# Week 16 - Applications of Integration

## 1. Area between Curves

- 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 $f(y) = y^2, g(y) = y^2 + 1, c = 0, d = 1$

- If treat them as functions of

## 2. Volumes

### 2.1. Sphere's Volume

- The cross-sectional area is $A(x) = \pi y^2 = \pi(r^2 - x^2)$
- Using the definition of volume with $a = -r$ and $b = r$, we have $\begin{aligned} V &= \int_{-r}^{r}A(x) dx = \int_{-r}^{r} \pi(r^2 - x^2) dx \\ &= 2\pi \int_0^r(r^2 - x^2)\ dx \\ &= 2\pi \big[r^2x - \frac{x^3}{3}\big]_0^r = 2\pi(r^3-\frac{r^3}{3}) \\ &= \frac{4}{3}\pi r^3 \end{aligned}$

### 2.2. Use Washer Method

- 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: $A(x) = \pi x^2 - \pi (x^2)^2 = \pi(x^2 - x^4)$ - Therefore we have $V = \int_0^1 A(x)dx = \int_0^1 \pi(x^2 - x^4) \ dx = \pi \big[\frac{x^3}{3} - \frac{x^5}{5}\big]_0^1 = \frac{2\pi}{15}$

### 2.3. Use Shells Method

- 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 $dx$:

- 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 $x - x^2$.
- The volume of the given solid is: $\begin{aligned} V &= \int_0^1 2 \pi (2 - x) (x - x^2) dx \\ &= 2 \pi \int_0^1 (x^3-3x^2 + 2x) dx \\ &= 2 \pi \big[\frac{x^4}{4} - x^3 + x^2\big]_0^1 = \frac{\pi}{2} \end{aligned}$

### 2.4. Disks and Washers versus Cylindrical Shells

- 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.

## 3. Arc Length

**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 $|P_{i-1}P_i|$: $L = \lim_{n \to \infty} \sum_{i=1}^n |P_{i-1}P_i|$ - If we let $\Delta y_i = y_i - y_{i-1}$, then $\begin{aligned} |P_{i-1}P_i| &= \sqrt{(x_i-x_{i-1})^2 + (y_i-y_{i-1})^2} = \sqrt{(\Delta x_i)^2 + (\Delta y_i)^2} \\ &= \Delta x_i \sqrt{1 + (\frac{\Delta y_i}{\Delta x_i})^2} \\ &= \sqrt{1 + (f'(x))^2}dx \end{aligned}$
- So $L = \int_a^b\sqrt{1 + [f'(x)]^2}dx$

- Suppose that a curve