Volume_AverageValue.html

Calculating Volume and Average Value

General - Volume by "Slicing"

Calculating the volume of a box is easy: V = (length)(width)(height). A more useful (for us) way of expressing that relationship is V = (area of base)(height).

A box is an special example of a 3 dimensional object called a cylinder. The most familiar everyday cylinder is a tin can. The volume of a tin can of radius r and height h is or V = (area of base)(height). Mathematically, a cylinder is any 3-dimensional object bounded by two parallel planes (either one of which can be called the base of the cylinder) and consisting of all the points lying on all the line segments perpendicular to and between the bases. Just like for the box or tin can, the volume of a general cylider is V = (area of base)(height).

To calculate the volume of irregularly shaped objects, the idea is to slice up the object into many small slices, each of which is approximately a cylinder. We calculate the appropximate volume of each slice by the formula V = (area of base)(height) and add them all up to approximate the whole volume. We make the approximation better and better by taking more and more slices (the more slices, the closer each slice is to a true cylinder -- see the "loaf of bread" illustration in the text, page 441). This process should sound familiar; it is essentially the same idea that was behind using rectangles to approximate irregularly shaped areas (i.e. areas between curves ).

So, imagine a 3-dimensional object S lying above the x-axis between x = a and x = b. Starting at x = a we cut up S into slices by slicing with planes perpendicular to the x-axis. Each slice has thickness (or "height") and has Volume approximately where A(x) is the cross-sectional area of S at x . The more slices we take, the smaller is for each slice and the closer each slice is to a true cylinder. It seems natural (nod your heads "of course") to then define the volume to be = and that is precisely what we do!

e.g. Find the volume of a ball of i) radius 2 m ii) radius r

Put the ball with its centre at the origin, so the x-axis forms a diameter. At any x [0,2], a cross section is a circle with radius and area = (and the volume of a "slice" at x is approximately ). Therefore, the volume is:

> V:=Int(Pi*(4-x^2),x=-2..2);value(%)*`square meters`;evalf(%);

ii) Repeat i) except with r instead of 2 to get the formula for the volume of a sphere of radius r:

> V:=Int(Pi*(r^2-x^2),x=-r..r);value(%)*`square meters`;

>

Volume by the "disk" and/or "washer method"

Solid of Revolution : S is a solid of revolution if it can be obtained by rotating a plane region about a line. The easiest solids of revolution to visualize are those obtained by rotating the plane region formed by the area under y= f(x) above [a,b] around the x-axis.

For example, rotate the area under between x = -1 and x = 1 to obtain a ball of radius1unit. So a sphere is really a solid of revolution. The volume of a solid of revolution is in principle easy to calculate since the base of a cross section perpendicular to the axis of revolution is always a circle. So, if the solid is lying with its axis of revolution along the x-axis then the volume of a representative slice at x is approximately V ~ = where is the curve that was rotated about the x-axis to obtain the solid.

So, suppose a solid S is obtained by rotating about the x-axis between x = a and x = b. The volume of the resulting solid of revolution is then . There is an analogous result if S is obtained by revolving about the y-axis between y = c and y = d: .

If f and g are positive functions satisfying on [a,b] and the region bounded by these curves is rotated about the x-axis then the resulting solid will have a hole in it and the volume of the hole must be subtracted from the volume obtained by rotating about the x-axis: = = . An analogous result holds for rotation about the y-axis.

e.g.

Let A be the region bounded by and x = 1. Calculate the volume of the solid that results when A is revolved about i) the x-axis and ii) the y-axis

> f:=sqrt(x):L:={[1,0],[1,1]}:with(plots):with(CalcP):

> plot1:=plot(f,x=0..1,y=-1..1,colour=black):plot2:=plot(L,colour=black):

> display([plot1,plot2],scaling=constrained);

> revolve(f,x=0..1);LeftDisk(f,x=0..1,10);

You can see from the above that when A is rotated about the x-axis a rounded, solid bowl is obtained (and cross-sections perpendicular to the x-axis will be solid circles). When A is rotated about the y-axis a deeply indented bowl with a round base is obtained (and a cross-section perpendicular to the y-axis will be an annulus (or washer)).

> V1:=Int(Pi*sqrt(x)^2,x=0..1);V1:=value(%)*`cubic units`;

ii) For the second volume, you must realiz the the outside radius is x = f(y) = 1 and the inside radius is x = . It follows that the new volume is:

> V2:=Int(Pi*(1-(y^2)^2),y=0..1);value(%)*`cubic units`;

e.g.

In the following example the plane region is revolved about a line other than an axis. This must be compensated for when specifying the radii.

# 13 p 448. Rotate the region bounded by and y =1 about the line y = 2 and find the volume of the resulting solid.

> f:=x^4:

> p1:=plot(f,x=-1..1,colour=black):p2:=plot(1,x=-1..1,colour=black):p3:=plot(2,x=-2..2,y=0..4):

> display([p1,p2,p3],scaling=constrained);

=> the solid resulting will be a napkin ring with the volume below: (Remember that the radii are calculated relevant to the line y = 2.)

> V:=Int(Pi*((2-f)^2-(2-1)^2),x=-1..1);value(%)*`cubic units`;

Volume by the "shell method"

This is impossible to explain in Maple! See blackboard .

e.g. (from text) Revolve the region bounded by and the x-axis about the y-axis and caculate the volume of the resulting solid.

> f:=2*x^2-x^3;plot(f,x=0..2,y=-0..2,colour=black);

Note: in order to use the "washer method" on the solid you would first have to solve for x. Even if you could do that (it is possible in this example to do that but it would be very difficult and very cumbersome) you would still have to distinguish between the outer and inner radii.

volume using "shells":

> V:=Int(2*Pi*x*f,x=0..2);value(%)*`cubic units`;

Just in case you want to use the disk/washer method on this one first look at what Maple yields when you try to solve for x:

> solve(y = 2*x^2-x^3,x);

Wow!!!!!!!!!!!!!!

e.g. (#5, pg 454) Find the volume generated by revolving the region bounded by , y = 0, x = 0, x = 1 about the y-axis.

> C:=exp(1)^(-x^2);

> plot({C,[[1,0],[1,exp(1)^(-1)]]},x=0..1.5,y=0..1.1,colour=black,title=region);

> V:=Int(2*Pi*x*exp(1)^(-x^2),x=0..1);

> value(V);evalf(%,3);

e.g.

Consider the region bounded by and for x > 0 and calculate the volume generated if the region is revolved about the line y = 2 two diffrent ways.

> plot({x,x^3,2},x=0..1.5,y=0..2.5,colour=black);

> Volume_disks:=Int(Pi*((2-x^3)^2-(2-x)^2),x=0..1);value(%);

> Volume_shells:=Int(2*Pi*(2-y)*(y^(1/3)-y),y=0..1);value(%);

The Average Value of a function

The "Hard Way:"

Suppose that the temperature in a city (in degrees Celsius) t hours after 9 A.M. is modelled by . Find the average temperature in the city for the 12 hour period from 9 A.M. to 9 P.M.

> T:=t->10+8*sin(Pi*t/12);

> plot(T(t),t=0..12,colour=black);

We can start by estimating the average temp. by sampling temperatures at regular intervals and taking the aritmetic mean (i.e. the "usual" average) of those readings. We can try sampling more and more readings and see what happens.

Here's the average of the hourly predictions from the model:

> estimate1:=Sum(T(i),i=0..12)/12;evalf(%,5);