Two Integration Problems

Date: 05/26/99 at 00:35:21
From: Sandi
Subject: Integration

I have a couple of questions involving integration:

1. The first is just straightforward integration of x[sqrt(1-x^2)].
   The answer given is 1/2[sin^-1 x + (1/4)x[sqrt(1-x^2)]] + c. 
   I just don't see how the answer is achieved.

2.  Find the volume of the torus formed by rotating the circle x^2 + 
    (y - R)^2 = r^2 (where r is less than or equal to R) about the
    X-axis. What happens when r = R?

    For this one, before you begin to talk about the shell or disc 
    method, I haven't learned either of these as yet, and don't know 
    what they are. The way I've learned to do it is 

    V = int  pi y^2 dx

Thank you,
Sandi

Date: 05/26/99 at 09:28:31
From: Doctor Rob
Subject: Re: Integration

1. The obvious thing to do is substitute x = sin(t), dx = cos(t)*dt, 
sqrt(1-x^2) = cos(t). Then you have to integrate sin(t)*cos^2(t)*dt. 
The integral of that is -cos^3(t)/3 + c = -(1-x^2)^(3/2) + c. This 
answer is different from yours above. Taking the answer above and 
differentiating got me (sqrt(1-x^2))/4 + 3/(8*sqrt(1-x^2)), not what 
you started with above. Something is wrong somewhere.

2. Solve the equation of the circle for y:

   y = R +- sqrt(r^2-x^2).

The formula you are asking me to use computes the volume of revolution 
of the area between the x-axis and the curve y = f(x). In this case, 
we want the volume of revolution of the area between two curves,

   y = R - sqrt(r^2-x^2),
   y = R + sqrt(r^2-x^2).

This can be found by using your formula twice, subtracting one volume 
from the other. The limits of integration will be -r <= x <= r, so the 
volume is

                  r
   V = Pi*INTEGRAL  (R + sqrt(r^2-x^2))^2 dx
                 -r

                      r
         - Pi*INTEGRAL  (R - sqrt(r^2-x^2))^2 dx,
                     -r

                  r
   V = Pi*INTEGRAL  (R + sqrt(r^2-x^2))^2 - (R - sqrt(r^2-x^2))^2 dx,
                 -r

                  r
   V = Pi*INTEGRAL  4*R*sqrt(r^2-x^2) dx.
                 -r

You can finish from here, I think.

- Doctor Rob, The Math Forum
  http://mathforum.org/dr.math/   

Date: 05/29/99 at 03:16:19
From: Sandi
Subject: RE: Integration

Thanks Dr Rob,

I've worked it through, but the answer in the back of the book is 
2pi^2 Rr^2 and I don't see how to get this answer.

Thank you for your response,
Sandi 

Date: 06/02/99 at 10:32:02
From: Doctor Rob
Subject: RE: Intergration (Dr Rob)

I suggest using the substitution x = r*sin(t) in the last integral 
above. Then sqrt(r^2-x^2) = r*cos(t), and dx = r*cos(t) dt. Also when 
x = -r, t = -Pi/2, and when x = r, then t = Pi/2. Then

                          Pi/2
   V = 4*Pi*R*r^2*INTEGRAL    cos^2(t) dt.
                         -Pi/2

Now the identity cos^2(t) = (1+cos[2*t])/2 can be used, and that can 
easily be integrated with respect to t, giving t/2 + sin[2*t]/4. 
Evaluating this between the limits Pi/2 and -Pi/2 gives Pi/2 for the 
value of the integral, so the final result is

   V = 4*Pi*R*r^2*(Pi/2) = 2*Pi^2*R*r^2,

as indicated in your book.

- Doctor Rob, The Math Forum
  http://mathforum.org/dr.math/