The area between the curves $y=f(x)$ and $y=g(x)$ and between $x=a$ and $x=b$ is $$A=\int_a^b |f(x) - g(x)| dx$$
For example, $f(x) = x^2, g(x) = 1, a = -1, b = 1$
we can simply get the integral function $\int_{-1}^{1} 1 - x^2 \\ dx$
Some regions are best treated by regarding x as a function of y. If a region is bounded by curves with equations $x=f(y)$, $x=g(y)$, $y=c$, and $y=c$, where f and g are continuous and $f(y) \ge g(y)$ for $c \le y \le d$, then its area is $$A=\int_c^d |f(y) - g(y)| dy$$
For example, $f(x)=\sqrt{x}, g(x) = \sqrt{2x-1}, a = 0, b = 1$
If treat them as functions of x, we need to split the area to two parts.
So we rewrite the functions to $f(y) = y^2, g(y) = \frac{y^2 + 1}{2}, c = 0, d = 1$
The region $\mathscr{R}$ enclosed by the curves $f(x) = x, g(x) = x^2$ is rotated about the x-axis. Find the volume of the resulting solid.
The curves $y = x \text{ and } y = x^2$ intersect at the points (0, 0) and (1, 1). The region between them, the solid of rotation, and a cross-section perpendicular to the x-axis are shown above. A cross-section in the plane $P_x$ has the shape of a washer (an annular ring) with inner radius $x^2$ and outer radius $x$, so we find the cross-sectional area by subtracting the area of the inner circle from the area of the outer circle:
To face a situation like below. If we slice perpendicular to the y-axis, we get a washer. But to compute the inner radius and the outer radius of the washer, we’d have to solve the cubic equation $f(x)$ for x in terms of y; that’s not easy.
For this situation, we use the method of cylindrical shells.
Instead of slicing the perpendicular to the x-axis:
Then, flatten as below:
Radius $x$, circumference $2 \pi x$, height $f(x)$, and thickness $\Delta x$ or $dx$:
For example: Find the volume of the solid obtained by rotating the region bounded by $y=x-x^2$ and $y=0$ about the line $x = 2$.
The figure below shows the region and a cylindrical shell formed by rotation about the line $x = 2$. It has radius $2 - x$, circumference $2\pi (2-x)$ , and height $x - x^2$.
If the region more easily described by top and bottom boundary curves of the form $y = f(x)$, or by left and right boundaries $x = g(y)$, use Washers.
If we decide that one variable is easier to work with than the other, then this dictates which method to use.
Draw a sample rectangle in the region, corresponding to a cross-section of the solid. The thickness of the rectangle, either $\Delta x$ or $\Delta y$, corresponds to the integration variable. If you imagine the rectangle revolving, it becomes either a disk (washer) or a shell.
Formula: If $f’$ is continuous on [a, b], then the length of the curve $y = f(x), a \le x \le b$, is $$L = \int_a^b \sqrt{1+[f’(x)]^2}dx$$
Proves:
Suppose that a curve C is defined by the equation $y = f(x)$, where $f$ is continuous and $a \le x \le b$. We obtain a polygonal approximation to C by dividing the interval [a, b] into n subintervals with endpoints $x_0, x_1, \ldots, x_n$ and equal width $\Delta x$. If $y_i = f(x_i)$, then the point $P_i(x_i, y_i)$ lies on C and the polygon with vertices $P_0, P_1, \ldots, P_n$, is an approximation of C.
We define the length L of the curve C as thee limit of the lengths $|P_{i-1}P_i|$: $$L = \lim_{n \to \infty} \sum_{i=1}^n |P_{i-1}P_i|$$
