Find The Volume of a Solid of Revolution

How to find the volume of a solid of revolution generated by revolving a region bounded by the graph of a function around one of the axes using definite integrals? The tutorial below with examples and detailed solutions is designed to answer the above question. A set of exercises with answers is presented at the end.


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Formulas to calculate the volume generated by revolving graphs of functions around one of the axes are given below.

1 - If f is a function such that f(x) >= 0 for all x in the interval [x1 , x2], the volume of the solid generated by revolving, around the x axis, the region bounded by the graph of f, the x axis (y = 0) and the vertical lines x = x1 and x = x2 is given by the integral

Volume = x1 x2 p [ f(x) ] 2 dx


volume, one function, formula


2 - If f and h are functions such that f(x) >= h(x) for all x in the interval [x1 , x2], the volume of the solid generated by revolving, around the x axis, the region bounded by the graphs of f and h, between x = x1 and x = x2 is given by the integral

Volume = x1 x2 p [ f(x) 2 - h(x) 2] dx


volume, 2 functions, formula




3 - If z is a function such that x = z(y) >= 0 for all y in the interval [y1 , y2], the volume of the solid generated by revolving, around the y axis, the region bounded by the graph of z, the y axis (x = 0) and the horizontal lines y = y1 and y = y2 is given by the integral

Volume = y1 y2 p [ z(y) ] 2 dy


volume, one function of y , formula
4 - If z and w are functions such that z(y) >= w(y) for all y in the interval [ y1 , y2 ], the volume of the solid generated by revolving the region bounded by the graphs of z and w, between y = y1 and y = y2, around the y axis , is given by the integral

Volume = y1 y2 p [ z(y) 2 - w(y) 2] dy


volume, 2 functions of y, formula

A - Find Volume of a Solid of Revolution Using Definite Integrals



Example 1: Find the volume of the solid generated by revolving the region bounded by the graph of y = x, y = 0, x = 0 and x = 2.(see figure below).

volume to be calculated in example 1
Solution to Example 1:

Solution 1: This problem may be solved using the formula for the volume of a right circular cone.

volume = (1/3)* p * (radius) 2 *height

= (1/3)* p (2) 2 *2

= 8p /3*

Solution 2:

We shall now use definite integrals to find the volume defined above. If we let f(x) = x according to 1 above, the volume is given by the definite integral

Volume = x1 x2 p [ f(x) ] 2 dx

= 0 2 p x 2 dx

= p 0 2 x 2 dx

= p [ x3 / 3 ] 0 2 = 8 p / 3

The first method works because y = x is a linear function and the volume generated is that of a right circular cone , however the second method work for shapes other than cones and will be used in the examples below.



Example 2: Find the volume of the solid generated by revolving the semicircle y = sqrt (r 2 - x 2) around the x axis, radius r > 0.

Solution to Example 2:

volume, example 2


The graph of y = sqrt (r 2 - x 2) is shown below and y > = 0 from x = -r to x = r. The volume is given by formula 1 as follows

Volume = x1 x2 p [ f(x) ] 2 dx

= -r r p [ sqrt (r 2 - x 2) ] 2 dx

= -r r p [ r 2 - x 2) ] dx

= -r r p [ r 2 - x 2) ] dx

= p [ r2x - x3 / 3 ] -r r

= p [ (r3 - r3 / 3) - (- r3 + r3 / 3)]

= (4 / 3) p r3

This is the very well known formula for the volume of the sphere. If you revolve a semi circle of radius r around the x axis, it will generate a sphere of radius r.

Example 3: Find the volume of the solid generated by revolving the shaded region about the y axis.

volume, example 3


Solution to Example 3:

The shaded region is bounded by the x axis, the line that passes through the points (0,0) and (1,1) and has the equation y = x, and the line that passes through the points (1,1) and (2,0) and has the equation y = -x + 2. Since the solid is generated by revolving through the y axis, we shall use formula 4 given above to find the volume as follows

Volume = y1 y2 p [ z(y) 2 - w(y) 2] dy


where x = z(y) = -y + 2 (solve y = -x + 2 for x) and x = w(y) = y (solve y = x for x)

= 0 1 p [ (-y + 2) 2 - y 2] dy

= ò0 1 p [ -4y + 4] dy

= p [ -2y2 + 4y ] 0 1

= 2p

Exercises:

(1) Find the volume of the solid generated when the region between the graphs of f(x) = x 2 + 2 and h(x) = x is revolved about the x axis and over the interval [0,1].

(2) Find the volume of the region generated when the region bounded by y = 2x, x = 0, y = 0 and y =2 is revolved about the y axis.

Answers to Above Exercises

(1) 26 Pi /5

(2) 2 pi /3

More references on integrals and their applications in calculus.


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Updated: 2 April 2013

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