Integral calculus
Applications of the definite integral
The volume of a solid of revolution
The volume of a sphere
The volume of a spherical segment
The volume of a cone
The volume of a solid of revolution
A solid figure generated by revolving a given curve around an axis of revolution is called the solid of revolution.
The volume generated by the region of the coordinate plane bounded by the segment of a curve y = f (x) between x = a, y = b and the x-axis, revolving around the x-axis, is shown in the left figure below.
Note that the infinitesimal volume of the cylinder representing an element of the integration,
dV = area ´ width = py2 ´ dx.
The volume generated by the segment of a curve x = g (y) between y = c and y = d, revolving around the y-axis, is shown in the right figure above.
The volume of a sphere
 Example:   Find the volume of a sphere generated by a semicircle revolving around the
x-axis.
 Solution:  Since the endpoints of the diameter, lying on the x-axis, are  - r and r, then

The volume of a spherical segment
Example:    Find the volume of a spherical segment generated by the portion of the right semicircle between  y = a and  y = a + h, revolving around the y-axis, as is shown in the below figure.
 Solution:  Since the right semicircle equation then
 or, by substituting  r12 = r2 - a2 and  r22 = r2 - (a + h)2
The volume of a cone
Example:  Find the volume of a right circular cone generated by the line (segment) passing through the origin and the point (h, r), where h denotes the height of the cone and r is the radius of its base, revolving around the x-axis, as shows the below figure.
Solution:  The equation of the generating line
 then,

Example:    Find the volume of a solid of revolution generated by a plane bounded by the segment of a curve   y = - x2 + 3x and the x-axis, revolving around the x-axis, as shows the below figure.
Solution:  The limits of the integration  - x2 + 3x = 0,
 x(-x + 3) = 0,    x1 = 0  and  x2 = 3  then,

Example:    Find the volume of a solid of revolution generated by one cycle of the cycloid   x = r (t - sin t)y = r (1 - cos t) and the x-axis, revolving around the x-axis, as shows the below figure.
Solution:  Since  y2 = r2(1 - cos t)2,   dx = r (1 - cos t)dt  the limits of the integration 0 < t < 2p, then
Example:    Find the volume of a solid of revolution generated by the arc of the sinusoid  y = sin x between   x = 0 and  x = p/2, revolving around the y-axis, as shows the below figure.
Solution:  Since curve rotates around the y-axis, we should apply the inverse of the sine, i.e., we use x = g (y) form or x = arcsin  y = sin-1y.  Thus, the limits of the integration, for  x1 = 0,   y1 = 0  and for  x2 = p/2y2 = 1  then, applying the integration by parts twice
Functions contents G