Finding the surface area of revolution of a curve involves using integration to sum the areas of infinitesimally thin bands generated as the curve spins around an axis. Essentially, you are calculating the integral of the product of the circumference traced by a point on the curve (2π times the radius) and an infinitesimal piece of the curve's arc length (ds).
Understanding the Core Concept
When a curve is revolved around an axis, it sweeps out a three-dimensional surface. To find the area of this surface, we can imagine dividing the curve into tiny segments. Each segment, when revolved, generates a small band or frustum (a cone with the top cut off). The surface area is the sum of the areas of all these bands. Using calculus, this sum becomes an integral.
The fundamental idea is:
Surface Area ≈ Σ (Circumference) * (Arc Length Segment)
This translates to an integral:
Surface Area = ∫ 2πr ds
Where:
- 2πr is the circumference of the circle traced by a point on the curve (r is the radius, which is the distance from the curve to the axis of revolution).
- ds is the differential of arc length, which represents an infinitesimally small piece of the curve.
Formulas for Surface Area of Revolution
The specific formula depends on whether you revolve the curve around the x-axis or the y-axis and whether the curve is defined as y = f(x) or x = g(y).
Revolving Around the X-axis
If the curve is given by y = f(x), from x = a to x = b, and you revolve it around the x-axis:
- The radius
ris the distance from the curve to the x-axis, which isy = f(x). - The arc length element
dsis√1 + (f′(x))² dx.
The surface area formula is:
$$ \text{Surface Area} = \int_a^b 2\pi f(x) \sqrt{1 + (f′(x))^2} \, dx $$
Revolving Around the Y-axis
If the curve is given by y = f(x), from x = a to x = b, and you revolve it around the y-axis:
- The radius
ris the distance from the curve to the y-axis, which isx. - The arc length element
dsis√1 + (f′(x))² dx.
The surface area formula is:
$$ \text{Surface Area} = \int_a^b 2\pi x \sqrt{1 + (f′(x))^2} \, dx $$
Alternatively, if the curve is given by x = g(y), from y = c to y = d, and you revolve it around the y-axis:
- The radius
ris the distance from the curve to the y-axis, which isx = g(y). - The arc length element
dsis√1 + (g′(y))² dy.
The surface area formula is:
$$ \text{Surface Area} = \int_c^d 2\pi g(y) \sqrt{1 + (g′(y))^2} \, dy $$
This latter formula is demonstrated in the provided reference.
Example Calculation (Revolving around the y-axis)
Let's look at the example from the reference to see how this is applied. The problem involves finding the surface area generated by revolving the curve defined by $x = g(y) = \frac{1}{3}y^3$ around the y-axis, for the range of y from 0 to 2.
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Identify the function and limits: The curve is $x = g(y) = \frac{1}{3}y^3$, and the limits are y = 0 to y = 2.
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Identify the axis of revolution: The y-axis.
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Determine the radius: Since revolving around the y-axis with x = g(y), the radius is $r = g(y) = \frac{1}{3}y^3$.
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Calculate the derivative g′(y):
$g′(y) = \frac{d}{dy}\left(\frac{1}{3}y^3\right) = \frac{1}{3} \cdot 3y^2 = y^2$. -
Calculate the square of the derivative (g′(y))²:
$(g′(y))² = (y^2)² = y^4$. -
Calculate the arc length element ds = √1 + (g′(y))² dy:
$ds = \sqrt{1 + y^4} \, dy$. -
Set up the integral using the formula ∫dc(2πg(y)√1+(g′(y))²)dy:
The integral is from y=0 to y=2:
$ \text{Surface Area} = \int_0^2 2\pi \left(\frac{1}{3}y^3\right) \sqrt{1 + y^4} \, dy $
$ \text{Surface Area} = 2\pi \int_0^2 \frac{1}{3}y^3 \sqrt{1 + y^4} \, dy $
$ \text{Surface Area} = \frac{2\pi}{3} \int_0^2 y^3 \sqrt{1 + y^4} \, dy $This matches the integral shown in the reference: $2\pi/3∫20(y3√1+y4)dy$.
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Evaluate the integral: The reference shows the steps for evaluating this integral using a u-substitution.
- Let $u = 1 + y^4$.
- Then $du = 4y^3 \, dy$, which means $y^3 \, dy = \frac{1}{4} du$.
- Change the limits of integration:
- When $y=0$, $u = 1 + 0^4 = 1$.
- When $y=2$, $u = 1 + 2^4 = 1 + 16 = 17$.
- Substitute into the integral:
$ \frac{2\pi}{3} \int_1^{17} \sqrt{u} \left(\frac{1}{4} du\right) $
$ = \frac{2\pi}{12} \int_1^{17} u^{1/2} \, du $
$ = \frac{\pi}{6} \int_1^{17} u^{1/2} \, du $ - Integrate $u^{1/2}$:
$ \int u^{1/2} \, du = \frac{u^{3/2}}{3/2} = \frac{2}{3} u^{3/2} $ - Evaluate the definite integral:
$ \frac{\pi}{6} \left[ \frac{2}{3} u^{3/2} \right]_1^{17} $
$ = \frac{\pi}{6} \left( \frac{2}{3} (17)^{3/2} - \frac{2}{3} (1)^{3/2} \right) $
$ = \frac{\pi}{6} \cdot \frac{2}{3} \left( (17)^{3/2} - 1 \right) $
$ = \frac{\pi}{9} \left( (17)^{3/2} - 1 \right) $
This exactly matches the result shown in the reference before the approximation: $\pi/9[(17)³ᐟ²−1]$.
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Calculate the approximate numerical value:
$ \frac{\pi}{9} \left( (17)^{3/2} - 1 \right) \approx \frac{3.14159}{9} (17 \cdot \sqrt{17} - 1) \approx \frac{3.14159}{9} (17 \cdot 4.123 - 1) \approx \frac{3.14159}{9} (70.091 - 1) \approx \frac{3.14159}{9} (69.091) \approx 24.118 \text{ units}^2 $.This matches the approximate value from the reference: ≈24.118units².
Summary of Formulas
Here is a quick reference table for the surface area of revolution formulas:
| Revolution Axis | Curve Definition | Formula |
|---|---|---|
| X-axis | y = f(x), a ≤ x ≤ b | $ \int_a^b 2\pi f(x) \sqrt{1 + (f′(x))^2} \, dx $ |
| X-axis | x = g(y), c ≤ y ≤ d | $ \int_c^d 2\pi y \sqrt{1 + (g′(y))^2} \, dy $ |
| Y-axis | y = f(x), a ≤ x ≤ b | $ \int_a^b 2\pi x \sqrt{1 + (f′(x))^2} \, dx $ |
| Y-axis | x = g(y), c ≤ y ≤ d | $ \int_c^d 2\pi g(y) \sqrt{1 + (g′(y))^2} \, dy $ |
The key steps are always to identify the axis of revolution, determine the appropriate function form (y=f(x) or x=g(y)), calculate the necessary derivative, and set up and evaluate the correct integral using the radius (distance from the curve to the axis) and the arc length element.
