Methods to find volume in calculus?

Answer 1

The two main methods for calculating volumes in calculus are by a pair of procedures collectively known as volumes of revolution. The main idea behind both is that you revolve some 2 dimensional shape around an axis and then figure out the area of this revolved object.

The first method is called the Method of Rings. If you treat each revolved piece as a circle at a given value of x you can say the area of the thin piece is πr² where r is the distance of the function from the axis of revolution. You can then integrate this area between the beginning and endpoints of the volume being revolved. For example, if you want to revolve y=x² around the x axis between x=0 and x=1, you would take the integral of π*(x²)² from 1 to 4 which would evaluate to π/5.

The other main method is called the method of cylinders and depends upon treating the revolved pieces of the function as infinitely thin cylinders instead of circles. The area of each of these revolved cylinders is 2π times their distance from the axis of rotation (a radius) times the value of the function at a point (the height). For example, to find the volume made by revolving the graph of y=x² around the y-axis from y=0 to y=1 would be the integral from 0 to 1 of 2π*x*x² which evaluates to π/2.

In other circumstances, it is also possible to calculate a volume not generated by a revolution. If you want the volume you get by projecting a given graph up to a the same height at every y value for a given x, you could calculate it by integrating the width of the graph at a given x value times its height at that x value. For example, if you want to find the volume of the region bounded by the equations y=√x, y=0, and x=1 that has a height equal to its x value, you would take the integral of the function (√x) times the height (x) from 0 to 1, which would give you a volume of 2/5.

With even more calculus knowledge you will eventually be able to use multiple integrals to find the volume of any region between any curves in three dimensional space, but the method of setting up the equations is often not easily apparent, and evaluating the integrals is usually even harder.