Calculus 2

Term resources

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

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

A=acf(x)g(x)dxA = \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=ab[f(x)g(x)]dx+bc[f(x)g(x)]dxA = \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)u(y) and v(y)v(y) be continuous functions such that u(y)v(y)u(y) \geq v(y) over an interval [c,d][c, d] along the y-axis. A horizontal strip has a width of u(y)v(y)u(y) - v(y). Thus, the integration formula for area is

A=cd[u(y)v(y)]dyA = \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=abA(x)dxV = \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=abπ[f(x)]2dxV = \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=abπ([f(x)]2[g(x)]2)dxV = \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=ab2πxf(x)dxV = \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 OO, called the origin or pole and a fixed ray emanating from the origin called the polar axis. We associate with each point PP in the plane an ordered pair of numbers P(r,θ)P(r, \theta), where rr (radial coordinate) is the distance from OO to PP and θ\theta (polar angle) is the angle measured from the polar axis.

Changing coordinates

To change from polar to Cartesian form use the formulas:

  • x=rcos(θ)x = r\cos(\theta)
  • y=rsin(θ)y = r\sin(\theta)

To change from Cartesian to polar form use the formulas:

  • r=x2+y2r = \sqrt{x^2 + y^2}
  • θ=arctan(yx)\theta = \arctan(\frac{y}{x})

Polar graphs

The graph of an equation in polar coordinates is the set of all points PP whose polar coordinates satisfy the given equation. Solve by constructing a table of values for rr and solving for θ\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 rr is given by: A=12r2θA = \frac{1}{2}r^2\theta

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

A=ab12[f(θ)]2dθA = \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=0r=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)y =f(x)

s=ab1+(dydx)2dxs = \int_{a}^{b} \sqrt{1 + (\frac{dy}{dx})^2} dx

for x=g(y)x =g(y)

s=cd1+(dgdy)2dys = \int_{c}^{d} \sqrt{1 + (\frac{dg}{dy})^2} dy

Surface area

for y=f(x)y =f(x)

s=ab2πy1+(dydx)2dxs = \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 dd in the direction of an applied constant force FF, the work WW done is:

W=FdW=Fd

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

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

Hooke's law states that the force FF on a spring is proportional to the displacement xx:

F(x)=kxF(x) = kx

where kk is the spring constant.

To model work using Hooke's formula, (1) solve for kk 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)=abkxdxF(x) = \int_{a}^{b} kx \: dx

If you are given work done WW instead of force FF, integration is required first to solve for kk

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

where aa and bb are the given displacement, then plug kk into the work done by a variable force formula as above.

Work done in emptying a tank

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

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

To calculate work done in emptying a tank, use the work done by a variable force equation where the lower limit aa is the minimum x-value of the liquid and the upper limit bb is the maximum x-value of the liquid. Remember that F(x)=dFF(x) = dF where dd is the distance between xx and its destination (largersmaller)(\text{larger} - \text{smaller}) and force FF is the volume of xx, a "slice" of liquid vxv_{x} times density ρ\rho i.e. (πr2dx)(ρ)(\pi r^2 dx)(\rho).

abd(vxρ)\int_{a}^{b} d(v_{x} \rho)

Note that the volume of a "slice" depends on the shape of the tank but always uses dxdx as a dimension. For example a "slice" in a cylindrical tank will have a volume of πr2dx\pi r^2 dx.

Hydrostatic force problems

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

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

Moments and center of mass

xˉ\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 ρ\rho denote density of the lamina.

  1. The mass of the lamina is
m=ρabf(x)dxm = \rho \int_{a}^{b} f(x) dx
  1. The moment MxM_{x} with respect to the x-axis is
Mx=ab(f(x)+g(x)2)(f(x)g(x))dxM_{x} = \int_{a}^{b} (\frac{f(x) + g(x)}{2})(f(x) - g(x))dx
  1. The moment MyM_{y} with respect to the y-axis is
My=abx(f(x)g(x))dxM_{y} = \int_{a}^{b} x(f(x) - g(x))dx
  1. The coordinates of the center of mass (xˉ,yˉ)(\bar{x}, \bar{y}) of the system are
xˉ=Mym\bar{x} = \frac{M_{y}}{m}
yˉ=Mxm\bar{y} = \frac{M_{x}}{m}

where MxM_{x} is moment about the x-axis and MyM_{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)f(x) = g(x)

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

  1. Calculate the total mass

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

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

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

  1. Compute the moments

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

where f(x)g(x)f(x) \geq g(x) on [a,b][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)=udv+vdud(uv) = u\,dv + v\, du.

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

Rewritten as

udv=uvvdu\int u\,dv = uv - \int v\,du

Trigonometric methods

Powers of Sine and Cosine

sinmxcosnxdx\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 dudu.

Powers of Secant and Tangent

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.