=1, showing a curve starting at the origin with a vertical tangent.

Try switching to \( x = g(y) \)

\[ y = x^{2/3} \] \[ x = y^{3/2} \]
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Now use:

\[ \int_{c}^{d} \sqrt{1 + \left( \frac{dx}{dy} \right)^2} \, dy \]

\( c = 0, d = 1 \) for this example

(\( y \) of starting point is 0, \( y \) of ending point is 1)

\[ x = y^{3/2} \quad \frac{dx}{dy} = \frac{3}{2} y^{1/2} \]

which always exists on \( 0 \le y \le 1 \).

\[ \int_{0}^{1} \sqrt{1 + \left( \frac{3}{2} y^{1/2} \right)^2} \, dy = \int_{0}^{1} \sqrt{1 + \frac{9}{4} y} \, dy \]

\( u = 1 + \frac{9}{4} y \) and so on

\[ \dots = \frac{1}{27} (13\sqrt{13} - 8) \approx 1.44 \]
PAGE 7

6.6 Surface Area of Solid of Revolution

Consider a function \( y = f(x) \) on the interval \( [a, b] \). We want to find the surface area of the solid formed when this curve is revolved around the x-axis.

Graph of y=f(x) from x=a to x=b and the resulting 3D solid of revolution around the x-axis.

Volume:

Calculated using disk or shell methods.

Surface area = ?

To find the surface area, we approximate the surface with thin strips. A rectangle in the 2D plane corresponds to a strip on the 3D surface.

Idea:

  • Find the area of one strip.
  • Then integrate to accumulate all strips.
Diagram showing a small rectangular segment on the curve and its corresponding circular strip on the solid.

If we cut and then unwrap a single strip, it looks like a curved band. Let \( d_1 \) be the diameter from radius \( r_1 \) and \( d_2 \) be the diameter from radius \( r_2 \).

As we shrink the interval, the curved strip will be approximately a rectangle.

Detailed view of a single strip being unwrapped into a nearly rectangular shape with length L and radii r1, r2.
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When unwrapped, the strip is approximately a rectangle with:

  • Width = \( 2\pi \cdot f(x) \) (circumference)
  • Length \( L \) = arc length of the segment
A thin rectangular strip representing the unwrapped surface element with width 2*pi*f(x).

From the previous section, the differential arc length is:

\[ L = \sqrt{1 + [f'(x)]^2} \, dx \]

Each strip has an area of:

\[ 2\pi \cdot f(x) \cdot \sqrt{1 + [f'(x)]^2} \, dx \]

Accumulating from \( x = a \) to \( x = b \), we get the formula for surface area:

\[ \int_{a}^{b} 2\pi f(x) \sqrt{1 + [f'(x)]^2} \, dx \]
around x-axis

Equivalent Form

For revolution around the y-axis (integrating with respect to y):

\[ \int_{c}^{d} 2\pi g(y) \sqrt{1 + [g'(y)]^2} \, dy \]
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Example: Surface Area of Revolution

Region bounded by \( y = \sqrt{x} \), \( x = 1 \), \( x = 4 \), \( y = 0 \) revolved around the x-axis.

Graph of the function y equals square root of x from x equals 1 to x equals 4 on the x-y plane.
A 3D sketch of the surface of revolution created by rotating the curve around the x-axis.

Surface area formula:

\[ \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f')^2} \, dx \]

Given \( f = \sqrt{x} = x^{1/2} \), then:

\[ f' = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}} \]

Calculation

\[ \int_{1}^{4} 2\pi \sqrt{x} \sqrt{1 + \left( \frac{1}{2\sqrt{x}} \right)^2} \, dx \]
\[ = 2\pi \int_{1}^{4} \sqrt{x} \sqrt{1 + \frac{1}{4x}} \, dx \]
\[ = 2\pi \int_{1}^{4} \sqrt{(x) \left( 1 + \frac{1}{4x} \right)} \, dx \]
\[ = 2\pi \int_{1}^{4} \sqrt{x + \frac{1}{4}} \, dx \]

Using substitution \( u = x + \frac{1}{4} \) and so on:

\[ = \dots = \frac{\pi}{6} (17\sqrt{17} - 5\sqrt{5}) \]