Integration can be used for much more than calculating areas. Whenever we have a function that describes a suitably calculable thing, the total of this can be calculated by the integral. Let’s see what needs to be summed to calculate the volume.
The simplest example is the case where some function f(x) “rotates around the x-axis” and forms a “solid of revolution”. For example, the square root function
If we look in the interval where x gets the values [0,4], a hill-shaped body on its side is formed as the function rotates. The volume of this is obtained by dividing the piece into slices in the x-direction, forming a “volume function” for the slice, and integrating it from zero to four. Let's draw such a slice of the picture
Now if we had the volume V(x) of the disc shown in the figure (as a function of x!), We would get the volume as an integral
The disc is drawn at a specific point (x) on the solid of revolution. At that point (x), the radius of the disk is the same as the value of the function f(x). Since it is a solid of revolution, the disc is in fact a circular cylinder with a base area of
If the thickness of the disc is x, then the volume of the disc is
Taking the limit value at which the width of the slice goes to zero gives the volume of the solid of revolution
That is, when the continuous function f(x) rotates about the x-axis, the volume of the solid of revolution formed between [a, b] is obtained as a volume integral
Sometimes the integration needs to be done with respect to some other variable (e.g. Rotation around the y-axis → integration with respect to y), in which case this is calculated similar to the area calculations for the y-axis made earlier [the function is first inverted to x(y)].
Please note that there is no need for the absolute value of the volume signs around the function, since the second power takes care of the integration of the function and will be positive. For this reason, there is also no need to study the zeros and divide the function into positive and negative pieces, as was done when calculating the area.
Example 1: The function f(x) rotates around the x-axis between [0,4]. Calculate the volume of the solid of revolution.
The body is not always a solid of revolution, or the rotating function is not always known. Even then, the volume is obtained using the same idea, as long as one knows how to form the “slice area”, i.e. the area function A(x).
Such volume integrals are explored in the “Applications” chapter.
More difficult tasks can be accomplished, for example, by using two different functions, each of which rotates around an axis.
Example 2: The area bounded by the functions f (x) = 2x + 1 and g (x) = 4 - x² and the x-axis rotates around the x-axis. Calculate the volume of the solid of revolution.
Draw the graph first
f(x) is an ascending line and g (x) is a downward-opening parabola. As the region rotates about the x-axis, the upper bound of the region is defined by f(x) at the beginning and g(x) at the end. That is, always the function whose absolute value is smaller. The integral must be calculated in two parts, between [0, B] and [B, 2], where B is the x-coordinate of the intersection of the curves and 2 is the positive zero of the function g(x) [you can solve this yourself].
The intersection points of the functions are obtained from the group of equations either with a CAS calculator or manually:
The point of intersection that we are interested in is therefore x = 1 and we get the integral
Next, you need to open the parentheses and calculate both integrals. There is a fourth power of x in the integral and the calculation is otherwise rather long, so just for fun, this is calculated with the Geogebra CAS calculator.
The volume is then
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