Calculating Volumes of Slicing

Let $f$ be continuous and nonnegative for $a\leq x\leq b$ and let $R$ denote the region bounded by the graph of $y=f(x),$ the $x$-axis, and the lines $x=a $ and $x=b.$ The volume of the solid obtained by rotating $R$ about the $x$-axis is
MATH

Example 1

Find the volume of the cone obtained by revolving about the $x$-axis the region bounded above by the graph of $f(x)=x/3$ and below by the $x$-axis for $0\leq x\leq 3$

Solution. Using formula
MATH
with $f(x)=x/3$ we obtain
MATH
Since this cone has height $h=3$ and base of radius $r=1,$ this result agrees with the formula from geometry for the volume of this cone:
MATH

Example 2

Find the volume of the solids obtained by rotating the region bounded by the graphs of $f(x)=\sqrt{x}$ and $g(x)=x^{2}$ about the $x$-axis.

Solution

The two graphs cross at $(0,0)$ and $(1,1)$ since the equation$\sqrt{x}=x^{2} $ implies $x=x^{4}$ or $x(1-x^{3})=0.$ Since $\sqrt{x}>x^{2}$ for $0<x<1,$ the region is bounded above by the graph of $f(x)=\sqrt{x}$ and below by the graph of $g(x)=x^{2}$, we can view the solid $S$ as the ``difference'' of two other solids. Let $S_{1}$ be the solid obtained by rotating the region bounded by the graph of $f(x)=\sqrt{x}$ about the x-axis for $0\leq x\leq 1$ and let $S_{2}$ be the solid obtained by rotating the region bounded by the graph of $g(x)=x^{2}$ about the $x$-axis for $0\leq x\leq 1.$ The the original solid $S$ is the solid that results when $S_{2}$ is removed from $S_{1}.$ The desired volume $V$ is, therefore, the difference between the volume of $S_{1}$ and the volume of $S_{2}.$ Using equation
MATH
twice, we obtain
MATH

As you may have observed, the volume in Example 2 could have been calculated by the single integral
MATH
In general, if the region $R$ is bounded above by the graph of $y=f(x)$ and below by the graph of $y=g(x)\geq 0$ and $a\leq x\leq b$, then the formula for volume is
MATH

Example 3

The region in the first quadrant bounded by the graph of $y=4-x^{2}$ and the coordinate axes is rotated about the $y$-axis. Find the volume of the resulting solid.

Solution

This problem is similar to that of Example 1 except that the role of $x$ and $y$ are reversed. Solving the given equation for $x$ as a function of $y$ gives $f(y)=\sqrt{4-y}.$ Since the rotation is about the $y$-axis, the integration will be with $y,$ and the limits of integration are from $y=0$ to $y=4.$ We obtain
MATH

The result of Example 3 generalizes as follows:

If the region $R$ is bounded by the graph of the continuous function $x=f(y)$ and the $y$-axis from $y=c$ to $y=d,$ then the volume of the solid obtained by rotating $R$ about the $y$-axis is
MATH

Solids of Known Cross-Sectional Area

If we ``slice'' a solid of revolution $S$ in the direction perpendicular to the axis of rotation, we obtain a circular disc. For example, if $S$ is obtained by revolving the region $R$ bounded by the graph of a nonnegative function $f$ about the x-axis for$a\leq x\leq b,$ the slice for which $x=x_{0}$ is a disc with radius $r=f(x_{0}).$ Consequently, its area $A(x_{0}) $ is MATH. In fact, we can interpret formula
MATH
for the volume of $S$ as the integral of the cross-sectional area function $A $ from $x=a$ to $x=b.$ Our next objective is to show that the volume of any solid cross-sectional area can be calculated in this way.

In particular, suppose that $S$ is the solid for which the area $A(x)$ of each cross-section perpendicular to the $x$-axis is known ($S$ need not be a solid of revolution). Moreover, suppose that $S$ extends from $x=a$ to $x=b$. We partition the interval $[a,b]$ into subintervals of equal length $\Delta x=(b-a)/n$ with endpoints MATH Using the partition, we subdivide solids $S$ into solids $S_{j}$ of thickness $\Delta x,$ where each $S_{j}$ corresponds to the interval $[x_{j-1},x_{j}]$. By selecting one number $t_{j}$ in each subinterval $[x_{j-1},x_{j}]$, we approximate the volume of $S_{j}$ by the volume $V_{j}$ of the cylinder with face area $A(t_{j})$; that is,
MATH

Summing the $V_{j}$ for $j=1,2,3,...,n$ yields an approximation to the volume $V$ of $S$
MATH

We assume that, as MATH , the Riemann sum on the right-hand side approaches the volume $V.$ Now, if $A$ is continuous for $a\leq x\leq b,$ the sum also approaches the definite integral
MATH
Thus, we obtain the formula for the volume of solids with known cross-sectional area.

Let $A(x)$ denote the area of the cross section of $S$ for $a\leq x\leq b$. If function $A$ is continuous on $[a,b],$ the volume $V$ of $S$ is
MATH

Notice this equation generalizes the familiar formula for the cylinder with base area $A$ and height $h:V=Ah.$ It also contains formula for the volume of a solid of revolution as a special case, since for such solids, MATH

Example 4

The base of a solid is a circle of radius $4$ cm. All cross sections perpendicular to a particular axis are squares. Find the volume of this solid.

Solution

If we impose an $xy$-coordinate system on the circular base so that the x-axis corresponds to the given axis, then the equation for the boundary of the base is
MATH
The equation for the upper semicircle is $y=\sqrt{16-x^{2}}$, and the equation for the lower semicircle is $y=-\sqrt{16-x^{2}}$. A cross section perpendicular to the x-axis will therefore intersect this circular base in a chord of length MATH Since this chord is one side of the square cross section, the area of the cross section is
MATH
The smallest and largest values of $x$ are, respectively, $-4$ and $4.$ The volume is therefore
MATH

Example 5

Find the volume of the solid generated by rotating the region bounded by the graph of $f(x)=\sqrt{4-x}$ and the $x$-axis for $0\leq x\leq 4$ about the line $y=-2.$

Solution

A cross section taken at location $x$ consists of a circle of radius MATH from which a smaller circle of radius $r=2$ has been removed. The cross-sectional area is therefore
MATH
We can now apply equation
MATH
to find that
MATH

This document created by Scientific WorkPlace 4.0.