shell method: region bounded by y=x^3, y=0, x=0, x=2 abt x=3

[Opus89]

I was assigned a problem dealing with the shell method by my professor. All I have to do is set the integral up, i don't have to solve it. I thought I had the right answer but after discussing the problem with a friend, I'm not so sure. The question is as follows:

Use the shell method to find the volume of the solid formed by rotating the region bounded by y=x^3, y=0, x=0, and x=2 about the line x=3.

I think I have a good picture of whats going on. I was going to let my function y= x^3 be my height and let my average radius be 3-x since there is a space between the function and the line it is going to revolve around. If there was no space between the function and the axis of revolution, my radius would have just been x, or so I thought.

I came up with the integral from 0 to 2 of (3-x)(x^3). My friend said this was incorrect and he said it should be (x)(3-x^3). These turn out to equal each other but one of them is the wrong way. Can anyone offer any insight? Thanks a bunch.

 

[galactus]

Re: shell method problem

Using shells:

\(\displaystyle 2{\pi}\int_{0}^{2}(3-x)x^{3}dx\)

You're correct in your set up.

Now, try it with washers and see if you get the same.

 

[Opus89]

Re: shell method problem

Hmm, I'm trying to see how you could use washer method for this one. I'm having a pretty big mental block here and I'm not seeing it for some reason.
It doesn't seem like there is anything to take the difference of. Any hints?

 

[skeeter]

washers ...

\(\displaystyle R = 3 - \sqrt[3]{y}\)

\(\displaystyle r = 1\)

\(\displaystyle V = \pi \int_0^8 (3 - \sqrt[3]{y})^2 - 1^2 \, dy\)