Find the volume of the solid of revolution, or state that it does not exist. #2

[abc1]

I'm having some trouble with this problem:

Find the volume of the solid of revolution, or state that it does not exist. The region bounded by f(x)= 6(4-x)^(-1/3) and the x-axis on the interval [0,4) is revolved avout the y-axis.

How would I be able to tell whether to use the shell, disk, or washer method?

 

[MarkFL]

We can use a function plotter to see the graph on the indicated interval:

View attachment 2238

Looks to me like I would try the shell method. Can you state the volume of an arbitrary shell?

 

[abc1]

How can I tell that I have to use the shell method?

Would the formula for this be 2pi times the integral from 0 to 4 of x* (6(4-x)^(-1/3) dx ?

So then, how would I take the limit of this? What test should I use?

 

[MarkFL]

How can I tell that I have to use the shell method?

Would the formula for this be 2pi times the integral from 0 to 4 of x* (6(4-x)^(-1/3) dx ?

So then, how would I take the limit of this? What test should I use?

You don't have to use the shell method, but can you see that the shell method allows for the computation of the volume as a single integral whereas the washer method will require an integral to be added to a cylinder? Either way would work though. In fact, I recommend to students to use more than one method both as a check as for the practice.

I always like to start by computing an element of the volume and then integrating. It appears that you do have the correct integral representing the volume. I think first though, I would use the substitution:

$$u=4-x$$

What do you have now?

 

[abc1]

Would it be 2pi times the integral from 0 to 4 of x* -(6(u)^(-1/3) du ?

 

[MarkFL]

Would it be 2pi times the integral from 0 to 4 of x* -(6(u)^(-1/3) du ?

You want to write that $x$ out front in terms of $u$...:D

 

[abc1]

so the limit as u approaches infinity from 4 to 0 of 2pi times the integral from 0 to 4 of (4-x)* -(6(u)^(-1/3) du since we have to rewrite the limits of integration as well because of the u-substitution?