Calculus II 07.02 Volume The Disk Method

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7.2 Volume The Disk Method

• Use the disk method to find the volume for a rotated solid.
• Use the washer method to find the volume for a rotated solid.
• Find the volume for a solid with known cross sections.

The Disk Method

Three-Dimensional Rotated-Solids, Solids of Revolution, are created by taking a closed figure and rotating it around one axis, the axis of revolution. The resulting shape is always symmetric along that axis. This fact allows us to find the volume using integrals. Some examples are axles, funnels, pills, bottles, and pistons, as shown in Figure 7.2.1.

Rotated Solids
Figure 7.2.1

Disk Volume \(\pi R^{2}w\) Figure 7.2.2

The simplest such solid is a right circular cylinder or disk, which is formed by revolving a rectangle about an axis as shown in Figure 7.2.2. The volume for this disk is $$\text{Disk Volume }= (\text{disk area})(\text{disk width}) = \pi R^{2}w $$ where \(R\) is the radius and \(w\) is the width. Divide the solid into rectangles, take each volume, then sum them to get the entire volume. For the smooth solid in Figure 7.2.4 create a representative rectangle in the plane region perpendicular to the axis. When this rectangle is revolved about the axis it generates a representative disk whose volume is $$ \Delta V = \pi R^{2}w \: \Delta x $$ Approximating the volume by \(n\) such disks with width \(\Delta x\) and radius \(R(x_{i})\) produces $$ \text{Volume}$$ $$\: \approx \sum_{i=1}^{n}\pi \left [ R(x_{i}) \right ]^{2}\:\Delta x $$ $$ = \pi \sum_{i=1}^{n} \left [ R(x_{i}) \right ]^{2}\:\Delta x $$ Disk Method Figure 7.2.3

This approximation improves because as \((\left \| \Delta \right \| \rightarrow 0)\:\text{and}\:(n \to \infty ) \) the error is reduced to 0. Using integrals the volume can be defined as

$$ \text{Volume } = \lim_{\left \| \Delta \right \| \rightarrow 0 } \pi \sum_{i=1}^{n} \left [ R(x_{i}) \right ]^{2}\:\Delta x = \pi \int_{a}^{b} \left [ R(x_{i}) \right ]^{2}\:dx $$

Schematically, the disk method looks like this.

Known Precalculus Formula		Representative Element		New Integration Formula
$$V=\pi R^{2}w \:\:\:\:$$	\( \rightarrow \)	$$\Delta V=\pi \left [ R(x_{i}) \right ]^{2}\:\Delta x \:\:\:\:$$	\( \rightarrow \)	$$V= \pi \int_{a}^{b} \left [ R(x_{i}) \right ]^{2}\:dx$$

Definition 7.2.1 The Disk Method

To find the volume for a rotated solid with the disk method, use either formula below. (See Figure 7.2.3.) To find the interval \([ a,b ]\) find where the functions meet the axis.

The Disk Method
Horizontal Revolution Axis	Vertical Revolution Axis
$$Volume = V = \pi \int_{a}^{b} \left [ R(x) \right ]^{2}\:dx $$	$$Volume = V = \pi \int_{c}^{d} \left [ R(y) \right ]^{2}\:dy$$
See Figures 7.2.4 and 7.2.5
Rotation about the horizontal axis. Figure 7.2.4	Rotation about the vertical axis. Figure 7.2.5

The simplest case is the graph for some function \(f\) rotated around the \(x\)-axis.

Example 7.2.1 Find the Volume for a Jelly Bean

Figure 7.2.6

Find the volume formed by revolving the region bounded by the graph $$ f(x)=\sqrt{\sin x} $$ and rotated around the \(x\)-axis. Solution From the representative rectangle in the upper graph in Figure 7.2.6, the radius for this solid is $$ R(x)=f(x)= \sqrt{\sin x} $$ and that the interval is \([ 0,\pi ]\). Plug this into The Disk Method and we get the volume \(V\) $$= \pi \int_{a}^{b} \left [ R(x) \right ]^{2}\:dx $$ Apply The Disk Method $$= \pi \int_{0}^{\pi} (\sqrt{\sin x})^{2}\:dx \:\:\:\: $$ Substitute \(\sqrt{\sin x}\) for \(R(x)\) $$= \pi \int_{0}^{\pi} \sin x\: dx$$ Simplify \(= \pi [ - \cos x ]_{0}^{\pi }\) Integrate \(= \pi (1+1)= 2\pi\) The Jelly Bean has the volume \(2\pi\).

Example 7.2.2 Using a Line That Is Not a Coordinate Axis

Figure 7.2.7

Form a solid and find its volume by rotating the region bounded by \(f(x)=2-x^{2}\) and \(g(x)=1\) about the line \(y=1\), as shown in Figure 7.2.7. Solution Equating \(f(x)\) and \(g(x)\) reveals that both graphs intersect the axis when \(x= \pm 1\). Subtract \(g(x)\) from \(f(x)\) to get the radius, \(R(x)\). \(R(x)\) \(=f(x)-g(x)\) \(=(2-x^{2})-1\) \(=1-x^{2}\) Integrate between -1 and 1. \(V\) $$= \pi \int_{a}^{b} \left [ R(x) \right ]^{2}\:dx $$ Apply The Disk Method $$= \pi \int_{-1}^{1} \left (1-x^{2} \right )^{2}\:dx $$ Substitute \((1-x^{2})^{2}\) for \(R(x)\). $$= \pi \int_{-1}^{1} \left (1-2x^{2}+x^{4} \right )\:dx $$ Simplify. $$= \pi \left [ x-\frac{2x^{3}}{3}+\frac{x^{5}}{5} \right ]_{-1}^{1}=\frac{16 \pi}{15}\:\:\:\: $$ Integrate

The Washer Method

Definition 7.2.2 The Washer Method

Figure 7.2.8

The disk method can be extended to cover Rotated-Solids with holes rotated about the axis by replacing the disk with a washer. The washer is formed by revolving a rectangle about the axis, as shown in Figure 7.2.8, with a gap between the rectangle and the axis. If \(r\) and \(R\) are the inner and outer radii for the washer and \(w\) is the washer's width, then the volume is $$ \text{Washer Volume }= \pi \left(R^{2}-r^{2} \right )w $$ just subtract the hole volume from the disk volume. This concept can be extended further by substituting graphs for rectangles over some interval and subtracting the outer region from the inner region as shown in Figure 7.2.9. The formula is $$ V= \pi \int_{a}^{b} \left [ R(x) \right ]^{2} - \left [ r(x) \right ]^{2} \:dx \:\:\:\: \color{red}{\text{The Washer Method }} $$ Figure 7.2.9

Example 7.2.3 Using the Washer Method

Figure 7.2.10

Find the volume enclosed by the graphs for \(y=\sqrt{x}\) and \(y=x^{2}\) rotated about the \(x\)-axis, as shown in Figure 7.2.10. Solution In Figure 7.2.10, you can see the outer and inner radii are as follows. \(R(x)=\sqrt{x} \:\:\:\: \) Outer radius \(r(x)=x^{2}\) Inner radius Integrating between 0 and 1 produces $$V$$ $$= \pi \int_{a}^{b} \left [ R(x) \right ]^{2} - \left [ r(x) \right ]^{2} \:dx $$ Apply The Washer Method $$= \pi \int_{0}^{1} \left [ (\sqrt{x}) \right ]^{2} - \left [ (x^{2}) \right ]^{2} \:dx \:\:\:\: $$ Substitute \(\sqrt{x}\) for \(R(x)\) and \(x^{2}\) for \(r(x)\) $$= \pi \int_{0}^{1} \left (x - x^{4} \right ) \:dx $$ Simplify $$= \pi \left [ \frac{x^{2}}{2}-\frac{x^{5}}{5} \right ]_{0}^{1} =\frac{3 \pi}{10}$$ Integrate

Example 7.2.4 Integrating with Respect to \(y\), The Two-Integral Case

In previous examples the rotation axis was horizontal and integrated with respect to \(x\). This example has a vertical rotation and requires integration with respect to \(y\) over two intervals.
Find the volume for a solid formed by revolving the region bounded by the graphs for

$$ y=x^{2}+1,\:y=0,\:x=0 \text{, and } x=1 $$

about the \(y\)-axis, as shown in Figure 7.2.12.

Mathematica renders the solid in Figure 7.2.11

Figure 7.2.12

Solution For the region shown in Figure 7.2.12, the outer radius is simply \(R=1\). There is no obvious formula that represents the inner radius. When \(0 \leqslant y \leqslant 1, r=0\), but when \(1 \leqslant y \leqslant 2\), \(r\) is determined by the equation \(y=x^{2}+1\), which implies that \(r=\sqrt{y=1}\).

$$r(y)=\left\{\begin{matrix} 0, & 0 \leqslant y \leqslant 1 \\ \sqrt{y-1}, & 1 \leqslant y \leqslant 2 \end{matrix} \right. $$

Using this definition for the inner radius, two integrals can be used to find the volume.

$$V$$	$$=\pi \int_{0}^{1} \left( 1^{2}-0^{2} \right)\:dy+\pi \int_{1}^{2} \left[ 1^{2}- \left(\sqrt{y-1} \right)^{2} \right ]\:dy \:\:\:\: $$	Apply The Washer Method
	$$=\pi \int_{0}^{1} 1\:dy+\pi \int_{1}^{2} \left(2-y \right) \:dy $$	Simplify
	$$=\pi \Large \left[ y \right]_{0}^{1} \Large + \pi \left[ 2y-\frac{y^{2}}{2} \right]_{1}^{2} $$	Integrate
	$$= \pi + \pi \left(4-2-2+ \frac{1}{2} \right) = \frac{3 \pi}{2} $$

Note that the first integral \(\pi \int_{0}^{1} 1\:dy\) represents the volume for a right circular cylinder with radius height 1. This volume can be determined without using calculus.

Example 7.2.5 Manufacturing

Figure 7.2.13

A manufacturer takes a metal sphere with radius 5 inches, as shown in Figure 7.2.13(a), and drills a 3 inch diameter hole straight through the center. What is the metal ring's volume? Solution Imagine the circle's cross-section inscribed on the \(xy\) plane with its center at the origin. The circle has the area \(x^{2}+y^{2}=5^{2}\). Next remove the volume occupied by the hole and everything below \(y=0\). The resulting shape is rendered in Figure 7.2.13(b). Now The Washer Method can be applied. Because the has has the radius 3 inches, let \(y=3\) and solve \(x^{2}+y^{2}=25\) to determine that the limits for integration are \(x=\pm4\). This makes the inner radius is \(r(x)=3\) and the outer radius is \(R(x)=\sqrt{25-x^{2}}\). $$V$$ $$=\pi \int_{-4}^{4} \left( \sqrt{25-x^{2}} \right)-(3)^{3}\:dx \:\:\:\: $$ Apply The Washer Method $$ = \pi \int_{-4}^{4} \left(16-x^{2} \right) \:dx $$ Simplify $$=\pi \left[ 16x-\frac{3^{3}}{3} \right]_{-4}^{4} = \frac{256 \pi}{3} \text{ cubic inches }$$ Integrate

Solids with Known Cross Sections

With the disk method, you can find the volume for a solid having a circular cross section whose area is \(A=\pi R^{2}\). This method can be generalized to solids with any shape, if there is a known a formula for the cross-section's area. Some common cross sections are squares, rectangles, triangles, semicircles, and trapezoids.

Definition 7.2.3 Volumes for Solids with Known Cross Sections

1. For cross sections of area taken perpendicular to the axis.
$$Volume = \int_{a}^{b} A(x)\:dx $$	See Figure 7.2.14(a).
2. For cross sections of area taken perpendicular to the axis.
$$Volume = \int_{c}^{d} A(y)\:dy $$	See Figure 7.2.14(b).

Figure 7.2.14

Example 7.2.6 Triangular Cross Sections

Figure 7.2.15

Find the volume for the triangular solid shown in Figure 7.2.15. The base for the solid is the region bounded by the lines $$ f(x)=1-\frac{x}{2},\: g(x)=-1+\frac{x}{2} \text{, and }x=0$$ The cross sections perpendicular to the axis are equilateral triangles. Solution The base and area for each triangular cross-section are as follows. $$Base$$ $$= \left(1-\frac{x}{2} \right)- \left(-1+\frac{x}{2} \right)=2-x $$ Base Length $$Area$$ $$=\frac{\sqrt{3}}{4}(base)^{2} $$ Equilateral triangle area $$A(x)$$ $$=\frac{\sqrt{3}}{4} \left( 2-x \right)^{2} $$ Cross-section area Because \(x\) ranges from 0 to 2, the solid's volume is $$V=\int_{a}^{b}A(x)\:dx=\int_{0}^{2}\frac{\sqrt{3}}{4}(2-x)^{2}\:dx=-\frac{\sqrt{3}}{4} \left[ \frac{ \left( 2-x \right)^{3}}{3} \right]_{0}^{2} =\frac{2\sqrt{3}}{3}$$

Example 7.2.7 Square Pyramid Volume

Figure 7.2.16

Prove that the volume for a square pyramid is $$ V=\frac{1}{3}hB $$ where \(h\) is the height and \(B\) is the area for the base. Solution As shown in Figure 7.2.16, you can pass a plane through the pyramid and parallel to the base at height \(y\) to form a square cross-section with sides \(b^{\prime }\) long. Repeating the process, you can show that $$ \frac{b^{\prime }}{b}=\frac{h-y}{h} \text{ or }b^{\prime }=\frac{b}{h}(h-y) $$ where the pyramid's base \(b\) long on each side. This makes the area formula $$ A(y)= \left( b^{\prime } \right)^{2}= \frac{b^{2}}{h^{2}} \left( h-y \right)^{2}.$$ Integrating between 0 and \(h\) produces $$V$$ $$ = \int_{0}^{h} A(y)\:dy $$ Apply The Cross-section area formula to the \(y\)-axis $$ = \int_{0}^{h} \frac{b^{2}}{h^{2}} \left( h-y \right)^{2}\:dy $$ Substitute the pyramid formula. $$ = \frac{b^{2}}{h^{2}} \int_{0}^{h} \left( h-y \right)^{2}\:dy $$ Simplify $$ = \left( \frac{b^{2}}{h^{2}} \right) \left[ \frac{ \left( h-y \right)^{3}}{3} \right]_{0}^{h} $$ Integrate $$ = \frac{b^{2}}{h^{2}} \left( \frac{h^{3}}{3} \right) = \frac{1}{3}hB $$ \(B=b^{2}\)

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Parent Article: Calculus II 07 Integration Applications