Volume of a TorusDate: 04/22/99 at 15:18:03 From: Michale S Subject: Finding the volume of a torus I am a senior in high school and I have a calculus question. I have a circle x^2+y^2 = 4 and I want to rotate that circle around the line X = -2. When that is done it forms a torus (a doughnut-like object). I would like to know how to find the volume of that using integrals. I first found the upper half of the circle y = sqroot(4-x^2). I found the area of that using the integral from -2 to 2 of sqrt(4-x^2) dX. After I did that I multiplied it by 2 to find the area of the whole circle (I want to do this problem by integration). I have read about rotating parabolas and other objects around an axis. It says multiply by pi, so should I multiply by pi or what should I do? The equation for the volume of a torus is v = .25*pi^2*(b^2-a^2)* (b-a). b = the radius of the big circle and a = the radius of the small circle making up the torus. Please give me some input as to what I should do. Thanks for your time. Date: 04/22/99 at 16:45:01 From: Doctor Rob Subject: Re: Finding the volume of a torus Thanks for writing to Ask Dr. Math. You can do this by summing up the volume of infinitesimally thin rings gotten by slicing the torus by planes perpendicular to the y-axis. The range of values of y is -2 <= y <= 2. The volume of such a ring is the difference of the volumes of two infinitesimally high cylinders. The radii of the two cylinders are R = 2 + sqrt(4-y^2) and r = 2 - sqrt(4-y^2). The height of each cylinder is h = dy. The volume of the larger cylinder is Pi*R^2*h and the volume of the smaller cylinder is Pi*r^2*h. The difference is Pi*(R^2-r^2)*h. Using the above values of R, r, and h, and summing over all y between -2 and 2 gives 2 V = INTEGRAL Pi*([2+sqrt(4-y^2)]^2 - [2-sqrt(4-y^2)]^2)*dy. -2 Now expand the squares, combine like terms, and integrate using the substitution y = 2*sin(u). Re-express everything in terms of y, and apply the limits of integration to get the answer. You can also do this by summing up the volume of infinitesimally thin cylindrical shells gotten by slicing the torus by concentric cylinders with axis the line x = -2. The radius of such a cylinder is 2 + x, and its height is 2*sqrt(4-x^2), where the range of values of x is -2 <= x <= 2. The area of the shell is the height times the circumference of the shell times its thickness dx. Then you get the integral 2 V = INTEGRAL 2*Pi*(2+x)*2*sqrt(4-x^2)*dx. -2 Factor the constant 4*Pi out in front of the integral, and expand what is left into two terms. One term can be integrated by using the substitution x = 2*sin(u), dx = 2*cos(u)*du, and the other by using the substitution 4 - x^2 = v, -2*x*dx = dv. Re-express everything in terms of x, and apply the limits of integration to get the answer. The answer should turn out the same for either method. I got an answer a little less than 160. This agreed with the formula on this page (scroll down to Circular or Ring Torus: http://mathforum.org/dr.math/faq/formulas/faq.ellipsoid.html - Doctor Rob, The Math Forum http://mathforum.org |
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