Consider various applications of integration such as computing volume, arc length, surface area, work, hydrostatic force, centroids of planar regions, and applications to business, economics, and life sciences.

Let $f(x)$ and $g(x)$ be continuous functions such that $f(x) \geq g(x)$ over an interval $[a, b]$. Let $\text{R}$ denote the region bounded above by $f(x)$, below by $g(x)$, on the left by $x = a$, and on the right by $x = b$. The area of $\text{R}$ is given by

$A = \int_{a}^{b} [f(x) - g(x)]dx$

**Area of compound regions**

If we want to look at regions bounded by the graphs of functions that cross one another, we modify the formula by using the absolute value function.

Let $f(x)$ and $g(x)$ be continuous functions over an interval $[a, b]$. Let $\text{R}$ denote the region between the graphs of $f(x)$ and $g(x)$, and bounded on the left by $x = a$, and on the right by $x = b$. The area of $\text{R}$ is given by

$A = \int_{a}^{c} |f(x) - g(x)|dx$

In practice, you must find the points of intersection of the curves and divide the integrals according to the *leading* and *trailing* curve.

$A = \int_{a}^{b} [f(x) - g(x)]dx + \int_{b}^{c} [f(x) - g(x)]dx$

**Area of regions, integrating along the y-axis**

Let $u(y)$ and $v(y)$ be continuous functions such that $u(y) \geq v(y)$ over an interval $[c, d]$ along the y-axis. A horizontal strip has a width of $u(y) - v(y)$. Thus, the integration formula for area is

$A = \int_{c}^{d} [u(y) - v(y)]dy$

Area methods can be modified to compute the volume of three-dimensional solids.

The **volume of a solid with a known cross-sectional area** perpendicular to the x-axis.

$V = \int_{a}^{b} A(x) dx$

The **disk method** is used to find a volume generated when a region is revolved about an axis perpendicular to an approximating strip.

$V = \int_{a}^{b} \pi [f(x)]^2 dx$

The **washer method** is used to find a volume generated when a region between two curves is revolved about an axis perpendicular to an approximating strip.

$V = \int_{a}^{b} \pi ([f(x)]^2 - [g(x)]^2) dx$

The **cylindrical shells method** is used to find a volume generated when a region is revolved about an axis parallel to an approximating strip.

$V = \int_{a}^{b} 2 \pi x \, f(x) dx$

In the **polar coordinate system**, points are plotted in relation to a fixed point $O$, called the origin or pole and a fixed ray emanating from the origin called the **polar axis**. We associate with each point $P$ in the plane an ordered pair of numbers $P(r, \theta)$, where $r$ (**radial coordinate**) is the distance from $O$ to $P$ and $\theta$ (**polar angle**) is the angle measured from the polar axis.

**Changing coordinates**

To change from polar to Cartesian form use the formulas:

- $x = r\cos(\theta)$
- $y = r\sin(\theta)$

To change from Cartesian to polar form use the formulas:

- $r = \sqrt{x^2 + y^2}$
- $\theta = \arctan(\frac{y}{x})$

The graph of an equation in polar coordinates is the set of all points $P$ whose polar coordinates satisfy the given equation. Solve by constructing a table of values for $r$ and solving for $\theta$.

Similar to Riemann sums in rectangular form, instead of using rectangular areas as the basic units being summed, we sum areas of circular sectors.

The area of a circular sector if radius $r$ is given by: $A = \frac{1}{2}r^2\theta$

Using this formula, a formula for finding the area enclosed by a polar curve is:

$A = \int_{a}^{b} \frac{1}{2}[f(\theta)]^2 d\theta$

- Find all simultaneous solutions of the given system of equations.
- Determine whether the pole $r=0$ like on the two graphs.
- Graph the curves to look for other points of intersection.

Arc length formula in integral form:

for $y =f(x)$

$s = \int_{a}^{b} \sqrt{1 + (\frac{dy}{dx})^2} dx$

for $x =g(y)$

$s = \int_{c}^{d} \sqrt{1 + (\frac{dg}{dy})^2} dy$

for $y =f(x)$

$s = \int_{a}^{b} 2\pi y \sqrt{1 + (\frac{dy}{dx})^2} dx$

The **work done by a constant force** if a body moves a distance $d$ in the direction of an applied constant force $F$, the work $W$ done is:

$W=Fd$

The **work done by a variable force** $F(x)$ in moving an object along the x-axis from $x=a$ to $x=b$ is given by:

$W=\int_{a}^{b} F(x) dx$

**Hooke's law** states that the force $F$ on a *spring* is proportional to the displacement $x$:

$F(x) = kx$

where $k$ is the **spring constant**.

To model work using Hooke's formula, (1) solve for $k$ then (2) use the *work done by a variable force* formula to solve for Hooke's formula using the displacement as the integral's lower and upper limits.

$F(x) = \int_{a}^{b} kx \: dx$

If you are given work done $W$ instead of force $F$, integration is required first to solve for $k$

$\int_{a}^{b} kx \: dx = W$

where $a$ and $b$ are the given displacement, then plug $k$ into the *work done by a variable force* formula as above.

**Work done in emptying a tank**

$\text{Force} = \text{Density} \times \text{Volume}$

*First*, set up a coordinate system; decide where to place $x = 0$.

To calculate work done in emptying a tank, use the *work done by a variable force* equation where the lower limit $a$ is the minimum x-value of the liquid and the upper limit $b$ is the maximum x-value of the liquid. Remember that $F(x) = dF$ where $d$ is the distance between $x$ and its destination $(\text{larger} - \text{smaller})$ and force $F$ is the volume of $x$, a "slice" of liquid $v_{x}$ times density $\rho$ i.e. $(\pi r^2 dx)(\rho)$.

$\int_{a}^{b} d(v_{x} \rho)$

Note that the volume of a "slice" depends on the shape of the tank but *always* uses $dx$ as a dimension. For example a "slice" in a cylindrical tank will have a volume of $\pi r^2 dx$.

**Hydrostatic force problems**

$\text{Force} = \text{Density} \times \text{Depth} \times \text{Area}$

First, set up a coordinate system i.e. let $x = 0$ be the waterline. Then let $\text{Depth}$ equal the distance between $x$ and its destination $(\text{larger} - \text{smaller})$. Remember that $\text{Area}$ is the area of $x$ (a "slice") and *not* the total surface area. Integrate from $x_{min}$ to $x_{max}$ of the fluid.

**Moments and center of mass**

$\bar{x}$ is the **center of mass**: the point where the fulcrum should be placed to make the system balance.

**Moments** measure the tendency of a body to rotate about an axis.

Let $\rho$ denote density of the lamina.

- The mass of the lamina is

$m = \rho \int_{a}^{b} f(x) dx$

- The moment $M_{x}$ with respect to the x-axis is

$M_{x} = \int_{a}^{b} (\frac{f(x) + g(x)}{2})(f(x) - g(x))dx$

- The moment $M_{y}$ with respect to the y-axis is

$M_{y} = \int_{a}^{b} x(f(x) - g(x))dx$

- The coordinates of the center of mass $(\bar{x}, \bar{y})$ of the system are

$\bar{x} = \frac{M_{y}}{m}$

$\bar{y} = \frac{M_{x}}{m}$

where $M_{x}$ is moment about the x-axis and $M_{y}$ is moment about the y-axis.

**Finding the centroid of a region bounded by two functions**

- Calculate the points of intersection to set the limits of integration

Solve for $f(x) = g(x)$

Then set $\int_{a}^{b}$

- Calculate the total mass

Assuming we are calculating for a lamina and $\rho = 1$ (See *area between two curves*)

$\int_{a}^{b} [f(x) - g(x)]dx$

where $f(x) \geq g(x)$ on $[a, b]$.

- Compute the moments

Distance to y-axis is $x$; distance to x-axis is the average of $f(x)$ and $g(x)$.

where $f(x) \geq g(x)$ on $[a, b]$.

- Solve for the coordinates of the center of mass.

**Integration by parts** is the inverse of the product rule for differentiation: $d(uv) = u\,dv + v\, du$.

Integrate both sides to find the formula for integration by parts: $\int uv = \int u\,dv + \int v\,du$.

Rewritten as

$\int u\,dv = uv - \int v\,du$

$\int \sin^m x \cos^n x dx$

For even powers, square the trig identity.

For odd powers, peel off one trig function to become $du$.

This process can be thought of as the reverse of adding fractional algebraic expressions, and it allows us to break up rational expressions into simpler terms.