Solids of revolution - volume between two curves

**blackberrybees** · February 23rd 2010, 08:59 PM

The problem asks to find the volume of the region bounded between y=x^2 and y=2-x, with the region being rotated around the vertical line x=3.

This problem seems rather complex, and I have dwelled on it longer than I care to admit. Because it is rotated around the vertical axis, I gather that I must slice horizontally and integrate with respect to y, but I can't seem to get any further.

My confusion is further exacerbated by the fact that, assuming integration with respect to y, when y is between 0 and 1, the inner and outter bounds of the shape are dictated solely by the line y=x^2.

Is there some way that I am overlooking to integrate with respect to x? I am so confused!

Thank you in advance for any assistance.

**drumist** · February 23rd 2010, 09:37 PM

It's possible to work the problem by "slicing" either vertically or horizontally, as long as you know the way to do it in each case. In this problem, it makes sense to slice vertically because you can define it as a single region instead of having to divide it into multiple regions, like you noticed.

Basically when dealing with rotation problems like this, the method used is determined by whether the axis of rotation (in this problem, x=3) and the direction you are slicing the region (in this problem, vertical) are parallel or perpendicular.

If they are parallel, you use the cylindrical shells method. Imagine what a single vertical "slice" of the region looks like when you revolve it about the line x=3. You get a cylinder shape. Try to define the surface area of that single cylindrical shell in terms of x.

$A = 2 \pi r h = 2 \pi (3-x) (2-x-x^2)$

Then, our volume will be:

$V = \int A \, dx = 2 \pi \int_{-2}^1 (3-x)(2-x-x^2) \, dx$

**blackberrybees** · February 23rd 2010, 10:14 PM

Thank you kindly, friend. Given the point that we are at in my calculus course, I was thinking that it was intended that we solve this without using the shells method, so I wasn't even considering it.

It seems, however, that this is the only sensible way to go about it.

**drumist** · February 23rd 2010, 10:25 PM

If you do it the other way, then you just have to define it into the two regions and add up the two separate volumes. For the one you asked about, between y=0 and y=1, you would define the upper bound as $x=\sqrt{y}$ and the lower bound as $x=-\sqrt{y}$ . Which makes the outside radius $3-(-\sqrt{y}) = 3+\sqrt{y}$ and the inside radius is $3-\sqrt{y}$ .

So, your resulting volume would be like:

$V=\pi \int_0^1 \left[ \left(3+\sqrt{y}\right)^2 - \left(3-\sqrt{y}\right)^2 \right] dy + \pi \int_1^4 \left[ \left(3+\sqrt{y}\right)^2 - (1+y)^2 \right] dy$

It looks a bit more messy but it's still workable.

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