Two Integration ProblemsDate: 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/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/