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 [ x^{3} / 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 [ r^{2}x  x^{3} / 3 ] _{r}^{r}
= p [ (r^{3}  r^{3} / 3)  ( r^{3} + r^{3} / 3)]
= (4 / 3) p r^{3}
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
= ò_{0}^{1}p [ 4y + 4] dy
= p [ 2y^{2} + 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.
