# 6.5 Physical applications  (Page 2/11)

 Page 2 / 11

We now approximate the density and area of the washer to calculate an approximate mass, ${m}_{i}.$ Note that the area of the washer is given by

$\begin{array}{cc}\hfill {A}_{i}& =\pi {\left({x}_{i}\right)}^{2}-\pi {\left({x}_{i-1}\right)}^{2}\hfill \\ & =\pi \left[{x}_{i}^{2}-{x}_{i-1}^{2}\right]\hfill \\ & =\pi \left({x}_{i}+{x}_{i-1}\right)\left({x}_{i}-{x}_{i-1}\right)\hfill \\ & =\pi \left({x}_{i}+{x}_{i-1}\right)\text{Δ}x.\hfill \end{array}$

You may recall that we had an expression similar to this when we were computing volumes by shells. As we did there, we use ${x}_{i}^{*}\approx \left({x}_{i}+{x}_{i-1}\right)\text{/}2$ to approximate the average radius of the washer. We obtain

${A}_{i}=\pi \left({x}_{i}+{x}_{i-1}\right)\text{Δ}x\approx 2\pi {x}_{i}^{*}\text{Δ}x.$

Using $\rho \left({x}_{i}^{*}\right)$ to approximate the density of the washer, we approximate the mass of the washer by

${m}_{i}\approx 2\pi {x}_{i}^{*}\rho \left({x}_{i}^{*}\right)\text{Δ}x.$

Adding up the masses of the washers, we see the mass $m$ of the entire disk is approximated by

$m=\sum _{i=1}^{n}{m}_{i}\approx \sum _{i=1}^{n}2\pi {x}_{i}^{*}\rho \left({x}_{i}^{*}\right)\text{Δ}x.$

We again recognize this as a Riemann sum, and take the limit as $n\to \infty .$ This gives us

$m=\underset{n\to \infty }{\text{lim}}\sum _{i=1}^{n}2\pi {x}_{i}^{*}\rho \left({x}_{i}^{*}\right)\text{Δ}x={\int }_{0}^{r}2\pi x\rho \left(x\right)dx.$

We summarize these findings in the following theorem.

## Mass–density formula of a circular object

Let $\rho \left(x\right)$ be an integrable function representing the radial density of a disk of radius $r.$ Then the mass of the disk is given by

$m={\int }_{0}^{r}2\pi x\rho \left(x\right)dx.$

## Calculating mass from radial density

Let $\rho \left(x\right)=\sqrt{x}$ represent the radial density of a disk. Calculate the mass of a disk of radius 4.

Applying the formula, we find

$\begin{array}{cc}\hfill m& ={\int }_{0}^{r}2\pi x\rho \left(x\right)dx\hfill \\ & ={\int }_{0}^{4}2\pi x\sqrt{x}dx=2\pi {\int }_{0}^{4}{x}^{3\text{/}2}dx\hfill \\ & =2\pi {\frac{2}{5}{x}^{5\text{/}2}|}_{0}^{4}=\frac{4\pi }{5}\left[32\right]=\frac{128\pi }{5}.\hfill \end{array}$

Let $\rho \left(x\right)=3x+2$ represent the radial density of a disk. Calculate the mass of a disk of radius 2.

$24\pi$

## Work done by a force

We now consider work. In physics, work is related to force, which is often intuitively defined as a push or pull on an object. When a force moves an object, we say the force does work on the object. In other words, work can be thought of as the amount of energy it takes to move an object. According to physics, when we have a constant force, work can be expressed as the product of force and distance.

In the English system, the unit of force is the pound and the unit of distance is the foot, so work is given in foot-pounds. In the metric system, kilograms and meters are used. One newton is the force needed to accelerate $1$ kilogram of mass at the rate of $1$ m/sec 2 . Thus, the most common unit of work is the newton-meter. This same unit is also called the joule . Both are defined as kilograms times meters squared over seconds squared $\left(\text{kg}·{\text{m}}^{2}\text{/}{\text{s}}^{2}\right).$

When we have a constant force, things are pretty easy. It is rare, however, for a force to be constant. The work done to compress (or elongate) a spring, for example, varies depending on how far the spring has already been compressed (or stretched). We look at springs in more detail later in this section.

Suppose we have a variable force $F\left(x\right)$ that moves an object in a positive direction along the x -axis from point $a$ to point $b.$ To calculate the work done, we partition the interval $\left[a,b\right]$ and estimate the work done over each subinterval. So, for $i=0,1,2\text{,…},n,$ let $P=\left\{{x}_{i}\right\}$ be a regular partition of the interval $\left[a,b\right],$ and for $i=1,2\text{,…},n,$ choose an arbitrary point ${x}_{i}^{*}\in \left[{x}_{i-1},{x}_{i}\right].$ To calculate the work done to move an object from point ${x}_{i-1}$ to point ${x}_{i},$ we assume the force is roughly constant over the interval, and use $F\left({x}_{i}^{*}\right)$ to approximate the force. The work done over the interval $\left[{x}_{i-1},{x}_{i}\right],$ then, is given by

find volume of solid about y axis and y=x^3, x=0,y=1
what is the power rule
how do i deal with infinity in limits?
f(x)=7x-x g(x)=5-x
Awon
5x-5
Verna
what is domain
difference btwn domain co- domain and range
Cabdalla
x
Verna
The set of inputs of a function. x goes in the function, y comes out.
Verna
where u from verna
Arfan
If you differentiate then answer is not x
Raymond
domain is the set of values of independent variable and the range is the corresponding set of values of dependent variable
Champro
what is functions
give different types of functions.
Paul
how would u find slope of tangent line to its inverse function, if the equation is x^5+3x^3-4x-8 at the point(-8,1)
pls solve it i Want to see the answer
Sodiq
ok
Friendz
differentiate each term
Friendz
why do we need to study functions?
to understand how to model one variable as a direct relationship to another variable
Andrew
integrate the root of 1+x²
use the substitution t=1+x. dt=dx √(1+x)dx = √tdt = t^1/2 dt integral is then = t^(1/2 + 1) / (1/2 + 1) + C = (2/3) t^(3/2) + C substitute back t=1+x = (2/3) (1+x)^(3/2) + C
navin
find the nth differential coefficient of cosx.cos2x.cos3x
determine the inverse(one-to-one function) of f(x)=x(cube)+4 and draw the graph if the function and its inverse
f(x) = x^3 + 4, to find inverse switch x and you and isolate y: x = y^3 + 4 x -4 = y^3 (x-4)^1/3 = y = f^-1(x)
Andrew
in the example exercise how does it go from -4 +- squareroot(8)/-4 to -4 +- 2squareroot(2)/-4 what is the process of pulling out the factor like that?
Andrew
√(8) =√(4x2) =√4 x √2 2 √2 hope this helps. from the surds theory a^c x b^c = (ab)^c
Barnabas
564356
Myong
can you determine whether f(x)=x(cube) +4 is a one to one function
Crystal
one to one means that every input has a single output, and not multiple outputs. whenever the highest power of a given polynomial is odd then that function is said to be odd. a big help to help you understand this concept would be to graph the function and see visually what's going on.
Andrew
one to one means that every input has a single output, and not multiple outputs. whenever the highest power of a given polynomial is odd then that function is said to be odd. a big help to help you understand this concept would be to graph the function and see visually what's going on.
Andrew
can you show the steps from going from 3/(x-2)= y to x= 3/y +2 I'm confused as to how y ends up as the divisor
step 1: take reciprocal of both sides (x-2)/3 = 1/y step 2: multiply both sides by 3 x-2 = 3/y step 3: add 2 to both sides x = 3/y + 2 ps nice farcry 3 background!
Andrew
first you cross multiply and get y(x-2)=3 then apply distribution and the left side of the equation such as yx-2y=3 then you add 2y in both sides of the equation and get yx=3+2y and last divide both sides of the equation by y and you get x=3/y+2
Ioana
Multiply both sides by (x-2) to get 3=y(x-2) Then you can divide both sides by y (it's just a multiplied term now) to get 3/y = (x-2). Since the parentheses aren't doing anything for the right side, you can drop them, and add the 2 to both sides to get 3/y + 2 = x
Melin
thank you ladies and gentlemen I appreciate the help!
Robert
keep practicing and asking questions, practice makes perfect! and be aware that are often different paths to the same answer, so the more you familiarize yourself with these multiple different approaches, the less confused you'll be.
Andrew
please how do I learn integration
they are simply "anti-derivatives". so you should first learn how to take derivatives of any given function before going into taking integrals of any given function.
Andrew
best way to learn is always to look into a few basic examples of different kinds of functions, and then if you have any further questions, be sure to state specifically which step in the solution you are not understanding.
Andrew
example 1) say f'(x) = x, f(x) = ? well there is a rule called the 'power rule' which states that if f'(x) = x^n, then f(x) = x^(n+1)/(n+1) so in this case, f(x) = x^2/2
Andrew
great noticeable direction
Isaac
limit x tend to infinite xcos(π/2x)*sin(π/4x)