   Chapter 8.4, Problem 60E

Chapter
Section
Textbook Problem

# Volume of a Torus In Exercises 55 and 56, find the volume of the torus generated by revolving the region bounded by the graph of the circle about the y -axis. ( x − h ) 2 + y 2 = r 2 ,   h > r

To determine

To calculate: The volume of the torus generated by revolving the region bounded by the graph of the given circle about the y axis.

Explanation

Given:

The given circle (xh)2+y2=r2,h>r.

Formula used:

The trigonometric identity is:

cos2θ=1sin2θ

If f(x) is an odd function.

Then, aaf(x)dx=0

Calculation:

Consider the followingcircle

(xh)2+y2=r2.

From the above equation, this circle has centred at (h,0) and has radius r.

Now use cylindrical method, to get x=hr and x=h+r as the boundary of integration.

The cylinder’s width is 2y=2r2(xh)2 and length is 2πx and the thickness is dx.

Now, the volume V, which is given as:

V=hrh+r4πxr2(xh)2dxV=4πhrh+rxr2(xh)2dx

and,

dV=4πxr2(xh)2

Hence, the total volume is V=244πxr2(xh)2dx.

As r2(xh)2 is of the form a2x2;

Let xh=rsinθ;

Differentiating it, gives;

dx=rcosθdθ

Now to change the limit of integration;

Substitute x=hr to get,

sinθ=1θ=π2

Substitute x=h+r to get,

sinθ=1θ=

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