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}.
x0 = a, x1 = a + \Delta x, x2 = a + 2\Delta x, …, xi = a + i\Delta x.

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{i=0}^{n-1} f(xi) \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{n \to \infty} \sum{i=0}^{n-1} f(xi) \Delta x = \int{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.

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(xi) - g(xi).
The area of the rectangle is (f(xi) - g(xi)) \Delta x.
The approximate area between the curves using n rectangles is given by the Riemann sum: \sum{i=0}^{n-1} (f(xi) - 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{n \to \infty} \sum{i=0}^{n-1} (f(xi) - g(xi)) \Delta x = \int_{a}^{b} (f(x) - g(x)) dx

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{a}^{b} -g(x) dx = -\int{a}^{b} g(x) dx. If g(x) is below the x-axis, then -g(x) makes the area positive.

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.

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{0}^{1} (f(x) - g(x)) dx = \int{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}.

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{-r}^{r} (f(x) - g(x)) dx = \int{-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.

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, & \text{if } 0 \leq x \leq 2 \newline x - \sqrt{x + 2}, & \text{if } 2 < x \leq 4 \end{cases}
The area between the curves is given by breaking up the integral into two parts:
A = \int{0}^{4} |\sqrt{x + 2} - x| dx = \int{0}^{2} (\sqrt{x + 2} - x) dx + \int_{2}^{4} (x - \sqrt{x + 2}) dx

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.

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{i=0}^{n-1} \pi [f(xi)]^2 \Delta x
Take the limit as n approaches infinity to get the exact volume:
V = \lim{n \to \infty} \sum{i=0}^{n-1} \pi [f(xi)]^2 \Delta x = \int{a}^{b} \pi [f(x)]^2 dx = \int_{a}^{b} \pi y^2 dx

Rotate y = x^2 over the interval [0, 1] about the x-axis.
The volume is given by: V = \int{0}^{1} \pi (x^2)^2 dx = \pi \int{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}.

Rotate y = \sqrt{x} over the interval [0, 1] about the x-axis.
The volume is given by: V = \int{0}^{1} \pi (\sqrt{x})^2 dx = \pi \int{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}.

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{0}^{2} \pi (\frac{x}{2})^2 dx = \pi \int{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}.

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{0}^{h} \pi (\frac{r}{h}x)^2 dx = \pi \frac{r^2}{h^2} \int{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.

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{0}^{1} \pi (\sqrt{y})^2 dy = \pi \int{0}^{1} y dy = \pi [\frac{y^2}{2}]_{0}^{1} = \frac{\pi}{2}.

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{0}^{1} \pi * 1^2 dy - \int{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}