Calculating Rotational Volume for Enclosed Area by Revolution Around Y-Axis

[BadatPhysicsguy]

Homework Statement

An area is enclosed by y=1/x, the positive x-axis and x=1 and x=2. Determine the volume of the rotational body that is created around the y-axis.

Homework Equations

The formula is pi*integrate from d to c for x^2 dx.

The Attempt at a Solution

Take a look here: http://www.wolframalpha.com/input/?i=y=1/x,+y=0,+x=1,+x=2&a=*C.1-_*NonNegativeDecimalInteger-

I want the rotational volume for the area that is above the the purple, below the blue and between yellow and green. So how do I find the integration "limits" d and c, and what is x^2 in this case?

Okay so the integration points should be 1/x and then I enter x=1 because that is the higher one in the graph, 1/1 = 1. So the upper limit is 1. The lower should be zero, because the problem statement says that it is enclosed by the positive x-axis, y=0. But if I do this, I will also get all of it that is to the right of the green line. If I use 1/2 as my other integration limit, (x=2), there will be a missing rectangle just below.

How do I do this?

 

[6c 6f 76 65]

Imagine seeing the rotated figure from above, it kinda looks like a volcano. Try to make a radius from the origin: one from the O to the yellow line, and one from O to the green line. How long will both of these radii be (as a function of x, then make the substitution [itex]y=\frac{1}{x}[/itex])?

 

[BadatPhysicsguy]

Imagine seeing the rotated figure from above, it kinda looks like a volcano. Try to make a radius from the origin: one from the O to the yellow line, and one from O to the green line. How long will both of these radii be (as a function of x, then make the substitution [itex]y=\frac{1}{x}[/itex])?

Hello and thank you for another answer. Tell me if I'm understanding this correctly. So looking from above, I see a volcano. From above, the radius of the "hole" is from x=0 to x=1? The part between yellow and green is the "rock" part of the volcano and then there is a steep fall from the edge of the green side down to the ground.

How do I express the radius as a function of x? Now with the volcano explanation my first instinct is that I want the total volume of the volcano minus the "cylinder" that is the hole. So the radius from the volcano hole to the first rock part (yellow) is 1. The volume of the cylinder is then 3.14(pi)*1^2*height. The height is as high as the yellow goes, 1. So the volume of the cylinder is simply 3.14 volume units.

Now what is the total volume of the volcano? That is where I hit a problem again. Because if I integrate from 1 to 0 "the bottom" the volcano doesn't have a steep fall, instead the foundation continues to infinity.

 

[6c 6f 76 65]

Hello and thank you for another answer. Tell me if I'm understanding this correctly. So looking from above, I see a volcano. From above, the radius of the "hole" is from x=0 to x=1? The part between yellow and green is the "rock" part of the volcano and then there is a steep fall from the edge of the green side down to the ground.

How do I express the radius as a function of x? Now with the volcano explanation my first instinct is that I want the total volume of the volcano minus the "cylinder" that is the hole. So the radius from the volcano hole to the first rock part (yellow) is 1. The volume of the cylinder is then 3.14(pi)*1^2*height. The height is as high as the yellow goes, 1. So the volume of the cylinder is simply 3.14 volume units.

Now what is the total volume of the volcano? That is where I hit a problem again. Because if I integrate from 1 to 0 "the bottom" the volcano doesn't have a steep fall, instead the foundation continues to infinity.

Seems like you're heading in the right direction! As you creatively thought removing the cylinder is the key, now in order to compute the whole volume divide it into:
- The volume of the "large thin cylinder", bounded by [itex]x=0[/itex], [itex]x=2[/itex], [itex]y=\frac{1}{2}[/itex] (is the radius constant here?)
- The volume of the remaining part bounded by [itex]y=\frac{1}{2}[/itex], [itex]y=1[/itex] and [itex]y=\frac{1}{x}[/itex] (is the radius constant here?)

EDIT: Maybe I caused some confusion when talking about the "radius", here I mean the length from the y-axis to where it meets some boundary

 

[BadatPhysicsguy]

Seems like you're heading in the right direction! As you creatively thought removing the cylinder is the key, now in order to compute the whole volume divide it into:
- The volume of the "large thin cylinder", bounded by [itex]x=0[/itex], [itex]x=2[/itex], [itex]y=\frac{1}{2}[/itex] (is the radius constant here?)
- The volume of the remaining part bounded by [itex]y=\frac{1}{2}[/itex], [itex]y=1[/itex] and [itex]y=\frac{1}{x}[/itex] (is the radius constant here?)

Hm, so the volume of the "large thin cylinder" is simple geometry, right? The radius is 2 (from 0->2) and the height is 1/2. So that would be 2^2*pi*1/2. So 2*pi or approx 6.28 is the total bottom part.

As for the remaining part, I have to integrate. It must be expressed as x^2, so yx=1 => 1/y=x and then square it => x^2=1/y^2. Pi*integrate from 1 to 1/2 for 1/y^2 gives 1*pi volume units.

So I have the part bounded by y=1/2 and y=1 and y=1/x that is 1*pi volume units. The part just below it is 2*pi but that includes a chunk of the hole. So I want to remove the chunk that has radius 1, height 1/2. That is 1^2 * 3.14 * 1/2 so half a pi. 3*pi-0.5*pi is 2.5*pi. But the answer is 2*pi. Either I have added half a pi too much somewhere, calculated wrong from the very start or forgot to remove half a pi. What am I missing?

Also, is there no single "easy" way to calculate this? Or does the calculation have to be divided?

 

[6c 6f 76 65]

Hm, so the volume of the "large thin cylinder" is simple geometry, right? The radius is 2 (from 0->2) and the height is 1/2. So that would be 2^2*pi*1/2. So 2*pi or approx 6.28 is the total bottom part.

Correct!

As for the remaining part, I have to integrate. It must be expressed as x^2, so yx=1 => 1/y=x and then square it => x^2=1/y^2. Pi*integrate from 1 to 1/2 for 1/y^2 gives 1*pi volume units.

Correct!

So I have the part bounded by y=1/2 and y=1 and y=1/x that is 1*pi volume units. The part just below it is 2*pi but that includes a chunk of the hole. So I want to remove the chunk that has radius 1, height 1/2. That is 1^2 * 3.14 * 1/2 so half a pi.

Almost correct, what is the height of the "inner cylinder"?

Also, is there no single "easy" way to calculate this?

Well, you can make it a little simpler by saying the volume you're interested in finding is composed by:
- Volume bounded by: [itex]x=1, x=2, y=0, y=\frac{1}{2}[/itex] then get [itex]V_1=\pi\int_0^\frac{1}{2} (2^2-1^2)dy = \frac{3\pi}{2}[/itex]
- Volume bounded by: [itex]x=1, x=2, y=\frac{1}{2}, y=\frac{1}{x}[/itex] then get [itex]V_2=\pi\int_\frac{1}{2}^1 \frac{1}{y^2}-1^2 dy = \frac{\pi}{2}[/itex]
[itex]V_1+V_2=2\pi[/itex], but I'm not sure whether you can compute it without decomposing it.