# Calculus 2

## Applications of integrals

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.

### Area between two curves

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

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

### Volume

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.

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

### Polar forms and area

#### The Polar coordinate system

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:

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

#### Polar graphs

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 $\theta$.

#### Polar area

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:

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

#### Intersection of polar-form curves

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

### Arc length and surface area

#### Arc length

Arc length formula in integral form:

for $y =f(x)$

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

for $x =g(y)$

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

#### Surface area

for $y =f(x)$

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

### Physical applications

#### Work

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.

1. The mass of the lamina is
$m = \rho \int_{a}^{b} f(x) dx$
1. The moment $M_{x}$ with respect to the x-axis is
$M_{x} = \int_{a}^{b} (\frac{f(x) + g(x)}{2})(f(x) - g(x))dx$
1. The moment $M_{y}$ with respect to the y-axis is
$M_{y} = \int_{a}^{b} x(f(x) - g(x))dx$
1. 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

1. Calculate the points of intersection to set the limits of integration

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

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

1. 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 $[a, b]$.

1. 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 $[a, b]$.

1. Solve for the coordinates of the center of mass.

## Integration techniques

### Integration by parts

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

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

### Trigonometric methods

#### Powers of Sine and Cosine

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

For even powers, square the trig identity.

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

### Method of partial fractions

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.