Find The Volume of a Solid of Revolution

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

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





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

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

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).

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:

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.

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.