# 6.2 Determining volumes by slicing  (Page 6/12)

 Page 6 / 12

Rather than looking at an example of the washer method with the $y\text{-axis}$ as the axis of revolution, we now consider an example in which the axis of revolution is a line other than one of the two coordinate axes. The same general method applies, but you may have to visualize just how to describe the cross-sectional area of the volume.

## The washer method with a different axis of revolution

Find the volume of a solid of revolution formed by revolving the region bounded above by $f\left(x\right)=4-x$ and below by the $x\text{-axis}$ over the interval $\left[0,4\right]$ around the line $y=-2.$

The graph of the region and the solid of revolution are shown in the following figure. (a) The region between the graph of the function f ( x ) = 4 − x and the x -axis over the interval [ 0 , 4 ] . (b) Revolving the region about the line y = −2 generates a solid of revolution with a cylindrical hole through its middle.

We can’t apply the volume formula to this problem directly because the axis of revolution is not one of the coordinate axes. However, we still know that the area of the cross-section is the area of the outer circle less the area of the inner circle. Looking at the graph of the function, we see the radius of the outer circle is given by $f\left(x\right)+2,$ which simplifies to

$f\left(x\right)+2=\left(4-x\right)+2=6-x.$

The radius of the inner circle is $g\left(x\right)=2.$ Therefore, we have

$\begin{array}{cc}\hfill V& ={\int }_{0}^{4}\pi \left[{\left(6-x\right)}^{2}-{\left(2\right)}^{2}\right]dx\hfill \\ & =\pi {\int }_{0}^{4}\left({x}^{2}-12x+32\right)dx\phantom{\rule{0.2em}{0ex}}\text{}\phantom{\rule{0.2em}{0ex}}={\pi \left[\frac{{x}^{3}}{3}-6{x}^{2}+32x\right]\phantom{\rule{0.2em}{0ex}}|}_{0}^{4}=\frac{160\pi }{3}\phantom{\rule{0.2em}{0ex}}{\text{units}}^{3}.\hfill \end{array}$

Find the volume of a solid of revolution formed by revolving the region bounded above by the graph of $f\left(x\right)=x+2$ and below by the $x\text{-axis}$ over the interval $\left[0,3\right]$ around the line $y=-1.$

$60\pi$ units 3

## Key concepts

• Definite integrals can be used to find the volumes of solids. Using the slicing method, we can find a volume by integrating the cross-sectional area.
• For solids of revolution, the volume slices are often disks and the cross-sections are circles. The method of disks involves applying the method of slicing in the particular case in which the cross-sections are circles, and using the formula for the area of a circle.
• If a solid of revolution has a cavity in the center, the volume slices are washers. With the method of washers, the area of the inner circle is subtracted from the area of the outer circle before integrating.

## Key equations

• Disk Method along the x -axis
$V={\int }_{a}^{b}\pi {\left[f\left(x\right)\right]}^{2}dx$
• Disk Method along the y -axis
$V={\int }_{c}^{d}\pi {\left[g\left(y\right)\right]}^{2}dy$
• Washer Method
$V={\int }_{a}^{b}\pi \left[{\left(f\left(x\right)\right)}^{2}-{\left(g\left(x\right)\right)}^{2}\right]dx$

Derive the formula for the volume of a sphere using the slicing method.

Use the slicing method to derive the formula for the volume of a cone.

Use the slicing method to derive the formula for the volume of a tetrahedron with side length $a.$

Use the disk method to derive the formula for the volume of a trapezoidal cylinder.

Explain when you would use the disk method versus the washer method. When are they interchangeable?

For the following exercises, draw a typical slice and find the volume using the slicing method for the given volume.

A pyramid with height 6 units and square base of side 2 units, as pictured here. 8 units 3

A pyramid with height 4 units and a rectangular base with length 2 units and width 3 units, as pictured here. determine the area of the region enclosed by x²+y=1,2x-y+4=0
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CALCULUS
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