Area of a volume of revolution

Date: Tue, 22 Nov 1994 08:05:27 -0800 (PST)
From: Uma Chandavarkar
Subject: Applications of the Integral

Hello! I am a senior at Monta Vista High School in Cupertino, California. 
I am currently enrolled in Calculus. I am a little confused about finding 
volumes when graphs are revolved around an axis. I would specifically 
like to know more about finding the volume of a disk and the volume of a 
washer. Thank you!!!!

                        - Uma Chandavarkar

Date: 23 Nov 1994 16:19:22 GMT
From: Ken "Dr." Math
Subject: Re: Applications of the Integral

Hello there Uma!

You've asked about a great problem, that of finding the area of a volume
of revolution.  Well, there are essentially two different ways to do it,
the shell method and the disk method.  I'll go over the disk method with you.

The first thing you need to do is draw a picture.  I'm serious about
this.  You simply can't learn to do these problems unless you draw
pictures of what you're doing.  

I'm afraid I can't really draw a picture in this message, but I can guide you 
through making one.  Let's take the region below the curve y = x^3 
(that means x cubed) and above the curve y = x^2, between the x values of 
3 and 5.  Draw the picture, and see what the region looks like.  We're going 
to revolve it around the x axis, and find out what the volume of the resulting 
space figure is.  So you should try to draw that too, if you can.

Let's take that space figure and chop it up vertically with a ginsu knife,
so that we get a bunch of little washers.  The thickness of the washers is
going to be REALLY SMALL, and we're going to call that thickness dx (as in
the dx that sits out to the side of an integral).  To find the volume of
the washer, we have to remember the formula for the volume of a disk: it's
Pi*r^2*h, right?  

What we'll do is find the volume of the disk made by rotating the region
below y=x^3, and then subtract off the region below y=x^2.  Does that make
sense in your picture?  

Since the radius of one of these disks is just the height of the function,
the volume of one thin disk under y=x^3 is going to be Pi * (x^3)^2 * dx. 
To get the volume of the whole region, we add up (i.e. integrate) the
volumes of all the disks.  So we'll have the integral from 3 to 5 of
Pi*x^6 dx.  I'll let you evaluate it.

But we still have to subtract the region below the curve y=x^2. 
What's the volume of this region?

I hope this helps!

Ken "Dr." Math