# Homework Help: Finding/sketching volume with the washer method

1. Mar 17, 2008

### silicon_hobo

[SOLVED] Finding/sketching volume with the washer method

1. The problem statement, all variables and given/known data
Consider the region bounded by the curves $$y=x^2+1$$ & $$y=3-x^2$$. a) Using the disk/washer method, find the volume of the solid obtained by rotating this region about the x-axis. b) Setup the integral for finding the volume of the solid obtained by rotating this region about the x-axis. In each case, draw a diagram, indicating a typical "strip" (and its slice).

2. Relevant equations

3. The attempt at a solution
I'm pretty sure the first one is right wrt x.
http://www.mcp-server.com/~lush/shillmud/inta1.6.JPG

Is this the correct way to set it up wrt y? Is there a simpler way?
http://www.mcp-server.com/~lush/shillmud/intb1.6.JPG

Also, what should I draw to "indicate a 'strip'"? I see the words strip, slab, and slice often used in a seemingly interchangeable fashion. What are they asking for here? Thanks for reading.

2. Mar 17, 2008

### driscoll79

Hello there.

Your setup wrt has one mistake; It should be: $$\pi\int{[(3-x^2)^2-(x^2+1)^2]dx}$$

As for the second part, ideally you would not want to set this up wrt y, for the following two reasons:

1) It changes your method for finding volume from the washer method to cylindrical shells (assuming that your "slice" is perpendicular to the y-axis).

2) Given my assumption about your slice in #1, you'd have to set up 3 integrals in order to do the problem, which is pretty insane.

As for your last question, a "slice" or "strip" is simply the distance from one curve to the other for any given value of your variable of integration. For instance, take the point x=0.5. Draw a line or a narrow rectangle from the curve $$y=3-x^2$$ to the curve $$y=x^2 + 1$$ at that point. That is your slice/strip. The width of that strip is $$dx$$.

Last edited: Mar 17, 2008
3. Mar 17, 2008

### silicon_hobo

Thanks for the reply. I believe I have corrected my answer wrt x based on your suggestion although the new answer of 16pi seems a little bit large for the size of the rotated region. Also, I'm still not sure how to set it up wrt y. I know it's the harder way to do it but I would prefer to learn (and the question does ask for it). Cheers.

Last edited: Mar 17, 2008
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