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Volumes by Slices

April 13th, 2009 |

Author: Merrin

We have seen how integrals can be used to calculate the area beneath a curve or revolved around an axis. Integrals can also be used to calculate the volumes of objects. We will consider how to calculate the volumes of different class of objects. For one class of volumes, you have a formula for the cross sectional area. Suppose you have a sphere then you know its cross sections are circles and you even have equations for each. The differential volume of a cross section is given by

$dV=A(x)dx$

From what we already know about integration it is not too much of a leap to write the total volume by cross sections as.

$\displaystyle V=\int _{a}^{b}A(x)dx$

A second class of volumes can be made by adding up the area enclosed revolving a function $latex y=f(x)$ about the x axis. At a point x, the differential volume by this method is

$dV=\pi y^{2} dx$

You see each section formed by revolution is just the volume of a disc of width dx. The volume by revolution about the x axis between a and b is

$\displaystyle V\int _{a}^{b}\pi y^{2} dx \quad y=f(x)$

A third class of volumes is what you get when you have the area bounded between two function revolved about the x axis. Since one function is lower than the other it cuts out a hole in the volume, and this is called finding the volume by the method of washers.

For revolution about the x axis, we have

$\displaystyle V=\int _{a}^{b}\pi \left[f(x)-g(x)\right]^{2} dx$

If the volume is oriented about the y axis then instead we have.

$\displaystyle V=\int _{a}^{b}\pi \left[s(y)-t(y)\right]^{2} dy$

There is also a formula for an offset axis of revolution. If you want to calculate the volume of revolution aroun a shifted x axis y = b then if you are doing this by the method of washers where $latex f(x)>h$ between a < x < b then you write

$\displaystyle V=\int _{a}^{b}\pi (f(x)-h)^{2} dx$

If you want to do a volume of revolution using cylindrical shells and have about an axis y = h then you would have for x > h.

$\displaystyle V=\int _{a}^{b}2\pi (x-h)ydx$

For revolutions about the y axis instead of the x axis you just permute the positions of x and y in the formula.

Example 1. Find the volume of a pyramid with base area B and height h by horizontal cross sections.

Solution 1. We align the pyramid upside down along the y axis and at the origin. If you take any y cross section of the pyramid then,

$A(y)=B\cfrac{y^{2} }{h^{2} }$

The area scales with the square of the height because it is proportional to the square of the dimension. Now we can calculate the volume of the pyramid by cross sections.

$\displaystyle V=\int _{0}^{h}A(y)dy =\int _{0}^{h}B\cfrac{y^{2} }{h^{2} } dy =\cfrac{Bh}{3}$

Example 2. Find the volume of a sphere of radius R by circular cross sections.

Solution 2. Take a circle centered at (0, 0) with radius R. A circular cross section which passes through the sphere can be found from.

${x^{2} +y^{2} =R^{2} } \\ {y=\pm \sqrt{R^{2} -x^{2} } }$

We wish to take the top branch of y and revolve it around the x axis to find the volume of revolution for a sphere.

$\displaystyle {\int _{}^{}\pi y^{2} dx } = {\int _{-R}^{R}\pi (R^{2} -x^{2} )dx=\left. \pi (R^{2} x-x^{3} /3)\right| _{-R}^{R} }$

Example 3. Find the volume of revolution about the y axis of a slanted line to generate a right circular cone of height h and top radius R.

Solution 3. If you can take the volume of revolution about the \x axis you can take the volume of revolution about the y axis by permuting the two variables x and y. We would then have an integral in terms of dy instead of dx.

The main problem in finding the volume of a cone is to find the equation of the line representing the edge of the cone. The disc method for a cylindrical cone is equivalent to integrating the volume by circular cross sections.

$y=x\cfrac{h}{R} \quad \quad \quad \quad \quad \quad x=y\cfrac{R}{h}$