Checking Answer: Shell Method

[SeekerOfDragons]

Wanting to check my answer to the following question:

Consider the region bounded by the graphs of y = 1/x, y = x^2, x = 2. Use the shell method to find the volume of the solid formed by revolving the region about the line x = 2. (do not evaluate the integral)

I found that the equations intersected at (1, 1) by setting 1/x = X^2, 1 = x^3, x = 1

Using that information I came up with an answer of the following:

2pi * integral (1, 2) [(x - 2)(x^2 - 1/x)] dx

hopefully I understood the shell method correctly? confirmation or correction would be appreciated.

r/

SoD

 

[chrisr]

I would say the radius of every shell is 2-x,
as opposed to the disc method,
since in that case you are calculating (pi)r[sup:56iqs6wk]2[/sup:56iqs6wk],
so it does not matter whether you'd use 2-x or x-2.

However, now you are calculating the curved surface area of a hollow cylinder or shell,
so 2(pi)r(height) requires the positive {length measurement} of the shell radius.

 

[SeekerOfDragons]

ah. ok so it should be (2 - x) not (x - 2). Semi confused as to why exactly that is the order it needs to be in. Would it be safe to say it's always the value (x or y) the shell is being rotated around minus x or y depending on which direction it's being rotated?

r/

SoD

 

[chrisr]

hi Seekerofdragons,

If you quickly sketch the two curves and the line,
you notice that your bounded region lies to the left of that line.

Now, as you rotate it about x=2, you need to calculate the radius of either all the discs
(for the washer method) or the vertical cylinders for the shell method.

X in both cases is to the left of 2, hence less than 2,
therefore the radius you use is 2-x as this gives you the radius length,
(x-2 will give you a negative reading),
which is ok if you change the sign on your final answer,
but far better to know you need the actual length of the radius.

With the shell method, you are calculating the curved surface area of a multitude of cylinders,
which is circle circumference times height.
The circle circumference requires the "length" (positive) measure of the radius.

For the discs, it doesn't matter as squaring of the radius takes place to calculate disc area,
((pi)r[sup:2v6bzc7s]2[/sup:2v6bzc7s] is used instead of 2(pi)r).

If you look back also at Glenn's way to calculate these, you'll see that he avoids negative radii
when using the shell method.
If you do use negative radius with the shells, you will have to negate the overall negative answer
after evaluating the integrals (they will return negative answers).

You could rotate a bounded region to the right of a line, say x=k..
In that case the radius would be x-k, where x is the x co-ordinate of a point on the bounded region,
so you need to know if your region is to the right or left of the axis of rotation.

 

[SeekerOfDragons]

ok. think I have it. if the function is to the left of the x-axis of rotation, it'd be (# - x) and if the function was to the right of the x-axis of rotation, it'd be (x - #)

How would it work if you were to rotate on a y-axis of rotation?
if the function was above the y-axis of rotation, would it be (y - #) and if the function was below the y-axis of rotation would it then be (# - Y)?

r/

SoD

 

[chrisr]

yes, that's it.

Let's suppose your bounded region in this example was bounded between x = -1 and x = -2,
you then want to rotate that around the line x = -2.
The radius of the discs or cylinders now will be x-(-2),
because the bounded region is to the right of x = -2, so x is always > -2.

This could be the situation if one of the curves was -1/x instead of 1/x.

 

[BigGlenntheHeavy]

\(\displaystyle Shell\ Method:\)

\(\displaystyle 2\pi\int_{1}^{2}\bigg(x^{2}-\frac{1}{x}\bigg)(2-x)dx\)

\(\displaystyle Disc \ Method:\)

\(\displaystyle \pi\int_{1}^{4}(y^{1/2}-2)^{2}dy+\pi\int_{1/2}^{1}\bigg(\frac{1}{y}-2\bigg)^{2}dy\)