OBJECTIVES

When you finish this section, you should be able to:

1. Use the slicing method to find the volume of a solid (p. 433)

IN WORDS

The volume \(V_{i}\) of a slice is \begin{eqnarray*} V_{i} &=& \hbox{(Area of cross section)}\hbox{ (Thickness} \\ &&\hbox{of the slice)} = A( u_{i}) \Delta x \end{eqnarray*}

Volume of a Solid Using the Slicing Method

If for each \(x\) in \([a,b]\), the area \(A(x)\) of the cross section of a solid is known and is continuous on \([a,b]\), then the volume \(V\) of the solid is \begin{align*} \bbox[5px, border:1px solid black, #F9F7ED]{ V=\int_{a}^{b}A(x)~{\it dx}}\tag{1} \end{align*}

NEED TO REVIEW?

Geometry formulas are discussed in Appendix A.2, p. A-15.

Using the Slicing Method to Find the Volume of a Cone

Use the slicing method to verify that the volume of a right circular cone having radius \(R\) and height \(h\) is \(V=\dfrac{1}{3}\pi R^{2}h.\)

Solution We position the cone with its vertex at the origin and its axis on the \(x\)-axis, as shown in Figure 39.

Figure 40

The cone extends from \(x=0\) to \(x=h\). The cross section at any number \(x\) is a circle. To obtain its area \(A\), we need the radius \(r\) of the circle. Embedded in Figure 39 are two similar triangles, one with sides \(x\) and \(r;\) the other with sides \(h\) and \(R,\) as shown in Figure 40. Because these triangles are similar (AAA), corresponding sides are in proportion. That is, \begin{eqnarray*} \dfrac{r}{x} &=&\dfrac{R}{h} \\[6pt] r &=&\dfrac{R}{h}x \end{eqnarray*}

So, \(r\) is a function of \(x\) and the area \(A\) of the circular cross section is \begin{equation*} A=A(x)=\pi [ r(x)] ^{2}=\pi\! \left( \dfrac{R}{h}x\right) ^{2}= \dfrac{\pi R^{2}}{h^{2}}x^{2} \end{equation*}

Since \(A\) is a continuous function of \(x\) (where the slice was made), we can apply the slicing method. The volume \(V\) of the right circular cone is \begin{equation*} V = \int_{a}^{b}A(x)~{\it dx}=\displaystyle \int_{0}^{h}\frac{\pi R^{2}}{h^{2}}x^{2}~{\it dx}=\frac{\pi R^{2}}{h^{2}}\!\displaystyle \int_{0}^{h}x^{2}~{\it dx}\\[11pt] = \frac{\pi R^{2}}{h^{2}}\left[ \frac{x^{3}}{ 3}\right] _{0}^{h}=\frac{\pi R^{2}h}{3} ~\hbox{cubic units} \end{equation*}

NOTE

A right circular cone is a solid of revolution, so the formula for its volume also can be verified using the disk method.

NOW WORK

Problem 13.

Using the Slicing Method to Find the Volume of a Solid

A solid has a circular base of radius \(3\) units. Find the volume \(V\) of the solid if every plane cross section that is perpendicular to a fixed diameter is an equilateral triangle.

Solution Position the circular base so that its center is at the origin, and the fixed diameter is along the \(x\)-axis. See Figure 41(a). Then the equation of the circular base is \(x^{2}+y^{2}=9\). Each cross section of the solid is an equilateral triangle with sides \(= 2y,\) height \( h,\) and area \(A=\sqrt{3}y^{2}\). See Figure 41(b). Since \(y^{2}=9-x^{2},\) the volume \(V_i\) of a typical slice is \[ V_{i}=\hbox{(Area of the cross section)(Thickness)}=A(x_{i})~\Delta x=\sqrt{3 }\left( 9-x_{i}^{2}\right) \Delta x \] as shown in Figure 41(c).

Figure 41

The volume \(V\) of the solid is \begin{eqnarray*} V&=&\int_{a}^{b}A(x)~{\it dx}=\int_{-3}^{3}\sqrt{3}(9-x^{2})~{\it dx} \underset{\underset{\underset{\color{#0066A7}{\hbox{an even function.}}}{\color{#0066A7}{\hbox{The integrand is}}}} {\color{#0066A7}{\uparrow }}} {=}2\sqrt{3}\int_{0}^{3}(9-x^{2})~{\it dx}=2\sqrt{3}\left[ 9x-\frac{ x^{3}}{3}\right] _{0}^{3}\\[-10pt] &=&36\sqrt{3}\hbox{ cubic units} \end{eqnarray*}

NEED TO REVIEW?

Even and odd functions are discussed in Section P.1, pp. 9-10, and the integrals of even and odd functions are discussed in Section 5.6, pp. 393-394.

NOW WORK

Problem 3.

Using the Slicing Method to Find the Volume of a Pyramid

Find the volume \(V\) of a pyramid of height \(h\) with a square base, each side of length \(b\).

Solution We position the pyramid with its vertex at the origin and its axis along the positive \(x\)-axis. Then the area \(A\) of a typical cross section at \(x\) is a square. Let \(s\) denote the length of the side of the square at \(x\). See Figure 42(a). We form two triangles: one with height \(x\) and side \(s\), the other with height \(h\) and side \(b.\) These triangles are similar (AAA), as shown in Figure 42(b). Then we have \begin{equation*} \dfrac{x}{s}=\dfrac{h}{b}\qquad\hbox{ or }\qquad s =\dfrac{b}{h}x \end{equation*}

Figure 42

So, \(s\) is a function of \(x\), and the area \(A\) of a typical cross section is \(A=s^{2}=\dfrac{b^{2}}{h^{2}}x^{2}\). See Figure 42(c). The volume \(V\) of a typical cross section is \(V=\dfrac{ b^{2}}{h^{2}}x^{2}\Delta x\). See Figure 42(d).

Since \(A\) is a continuous function of \( x\), where the slice occurred, we have \begin{equation*} V=\int_{0}^{h}A(x) \,{\it dx}=\displaystyle \int_{0}^{h}\dfrac{b^{2}}{h^{2}} x^{2}{\it dx}= \frac{b^{2}}{h^{2}}\int_{0}^{h}x^{2}{\it dx}=\frac{b^{2}}{h^{2}} \left[ \frac{x^{3}}{3} \right] _{0}^{h}=\frac{1}{3}b^{2}h\hbox{ cubic units} \end{equation*}

NOW WORK

Problem 5.

Using the Slicing Method to Find the Volume of a Solid

Find the volume \(V\) of the solid whose base is the region bounded by the graphs of \(y=\sqrt{x}\) and \(y=\dfrac{1}{8}x^{2}\), if slices perpendicular to the base along the \(x\)-axis have cross sections that are squares.

Figure 43

Solution We begin by graphing the region bounded by the graphs of \(y=\sqrt{x}\) and \(y=\dfrac{1}{8}x^{2}\), as shown in Figure 43. The points of intersection of the two graphs are found by solving the equation \begin{eqnarray*} \sqrt{x} &=&\dfrac{1}{8}x^{2} \\[2pt] x &=&\dfrac{1}{64}x^{4} \\[2pt] x^{4}-64x &=&0 \\[2pt] x( x^{3}-64) &=&0 \\[2pt] x &=&0 \quad\hbox{or}\quad x=4 \end{eqnarray*}

The two graphs intersect at the points \(\left( 0,0\right)\) and (4,2).

The solid with slices perpendicular to the base along the \(x\)-axis that are squares is shown in Figure 44(a). See Figure 44(b). A slice perpendicular to the \(x\)-axis at \(x_{i}\) is a square with side \( s_{i}=\displaystyle\sqrt{x_{i}}-\dfrac{1}{8}x_{i}^{2}.\) The area \(A_{i}\) of the cross section at \(x_{i}\) is \[ A_{i}=s_{i}^{2}=\left( \sqrt{x_{i}}-\dfrac{1}{8} x_{i}^{2}\right) ^{2} \]

Figure 44

See Figure 44(c). The volume \(V_{i}\) of a typical slice is \begin{eqnarray*} V_{i}&=& \hbox{(Area of the cross section)(Thickness of the slice)}\\[3pt] &=&A (x_{i}) \Delta x=\left( \sqrt{x_{i}}-\dfrac{1}{8}x_{i}^{2}\right) ^{2}\Delta x \end{eqnarray*}

The volume \(V\) of the solid is \begin{eqnarray*} V &=&\int_{0}^{4}A(x) {\it dx}=\int_{0}^{4}\left( \sqrt{x}-\dfrac{1}{8} x^{2}\right) ^{2}{\it dx}=\int_{0}^{4}\left( x-\dfrac{1}{4}x^{5/2}+\dfrac{1}{64} x^{4}\right) {\it dx} \notag \\[5pt] &=&\left[ \dfrac{x^{2}}{2}-\dfrac{1}{14}x^{7/2}+\dfrac{1}{320}x^{5}\right] _{0}^{4}=8-\dfrac{128}{14}+\dfrac{1024}{320}=\dfrac{72}{35}\hbox{ cubic units } \end{eqnarray*}

NOW WORK

Problem 7.