Expert: Paul Klarreich - 12/17/2006

Question
I'm in a class of Math Analysis/ Calculus Honors and I've already tried this problem the conventional way: finding the radius, then finding the area, then finding the integral of the squared radius to the two set boundaries between zero and p.  But it doesn't work.

Find the volume of the solid that is generated by rotating about the indicated axis the region bounded by the given curves.

y = sin x, x = 0, x = p, y = 0.

If you could somehow explain this to me with a graph, I would be very thankful!  But I completely understand if that's impossible.  Thank you!

Answer
Questioner:  Joyce
Category:  Calculus
 
Subject:  volume using integrals
Question:  I'm in a class of Math Analysis/ Calculus Honors and I've already tried this problem the conventional way: finding the radius, then finding the area, then finding the integral of the squared radius to the two set boundaries between zero and p.  But it doesn't work.

>> I don't see why it shouldn't work.  Look through the solution below and see if you can find the sticking point.

Find the volume of the solid that is generated by rotating about the indicated axis the region bounded by the given curves.

y = sin x, x = 0, x = p, y = 0.

If you could somehow explain this to me with a graph, I would be very thankful!  But I completely understand if that's impossible.  Thank you!
..................................................
Hi, Joyce,

I don't think I can send you a graph -- there just isn't any way to do it on this site.  But we can still try the example.

There are a couple of issues here:

1. Exactly what is the value of p?  Did you mean to write 'pi'?  I think you did, because that makes the problem a little better, so I will assume that.

2. What is the indicated axis?  You didn't exactly say, so I am going to assume that when you wrote "y = 0" you meant "about the x-axis", since y=0 IS the x-axis.

So your example says (I assume):

Find the volume of the solid generated by rotating the region bounded by:

y = sin x, x = 0, x = pi

about the x-axis.
................
For these Applications of Integration (the title of your chapter, I'm sure) the basic scheme is the same:

A. Find one sample element of (area, arc length, mass, volume, etc).

B. Integrate that sample element over the indicated interval.

Here you have one arch of  y = sin x, rotated about the x-axis, producing a fat football-like object.

I shall use the 'disk' method.  Your sample element will be a disk, which is basically a very flattened cylinder.  

Volume formula for a cylinder,

V = pi r^2 h

In this example:

A. the radius will be the height of the curve, which is the function value:

r = y = sin x

B. the height (thickness?) of the cylinder is  dx.

The volume element is:

dV = pi sin^2 x dx

and you integrate:

{pi
|   pi sin^2 x dx
}0

WARNING: THE FOLLOWING DISCUSSION MAY CONTAIN FRACTIONS AND OTHER MATERIAL INAPPROPRIATE FOR CERTAIN COMPUTING SYSTEMS.  VIEW IT IN A FIXED-SIZE FONT, SUCH AS COURIER.

To integrate sin^2 x, we use the old 'half-angle' trick:
         1 - cos 2x
sin^2 x = -----------
             2

{pi     1 - cos 2x  
|   pi ------------ dx
}0          2
            sin 2x
= (pi/2)(x - -------)  from  x = 0 to x = pi
               2
               sin 2pi         sin 0
= (pi/2)[ (pi - -------) - (0 - ------) ]
                  2              2
             pi^2
= pi/2 [pi] = ----
              2