How Do You Calculate the Volume of a Rotating Body Around Different Axes?

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SUMMARY

The discussion focuses on calculating the volume of a rotating body formed by the intersection of the curve y=1/x and the line y=2.5-x, specifically around the x-axis and y-axis. For rotation around the x-axis, the volume is calculated using the integral formula π∫(1/x)² - (2.5-x)² dx from x=0.5 to x=2, resulting in a volume of 1.125π. For the y-axis, the discussion emphasizes the need to find the total volume under the purple line (y=2.5-x) and subtract the volume under the blue curve (y=1/x), highlighting the correct use of integration limits and variables.

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  • Familiarity with the concepts of curves and intersections
  • Knowledge of the formulas for rotation around the x-axis and y-axis
  • Proficiency in using integration techniques
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Homework Statement


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

Homework 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.

The Attempt at a Solution


I'm going to 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 going to 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?
 
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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
 
6c 6f 76 65 said:
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

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?
 
BadatPhysicsguy said:
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?
Exactly!
 
BadatPhysicsguy said:

Homework 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.
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 ##
 

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