Homework Help: Rotation around an axis

1. May 4, 2015

1. The problem statement, all variables and given/known data
The curve y=1/x and the line y=2.5-x enclose an area together. Determine the exact volume of the rotating body that is formed when this field rotates about
a) The x-axis and b) The y-axis

2. Relevant equations
The formula for rotation around the x-axis is pi*integrate from b to a y^2 dy..
The formula for rotation around the y-axis is pi*integrate from d to c x^2 dx.

3. The attempt at a solution
I'm gonna start with a). So what I've learned is that around the x-axis, first I have to find the points from where I'm gonna integrate (i.e. integrate from z to y and so on). To get these two points, I'm going to put the curve equal to the line so I know where they intersect.
2.5-x=1/x => 2.5x-x^2=1 => x^2-2.5x+1 = 0 and I get x(1) = 2 and x(2) = 0.5.

So if I'm going to find out the answer, I'll also use (1/x)^2 - (2.5-x)^2. So the total would be:

pi * integrate from 2 to 0.5 for (1/x)^2 - (2.5-x)^2 which gives me the correct answer 1.125*pi.

Okay, so how do I do it for b then?
Take a look here: http://www.wolframalpha.com/input/?i=y=1/x,+y=2.5-x
I want to calculate the volume for the rotation that is both under the purple and blue line. But how do I get that? It is obvious that the "integration limits" are from 2.5 to 0 from that graph. But how do I express it in a formula?

If I do the same as earlier, (1/x) - (2.5-x) Then doesn't that area under the purple line eliminate all of it that is under the blue and then some?

2. May 4, 2015

6c 6f 76 65

Think of it this way:
You'll get the volume of the figure if you first find the volume under the purple line and the remove the volume $between$ the purple line and the blue curve

3. May 4, 2015

So first I'd take the total volume under the purple line. But how do I get the volume that is between the purple line and the blue curve? Purple minus blue and then the total volume under purple minus whatever I got from the second one?

4. May 4, 2015

Exactly!

5. May 4, 2015

SammyS

Staff Emeritus
In this and your other recent thread, you have the form of the integral wrong.

For the rotation about the x-axis, y is a function of x, and the integration variable is x, not y, so that should be dx.

The integral is: $\displaystyle\ \pi\int_{x_1}^{x_2} y^2\, dx$