How Do You Calculate the Volume of Mountains and Polluted Air Around a City?

[mr3369]

Homework Statement

the function is y = -x^2+6x -8

suppose a city is surrounded by a ring of mountains and these mountains can be illustrated by rotating the above function around the y-axis. Find the volume of the Earth that makes up these mountains.

Suppose the city suffers from air pollution and wants to build a system to reduce this air pollution. However, in order to do so, it needs to find the volume of polluted air. This volume is the region between the y-axis and the function, rotated about either x or y axis. The picture for the polluted air is attached.

Homework Equations

the function is y = -x^2+6x -8

The Attempt at a Solution

For the first part of finding the volume of earth, I integrated the function, using the cylindrical shell method. And I got the answer as 8 pi.
I used the radius as r = x, height as h = -x^2+6x -8,
So I integrated 2pi*r*h, from 2 to 4 which are the x intercepts.

For the 2nd part, I made the function in terms of y, So first I did completing the square and got y=-(x-3)^2 +1
So that gives me x = sqrt(1-y) +3
I used x as the radius, and using disk method , revolved the function around the y-axis.
So radius = sqrt(1-y) +3 ,
Then i integrated pi(r^2) from 0 to 3.
However, my answer gives me root(-2) which isn't possible. So I am stuck in this 2nd part.

 

[Simon Bridge]

Best way is to sketch the function first ... then shade the region to be integrated.
http://en.wikipedia.org/wiki/Solid_of_revolution

For the second part ... you may have to divide the region up into parts you can integrate over and add op the bits. Start out putting it as x(y) if you like, or you can figure a way to find the volume that is not part of what you want and subtract it off a greater volume that includes both.

 

[haruspex]

Then i integrated pi(r^2) from 0 to 3.

You might want to check the range again. What are you integrating wrt?

 

[Simon Bridge]

I don't think I explained myself properly last time - I'll revisit:

Suppose the city suffers from air pollution and wants to build a system to reduce this air pollution. However, in order to do so, it needs to find the volume of polluted air. This volume is the region between the y-axis and the function, rotated about either x or y axis.

x or y axis?

Anyway - the area of interest is between the line ##x=0## and the curve ##y=-x^2+6x-8## ... however: y(x) is an inverted parabola ... so for a particular value of y there are two possible values of x. y(x) also has a maximum - so there is an upper limit to any integration along the y axis.

Rotating this shape about the y-axis should create a trumpet-shape with the point at y=y(0) (if the inner surface is the one intended). It is unclear what would be intended by using the outer surface; nor by rotating about the x-axis, since y(x) extends above and below it. That picture of the pollution would have been handy.

This is why I urged you to sketch the situation.

For the 2nd part, I made the function in terms of y, So first I did completing the square and got y=-(x-3)^2 +1
So that gives me x = sqrt(1-y) +3

What about the negative square-root: does it make a difference? Which side of the parabola is determining the length of the radius?

To do the method of disks - you are thinking of a horizontal disk, thickness dy, centered at (0,y), with radius x(y) so the volume of the disk at y is ##dV=\pi x^2(y)dy##. The trick is to get the correct x(y) and the correct limits for the integration.

I used x as the radius, and using disk method , revolved the function around the y-axis.
So radius = sqrt(1-y) +3 ,
Then i integrated pi(r^2) from 0 to 3.
However, my answer gives me root(-2) which isn't possible. So I am stuck in this 2nd part.

How did you choose those limits?