Volumes of Solids of Revolution / Method of Cylinders

jeepers101

I'm trying to solve a problem, which has been solved here as a method of rings problem

http://tutorial.math.lamar.edu/Classes/CalcI/VolumeWithRings.aspx

Its example 3 I'm looking at.

My problem is trying to express the curve in terms of y. I'm assuming it necessary. Is this possible?

sponsoredwalk

What curve do you mean?

Example 3 asks you to determine the volume of the solid obtained by
rotating the region bounded by y = x² - 2x & y = x about the line y = 4
so if you draw the curves



you see that the outer curve has to be the blue curve & the inner curve
has to be the red curve. Of course because you're rotating around the
x-axis (obviously along the line y = 4, not y = 0 as standard) you have to express
the outer radius of the ring (disk) as (skipping small details)
Πr² = Πy² = Π[f(x)]² = Π[4 - (x² - 2x)]²
which has y as a function of x, and the inner radius similarly.

I don't know what else to say unless you provide more detail.

jeepers101

Thanks for the reply.

For clarity I will call the blue curve 'the curve' and the red curve 'the line'.

Okay for the method of cylinders I need to find the general distance between the curve and the line, which will be the height.

This is found as x² - 2x - x = x² - 3x

Now I found the radius as 4 - y (although I'm not quite sure if this is correct due to the fact that my curve crosses the x axis- does this make a difference?)

This is where my problem lies. I need to express the height in terms of y so I can integrate the area of the cylinder.

MathsManiac

No, don't get an expression for the distance between the blue curve and the red line.

You need to get the distance from each of them to the line you are rotating about.

If you can visualise a slice of the solid of revolution, you'll see that it's like a ring. The volume you're looking for is made up of these rings, each of which has an outer radius (corresponding to the curve) and an inner radius (corresponding to the line).

That is, get R = 4-y for the curve and get r = 4-y for the line.

You then need to integrate Pi*(R^2)*dx - Pi*(r^2)*dx, from 0 to 3.

jeepers101

Yes but that is the method of rings. I understand that. Is it possible to do it using the method of cylinders?

See example 4

http://tutorial.math.lamar.edu/Classes/CalcI/VolumeWithCylinder.aspx

jeepers101

Basically, I'm looking to express y = x² - 2x in terms of y. I think that would be the only way of solving it using the cylinder method.

MathsManiac

The method you linked to would be quite a bizarre way to try to solve this problem, because it makes life far more complicated.

You would need to express x in terms of y for each of the curve and the line for a start. Then subtract the two expressions to get the horizontal distance between them and then proceed to rotate around the line y=4.

It's further complicated by the fact that, at the lower portion, the horizontal line that you're rotating is not bounded by the line and the curve, but by the curve at both ends.

So you'd need to do it in two separate bits.

jeepers101

Yes I appreciate I'm being awkward. I know its easier to do it the other way but I'm just trying to challenge myself.

So if I do it in two sections I'll still need to express the curve in terms of y. Is this possible?

sponsoredwalk

Alright I understand what you're trying to do now.

When you say you want to express the height in terms of y what you
mean is that you want the blue curve to be the top & you want to
subtract the red line from it, to get the height of the object you want to
rotate about the line y = 4.

The problem with taking this approach is that the blue 'curve' cannot
be inverted. Notice that the blue curve is no longer a function if
interpreted as a function of y, at the very least you can say it has
become a multiple valued function or you can interpret it as a relation but
you wont get anything nice out of that. What you could do is find the
minimum value of the curve by differentiating or whatever & then
interpret the curve as two separate functions on restricted domains,
but then you'd probably want to turn it into three separate functions in
order to deal with the subtraction issues you're going to run into.
If you want to go down this route then you're going to have to decide
on how to interpret each part of the curve as you invert it.

I'm all for making things more complicated as a means to teach yourself
how to deal with situations from different perspectives & being able to
cope with a concept from multiple viewpoints but when you start getting
issues like not having a function anymore that's where I stop. I think
it's a good idea to analyze revolution problems using as many techniques
as you can but to stop when you start dealing with situations where you
don't have functions anymore...

jeepers101

Okay. Thanks a million. Much appreciated.

MathsManiac

It may not be quite as intractable as all that. You can find the inverse relation of the function relatively easily, since it's a quadratic. The solution is x = 1 (+/-) sqrt(y+1).

For the lower section of the area concerned, the length of a horizontal line bounded at each end by the blue curve is 2*sqrt(y+1).

So, you're integrating 2*Pi*radius*width = 4*Pi*(4-y)*sqrt(y+1)dy from y=-1 to y=0. I make that 176*Pi/15.

For the upper section, the horizontal line has length 1+sqrt(y+1) - y. So you're integrating 2*Pi*(4 - y)*(1 + sqrt(y+1) - y) from y=0 to y=3. I make this bit 283*Pi/15.

Adding these gives 153*pi/5, which (happily) is the same answer as the one you get by the regular method.

jeepers101

Yeah I tried this again after your last post and I was able to get the upper area okay. The lower bit, though, I couldn't get.

How did you find the bounded width of the lower graph to be 2*sqrt(y+1)?

MathsManiac

You can solve for x in terms of y by treating y = x^2 - 2x as the quadratic equation: x^2 - 2x - y = 0. Solve that by either completing the square or using the formula for a quadratic equation to get x = 1 (+/-) sqrt(y+1). That is, the two values of x that correspond to a particular value of y are: x = 1 - sqrt(y+1) and x = 1 + sqrt(y+1). Subtract one from the other to get 2*sqrt(y+1).

jeepers101

Ahhh... very clever. Thanks again mathsmaniac.