周次: 3 日期: March 9, 2022 节次: 1
A curve sweeps out a surface in the space around is called a surface of revolution. The “area” of this surface depends on the length of the rope and the distance of each of its segments from the axis of revolution. In this section, we calculate the area of surface of revolution. Two theorems are given for volumes and areas.
Outline
Basic Principle for the Surface Area
Case 1. Cylindrical surface
The lateral (侧) surface area of cylindrical surface is defined as
$$ \text{Lateral Surface Area:}\A=2\pi y \Delta x $$
Case 2. Frustum of a cone
The lateral surface area of a frustum of a cone is defined as
$$ \text{Lateral Surface Area:}\A = 2\pi y^* \Delta s \= \frac12(2\pi y_1 + 2\pi y_2)\Delta s $$
**Note**
The frustum has the same surface area as that of a rectangle with side lengths $\Delta s$ and $2\pi y^*$.
Surface Area for Revolving the Graph of a function
Suppose $y = f(x)$ , $a\le x\le b$ is a non-negative continuous function. The surface sweeps out by revolving about the $x$-axis.
We partition the interval $[a,b]$ in the usual way. The partition subdivide the graph into short arcs.
A typical arc $PQ$ sweeps out a frustum of a cone. The surface area of the frustum is $2\pi y^* L$, where $y^*$ is the average height of the line segment joining $P$ and $Q$, and $L$ is the length of the arc.
$$ \text{Frustum surface area} \approx 2\pi \cdot \frac{f(x_{k-1}) + f(x_k)}{2}\cdot \sqrt{(\Delta x_k)^2 + (\Delta y_k)^2}\=\pi(f(x_{k-1}) + f(x_k))\sqrt{(\Delta x_k)^2 + (f(x_k)-f(x_{k-1}))^2} $$
If the function $f(x)$ is differentiable, then by the Mean Value Theorem, there is a point $c_k\in(x_{k-1},x_k)$ such that
$$ f(x_k) - f(x_{k-1}) = f'(c_k) \Delta x_k. $$
Then the area of surface of revolution is approximately
$$ \text{Surface area}\approx \sum_{k=1}^n \pi (f(x_{k-1}) + f(x_k)) \sqrt{1 + f'(c_k)^2}\Delta x_k\ = \sum_{k=1}^n 2\pi f(\xi_k) \sqrt{1+f'(c_k)^2}\Delta x_k. $$
This is not a Riemann sum. However, a theorem from advanced calculus assures as $||P||\to 0$, the sum converges to the integral
$$ \int_a^b 2\pi f(x) \sqrt{1+f'(x)^2}dx. $$
Definition. Surface Area for Revolution About the $x$-Axis
If the function $f(x)\ge 0$ is continuously differentiable on $[a,b]$, the area of the surface generated by revolving the curve $y=f(x)$ about the $x$-axis is
$$ S = \int_a^b 2\pi f(x) \sqrt{1+f'(x)^2}dx = \int_a^b 2\pi y \sqrt{1+y'(x)^2}dx. $$
Example 1. Applying the Surface Area Formula
Find the area of the surface generated by revolving the curve $y=2\sqrt{x}$, $1\le x \le 2,$ about the $x$-axis.
**Solution ($\frac{8\pi}{3}(3\sqrt3 - 2\sqrt2 )$)**
Revolution About the $y$-Axis
Definition. Surface Area for Revolution About the $y$-Axis
If the function $x = g(y)\ge 0$ is continuously differentiable on $[c,d]$, the area of the surface generated by revolving the curve $x=g(y)$ about the $y$-axis is
$$ S = \int_c^d 2\pi g(y) \sqrt{1+g'(y)^2}dy = \int_c^d 2\pi x \sqrt{1+x'(y)^2}dy. $$
Example 2. Revolution about the $y$-Axis
The line segment $x=1-y$, $0\le y\le 1$ is revolved about the $y$-axis to generate the cone. Find its lateral surface area.
Solution ($\sqrt2 \pi$)
Parametrized Curves
For general parametrized curve $x=f(t)$, $y=g(t)$, we could derive the surface area formula from the above formulae. Recall that $\sqrt{1+y'(x)^2 }dx = ds = \sqrt{f'(t)^2 + g'(t)^2}dt$. We could get the following formula.
Definition. Surface Area of Revolution for Parametrized Curve
If a smooth curve $x=f(t)$, $y=g(t)$, $a\le t\le b$, is traversed exactly once as $t$ increases from $a$ to $b$, then the areas of the surfaces generated by revolving the curve about the coordinate axes are as follows.
- Revolution about the $x$-axis ($y\ge 0$)
$$ S = \int_a^b 2\pi y(t) \sqrt{x'(t)^2+y'(t)^2}dt. $$
- Revolution about the $y$-axis ($x\ge 0$)
$$ S = \int_a^b 2\pi x(t)\sqrt{x'(t)^2 + y'(t)^2}dt. $$
Example 3. Applying Surface Area Formula
The standard parametrization of the circle of radius $1$ centered at the point $(0,1)$ in the $xy$-plane is
$$ x=\cos(t),\quad y=1+\sin(t),\quad 0\le t\le 2\pi. $$
Use this parametrization to find the area of the surface swept out by revolving the circle about the $x$-axis.
Solution ($4\pi^2$)
The Differential Form
Note that the differential of arc-length has appeared in the formula. Regardless of axis of revolution, the surface area has the form \
$$ S = \int 2\pi \text{(radius)}\text{(band width)} =\int 2\pi \rho, ds $$
In any particular problem, express the radius function $\rho$ and the arc-length differential $ds$ in terms of a common variable and supply limits of integration for that variable.
Example 4. Using the Differential Form for Surface Areas
Find the area of the surface generate by revolving the curve $y=x^3$, $0\le x\le 1/2$, about the $x$-axis.
Solution($\frac{\pi}{27}(\frac{125}{64}-1)$)
Cylindrical Versus Conical Bands
Why not find the surface area by approximating with cylindrical bands instead of frustum? As we do so, we got a much simpler integral formula,
$$ S = \int_a^b 2\pi f(x)dx. $$
However, this formula is incorrect!
The Theorems of Pappus
In the third century, a Greek named Pappus discovered two formulas related with centroids to surfaces and solids of revolution. The formulas provide shortcuts to a number of otherwise length calculations.
Theorem 1. Pappus’s Theorem for Volumes
If a plane region is revolved once about a line in the plane that does not cut through the region’s interior, then the volume of the solid it generates is
$$ \text{Volume} = \text{(Region’s Area)}\times \text{(Distance traveled by the region’s centroid)}\ = 2\pi \rho , A. $$
Proof
Example 5. Volume of a Torus
The volume of the torus (doughnut) generated by revolving a circular disk of radius $a$ about an axis in its plane at a distance $b\ge a$ from its center is $V = 2\pi b (\pi a^2)$.
Solution
Example 6. Locate the Centroid of a Semicircular Region
Solution
Theorem 2. Pappus’s Theorem for Surface Areas
If an arc of a smooth plane curve is revolved once about a line in the plane that does not cut through the arc’s interior, then the area of the surface generated by the arc is
$$ \text{Area} = \text{(Length of the arc)}\times \text{(Distance traveled by the arc’s centroid)}\ = 2\pi \rho, L. $$
Proof
Example 7. Surface Area of a Torus
Solution