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### Volume of a Torus

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Date: 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.

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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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Associated Topics:
High School Calculus

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