Find Volume of Rotated Region Bounded by y=x, y=sqrt(x)

[merced]

Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line. Sketch the region, the solid, and a typical disk or washer.

y = x
y = \sqrt{x}
rotate about y = 1

http://img461.imageshack.us/img461/5879/math10sp.jpg

=http://img161.imageshack.us/img161/5729/math23gk.th.jpg

So, I am integrating with respect to x.
Area = \int^1_{0}[(f(x))^2-(g(x))^2]dx

I can't figure out how to get f(x) and g(x). I would think that they are simply f(x) = x and g(x) = \sqrt{x}.

The book gives f(x) = 1 - x and g(x) = 1 - \sqrt{x}

I don't understand how that works.

 

[Curious3141]

I found the easiest way to do it is to demarcate the element of the volume of revolution.

In this case, one element is like a small piece of a cylindrical annulus (which consists of a cylinder with a centered cylindrical hole burrowed within it).

The cylindrical annulus has outer radius given by (1 - y_1) = 1 - x and inner radius given by (1 - y_2) = 1 - \sqrt{x}. The radii have the "one minus" because you are rotating about y = 1 and not the x-axis. From geometry you know that the volume of a cylindrical annulus is Volume = \pi(r_o^2 - r_i^2)h where r_o is external radius, r_i is internal radius and h is height. Apply the same logic to the element of volume here and you'll get the integrand below.

The complete volume can be evaluated as the integral

V = \int_0^1 \pi ((1 - y_1)^2 - (1 - y_2)^2) dx = \int_0^1 \pi ((1 - x)^2 - (1 - \sqrt{x})^2) dx. The bounds are of course derived from seeing where x = \sqrt{x}

That is in essence the same thing your book has given, in this case, f(x) = (1-x) and g(x) = (1 - \sqrt{x})

 

[merced]

Thanks,

I think I'm starting to understand why you get the 1 - x and 1 - \sqrt{x}. Is it that you have to treat the radius as a function? Why can't you pretend that you are rotating over the origin and take the radius from there?

I also don't understand why you have to subtract the function...in another example where you rotate over y = -1, you add 1 to the function.

I don't understand the difference.

 

[Curious3141]

Thanks,

I think I'm starting to understand why you get the 1 - x and 1 - [tex
]\sqrt{x}[/tex]. Is it that you have to treat the radius as a function? Why can't you pretend that you are rotating over the origin and take the radius from there?

I also don't understand why you have to subtract the function...in another example where you rotate over y = -1, you add 1 to the function.

I don't understand the difference.

You can't "pretend" you're rotating over the x-axis (you meant this when you said "origin" didn't you?), because the axis of rotation makes a big deal to the calculation. It's not a simple question of rotating about the x-axis then translating the volume. Just cut the required area at x = 1 and see how much the "curvey part" you're rotating differs from the "curvey part" you're rotating when you use the x-axis to rotate the object. There's no easy way to get from the solid of revolution formed from rotating about the x-axis to the one formed by rotating about y = 1.

Here we're taking (1 - y) as the radius. Now when rotating about y = -1, you replace the "1" with "-1" giving you (-1-y).

That seems different from (1+y) doesn't it? But when you square it to do the volume, you get the same thing : (-1-y)^2 = (1+y)^2. Simple, no?

 

[merced]

Ok, but what if you moved the function down 1 so that it is over the x-axis and then rotated it?

letting f(x) = x - 1 and g(x) = \sqrt{x} - 1.

 

[Curious3141]

Ok, but what if you moved the function down 1 so that it is over the x-axis and then rotated it?

letting f(x) = x - 1 and g(x) = \sqrt{x} - 1.

Yes, that would be fine. Verify that it works out the same way when you square it, e.g. (x-1)^2 = (1-x)^2.

 

[merced]

Thanks, I understand now!