Applications of Integration: Areas Between Curves and Volumes of Revolution

Riemann Sums and Definite Integrals

• Revising Riemann sums and the definition of the definite integral.
• Assume a continuous function $f(x)$ in the domain $[a, b]$.
• Take an equally spaced partition of $[a, b]$. For some integer $n$, $\Delta x = \frac{b - a}{n}$.
• $x<em>0 = a$, $x</em>1 = a + \Delta x$, $x<em>2 = a + 2\Delta x$, …, $x</em>i = a + i\Delta x$.

Example: Partitioning the Interval [0, 8]

• Interval: $[0, 8]$, $n = 4$
• $\Delta x = \frac{8 - 0}{4} = 2$
• $x<em>0 = 0$, $x</em>1 = 0 + 2 = 2$, $x<em>2 = 0 + 2(2) = 4$, $x</em>3 = 6$, $x_4 = 8$
• Breaks the interval $[0, 8]$ into sub-intervals of length 2.

Definite Integral as Area Under a Curve

• Area under the curve $y = f(x)$ between $a$ and $b$.
• Break the interval $[a, b]$ into $n$ sub-intervals of equal length $\Delta x$.
• Approximate the area using rectangles. The height of the $i$-th rectangle is $f(x_i)$.
• Area of one rectangle: $f(x_i) \Delta x$.
• Riemann sum: $\sum<em>{i=0}^{n-1} f(x</em>i) \Delta x$. This is an approximation of the area under the curve.
• The actual area $A$ is the limit of the Riemann sum as $n$ approaches infinity:
  $A = \lim<em>{n \to \infty} \sum</em>{i=0}^{n-1} f(x<em>i) \Delta x = \int</em>{a}^{b} f(x) dx$
• The definite integral represents the area under the curve $y = f(x)$ between the lines $x = a$ and $x = b$.

Area Between Two Curves

• Two curves $f(x)$ and $g(x)$ defined on the interval $[a, b]$, where $f(x) \geq g(x)$ for all $x$ in $[a, b]$ (i.e., $f$ is above $g$).
• Break the interval $[a, b]$ into sub-intervals of equal length $\Delta x$.
• The height of a representative rectangle between the curves is $f(x<em>i) - g(x</em>i)$.
• The area of the rectangle is $(f(x<em>i) - g(x</em>i)) \Delta x$.
• The approximate area between the curves using $n$ rectangles is given by the Riemann sum: $\sum<em>{i=0}^{n-1} (f(x</em>i) - g(x_i)) \Delta x$
• The definite integral representing the area between the curves is the limit of the above Riemann sum as $n$ approaches infinity: $A = \lim<em>{n \to \infty} \sum</em>{i=0}^{n-1} (f(x<em>i) - g(x</em>i)) \Delta x = \int_{a}^{b} (f(x) - g(x)) dx$

Special Cases

• If $g(x) = 0$, then $A = \int_{a}^{b} f(x) dx$, which is the area under the curve $y = f(x)$ and above the x-axis.
• If $f(x) = 0$, then $A = \int<em>{a}^{b} -g(x) dx = -\int</em>{a}^{b} g(x) dx$. If $g(x)$ is below the x-axis, then $-g(x)$ makes the area positive.

Example: Area with Negative Function

• Consider $y = \sin(x)$ and the interval $[\pi, 2\pi]$.
• Since $\sin(x)$ is negative in this interval, the area under the curve is considered negative.
• To find the positive area, we can consider $y = -\sin(x)$, which reflects the function about the x-axis, making it positive.
• The integral $\int_{\pi}^{2\pi} -\sin(x) dx$ gives the positive area under the curve $y = -\sin(x)$ which corresponds to the absolute value of the signed area from the original function.

Example: Enclosed Area Between Two Curves

• Find the area enclosed between $f(x) = 4x - x^2$ and $g(x) = 3x^2$.
• First, find the intersection points by setting $f(x) = g(x)$. $4x - x^2 = 3x^2 \implies 4x^2 - 4x = 0 \implies 4x(x - 1) = 0$. The functions intersect when $x = 0$ and $x = 1$.
• Determine which function is greater on the interval $[0, 1]$. At $x = \frac{1}{2}$, $f(\frac{1}{2}) = 4(\frac{1}{2}) - (\frac{1}{2})^2 = 2 - \frac{1}{4} = \frac{7}{4}$ and $g(\frac{1}{2}) = 3(\frac{1}{2})^2 = \frac{3}{4}$. Since f(\frac{1}{2}) > g(\frac{1}{2}), we have $f(x) \geq g(x)$ on $[0, 1]$.
• The area is given by the integral: $A = \int<em>{0}^{1} (f(x) - g(x)) dx = \int</em>{0}^{1} (4x - x^2 - 3x^2) dx = \int_{0}^{1} (4x - 4x^2) dx$
• Evaluating the integral: $A = [2x^2 - \frac{4}{3}x^3]_{0}^{1} = 2(1)^2 - \frac{4}{3}(1)^3 - (0) = 2 - \frac{4}{3} = \frac{2}{3}$.

Example: Area of a Circle

• Find the area of a circle with radius $r$ centered at the origin using integration.
• Consider the functions $f(x) = \sqrt{r^2 - x^2}$ and $g(x) = -\sqrt{r^2 - x^2}$, representing the upper and lower halves of the circle, respectively.
• We know the radius of the circle is $r$, the intersection happens at $x = -r$ and $x = r$.
• The area is given by the integral: $A = \int<em>{-r}^{r} (f(x) - g(x)) dx = \int</em>{-r}^{r} (\sqrt{r^2 - x^2} - (-\sqrt{r^2 - x^2})) dx = 2\int_{-r}^{r} \sqrt{r^2 - x^2} dx$
• The integral can be solved using the substitution $x = r\cos(u)$. (See Lecture 27 for details).
• The result of the integral is: $A = \pi r^2$.

Example: Absolute Value and Piecewise Functions

• Functions: $f(x) = \sqrt{x + 2}$ and $g(x) = x$ on the domain $[0, 4]$.
• Define a new function $h(x) = |f(x) - g(x)| = |\sqrt{x + 2} - x|$.
• Determine when $f(x) \geq g(x)$ and when f(x) < g(x).
• Find the intersection point: $\sqrt{x + 2} = x \implies x + 2 = x^2 \implies x^2 - x - 2 = 0 \implies (x - 2)(x + 1) = 0$
• The intersection is at $x = 2$ (since we are considering $x \geq 0$).
• Between $0$ and $2$, $\sqrt{x + 2} \geq x$, and between $2$ and $4$, x > \sqrt{x + 2}.
• Thus, the piecewise function $h(x)$ is defined as:
  $h(x) = \begin{cases} \sqrt{x + 2} - x, &amp; \text{if } 0 \leq x \leq 2 \newline x - \sqrt{x + 2}, &amp; \text{if } 2 &lt; x \leq 4 \end{cases}$
• The area between the curves is given by breaking up the integral into two parts:
  $A = \int<em>{0}^{4} |\sqrt{x + 2} - x| dx = \int</em>{0}^{2} (\sqrt{x + 2} - x) dx + \int_{2}^{4} (x - \sqrt{x + 2}) dx$

Volumes of Revolution by the Disc Method

• Consider a curve $y = f(x)$ defined on the interval $[a, b]$ and rotate the curve about the x-axis.
• This rotation generates a solid of revolution, and we want to find its volume.

Method

• Divide the interval $[a, b]$ into $n$ sub-intervals of equal length $\Delta x$.
• Consider a representative rectangle with height $f(x_i)$ and width $\Delta x$.
• When this rectangle is rotated about the x-axis, it forms a disc (a cylinder with a small height) with radius $f(x_i)$ and thickness $\Delta x$.
• The volume of this disk is given by: $\pi [f(x_i)]^2 \Delta x = \pi y^2 \Delta x$
• Approximate the total volume by summing the volumes of all the discs: $V \approx \sum<em>{i=0}^{n-1} \pi [f(x</em>i)]^2 \Delta x$
• Take the limit as $n$ approaches infinity to get the exact volume:
  $V = \lim<em>{n \to \infty} \sum</em>{i=0}^{n-1} \pi [f(x<em>i)]^2 \Delta x = \int</em>{a}^{b} \pi [f(x)]^2 dx = \int_{a}^{b} \pi y^2 dx$

Example: Rotating $y = x^2$ about the x-axis

• Rotate $y = x^2$ over the interval $[0, 1]$ about the x-axis.
• The volume is given by: $V = \int<em>{0}^{1} \pi (x^2)^2 dx = \pi \int</em>{0}^{1} x^4 dx$
• Evaluating the integral: $V = \pi [\frac{1}{5}x^5]_{0}^{1} = \pi (\frac{1}{5}(1)^5 - 0) = \frac{\pi}{5}$.

Example: Rotating $y = \sqrt{x}$ about the x-axis

• Rotate $y = \sqrt{x}$ over the interval $[0, 1]$ about the x-axis.
• The volume is given by: $V = \int<em>{0}^{1} \pi (\sqrt{x})^2 dx = \pi \int</em>{0}^{1} x dx$
• Evaluating the integral: $V = \pi [\frac{1}{2}x^2]_{0}^{1} = \pi (\frac{1}{2}(1)^2 - 0) = \frac{\pi}{2}$.

Example: Cone

• Given the line $y = (1/2)x$ between 0 and 2, calculate the volume that is obtained after rotating the line about the x axis
• $V = \int<em>{0}^{2} \pi (\frac{x}{2})^2 dx = \pi \int</em>{0}^{2} \frac{x^2}{4} dx = \frac{\pi}{4} \int_{0}^{2} x^2 dx$
• $V = \frac{\pi}{4} [\frac{x^3}{3}]_{0}^{2} = \frac{\pi}{4} (\frac{8}{3} - 0) = \frac{2\pi}{3}$.

Deriving the Volume Formula for a Cone

• Consider the line $y = \frac{r}{h}x$ over the interval $[0, h]$. When rotated about the x-axis, it forms a cone.
• At $x = h$, $y = \frac{r}{h}(h) = r$, so the cone has height $h$ and radius $r$.
• The volume is given by: $V = \int<em>{0}^{h} \pi (\frac{r}{h}x)^2 dx = \pi \frac{r^2}{h^2} \int</em>{0}^{h} x^2 dx$
• Evaluating the integral: $V = \pi \frac{r^2}{h^2} [\frac{x^3}{3}]_{0}^{h} = \pi \frac{r^2}{h^2} (\frac{h^3}{3} - 0) = \frac{1}{3}\pi r^2 h$.

Rotation around the y-axis

• Consider the volume obtained by rotating the formula about the y-axis.
• Rotate $y = x^2$ about the y-axis over the interval $[0, 1]$. Now use $x = \sqrt{y}$
• $V = \int<em>{0}^{1} \pi (\sqrt{y})^2 dy = \pi \int</em>{0}^{1} y dy = \pi [\frac{y^2}{2}]_{0}^{1} = \frac{\pi}{2}$.

Volume Between Curves by Rotation Around the y-axis

• $x= \sqrt{y}$ from $y=x^2$ and the line $x = 1$ rotate the defined region around the y-axis to define the volume.
• $V = \int<em>{0}^{1} \pi * 1^2 dy - \int</em>{0}^{1} \pi (\sqrt{y})^2 dy$
• This volume takes the volume from a filled circular cylinder and cuts out the portion that we don't want which is defined by $x= \sqrt{y}$