Determining Volumes by Slicing: Learn It 3

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Determining Volumes by Slicing: Learn It 3

The Disk Method

When we use the slicing method with solids of revolution, it is often called the disk method because, for solids of revolution, the slices used to over approximate the volume of the solid are disks.

To see this, consider the solid of revolution generated by revolving the region between the graph of the function $f(x)={(x-1)}^{2}+1$ and the $x\text{-axis}$ over the interval $\left[-1,3\right]$ around the $x\text{-axis}\text{.}$

This figure has four graphs. The first graph, labeled “a” is a parabola f(x)=(x-1)^2+1. The curve is above the x-axis and intersects the y-axis at y=2. Under the curve in the first quadrant is a vertical rectangle starting at the x-axis and stopping at the curve. The second graph, labeled “b” is the same parabola as in the first graph. The rectangle under the parabola from the first graph has been rotated around the x-axis forming a solid disk. The third graph labeled “c” is the same parabola as the first graph. There is a shaded region bounded above by the parabola, to the left by the line x=-1 and to the right by the line x=3, and below by the x-axis. The fourth graph labeled “d” is the same parabola as the first graph. The region from the third graph has been revolved around the x-axis to form a solid. — Figure 9. (a) A thin rectangle for approximating the area under a curve. (b) A representative disk formed by revolving the rectangle about the $x\text{-axis}\text{.}$ (c) The region under the curve is revolved about the $x\text{-axis},$ resulting in (d) the solid of revolution.

We already used the formal Riemann sum development of the volume formula when we developed the slicing method. We know that

V = \int_{a}^{b} A (x) d x

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

The only difference with the disk method is that we know the formula for the cross-sectional area ahead of time; it is the area of a circle.

the disk method

Let $f(x)$ be continuous and nonnegative.

Define $R$ as the region bounded above by the graph of $f(x),$ below by the $x\text{-axis,}$ on the left by the line $x=a,$ and on the right by the line $x=b.$

Then, the volume of the solid of revolution formed by revolving $R$ around the $x\text{-axis}$ is given by:

V = \int_{a}^{b} π {[f (x)]}^{2} d x .

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

Returning to our example, the volume is given by:

\begin{array}{cc} V & = \int_{a}^{b} π {[f (x)]}^{2} d x \\ = \int_{- 1}^{3} π {[{(x - 1)}^{2} + 1]}^{2} d x = π \int_{- 1}^{3} {[{(x - 1)}^{4} + 2 {(x - 1)}^{2} + 1]}^{2} d x \\ = π {[\frac{1}{5} {(x - 1)}^{5} + \frac{2}{3} {(x - 1)}^{3} + x] |}_{- 1}^{3} = π [(\frac{32}{5} + \frac{16}{3} + 3) - (- \frac{32}{5} - \frac{16}{3} - 1)] = \frac{412 π}{15} {units}^{3} . \end{array}

$\begin{array}{cc}\hfill V& ={\displaystyle\int }_{a}^{b}\pi {\left[f(x)\right]}^{2}dx\hfill \\ & ={\displaystyle\int }_{-1}^{3}\pi {\left[{(x-1)}^{2}+1\right]}^{2}dx=\pi {\displaystyle\int }_{-1}^{3}{\left[{(x-1)}^{4}+2{(x-1)}^{2}+1\right]}^{2}dx\hfill \\ & =\pi {\left[\frac{1}{5}{(x-1)}^{5}+\frac{2}{3}{(x-1)}^{3}+x\right]|}_{-1}^{3}=\pi \left[(\frac{32}{5}+\frac{16}{3}+3)-(-\frac{32}{5}-\frac{16}{3}-1)\right]=\frac{412\pi }{15}{\text{units}}^{3}.\hfill \end{array}$

Use the disk method to find the volume of the solid of revolution generated by rotating the region between the graph of $f(x)=\sqrt{x}$ and the $x\text{-axis}$ over the interval $\left[1,4\right]$ around the $x\text{-axis}\text{.}$

The graphs of the function and the solid of revolution are shown in the following figure.

This figure has two graphs. The first graph labeled “a” is the curve f(x) = squareroot(x). It is an increasing curve above the x-axis. The curve is in the first quadrant. Under the curve is a region bounded by x=1 and x=4. The bottom of the region is the x-axis. The second graph labeled “b” is the same curve as the first graph. The solid region from the first graph has been rotated around the x-axis to form a solid region. — Figure 10. (a) The function $f(x)=\sqrt{x}$ over the interval $\left[1,4\right].$ (b) The solid of revolution obtained by revolving the region under the graph of $f(x)$ about the $x\text{-axis}.$

We have

\begin{array}{cc} V & = \int_{a}^{b} π {[f (x)]}^{2} d x \\ = \int_{1}^{4} π {[\sqrt{x}]}^{2} d x = π \int_{1}^{4} x d x \\ = {\frac{π}{2} x^{2} |}_{1}^{4} = \frac{15 π}{2} . \end{array}

$\begin{array}{cc}\hfill V& ={\displaystyle\int }_{a}^{b}\pi {\left[f(x)\right]}^{2}dx\hfill \\ & ={\displaystyle\int }_{1}^{4}\pi {\left[\sqrt{x}\right]}^{2}dx=\pi {\displaystyle\int }_{1}^{4}xdx\hfill \\ & ={\frac{\pi }{2}{x}^{2}|}_{1}^{4}=\frac{15\pi }{2}.\hfill \end{array}$

The volume is $\frac{(15\pi )}{2}$ units³.

So far, our examples have all concerned regions revolved around the $x\text{-axis,}$ but we can generate a solid of revolution by revolving a plane region around any horizontal or vertical line.

the disk method for solids of revolution around the $y$ -axis

Let $g(y)$ be continuous and nonnegative.

Define $Q$ as the region bounded on the right by the graph of $g(y),$ on the left by the $y\text{-axis,}$ below by the line $y=c,$ and above by the line $y=d.$

Then, the volume of the solid of revolution formed by revolving $Q$ around the $y\text{-axis}$ is given by:

V = \int_{c}^{d} π {[g (y)]}^{2} d y .

$V={\displaystyle\int }_{c}^{d}\pi {\left[g(y)\right]}^{2}dy.$

In the next example, we look at a solid of revolution that has been generated by revolving a region around the $y\text{-axis}\text{.}$ The mechanics of the disk method are nearly the same as when the $x\text{-axis}$ is the axis of revolution, but we express the function in terms of $y$ and we integrate with respect to $y$ as well.

Let $R$ be the region bounded by the graph of $g(y)=\sqrt{4-y}$ and the $y\text{-axis}$ over the $y\text{-axis}$ interval $\left[0,4\right].$ Use the disk method to find the volume of the solid of revolution generated by rotating $R$ around the $y\text{-axis}\text{.}$

Figure 11 shows the function and a representative disk that can be used to estimate the volume. Notice that since we are revolving the function around the $y\text{-axis,}$ the disks are horizontal, rather than vertical.

This figure has two graphs. The first graph labeled “a” is the curve g(y) = squareroot(4-y). It is a decreasing curve starting on the y-axis at y=4. Between the curve and the y-axis is a horizontal rectangle. The rectangle starts at the y-axis and stops at the curve. The second graph labeled “b” is the same curve as the first graph. The rectangle from the first graph has been rotated around the y-axis to form a horizontal disk. — Figure 11. (a) Shown is a thin rectangle between the curve of the function $g(y)=\sqrt{4-y}$ and the $y\text{-axis}\text{.}$ (b) The rectangle forms a representative disk after revolution around the $y\text{-axis}\text{.}$

The region to be revolved and the full solid of revolution are depicted in the following figure.

This figure has two graphs. The first graph labeled “a” is the curve g(y) = squareroot(4-y). It is a decreasing curve starting on the y-axis at y=4. The region formed by the x-axis, the y-axis, and the curve is shaded. This region is in the first quadrant. The second graph labeled “b” is the same curve as the first graph. The region from the first graph has been rotated around the y-axis to form a solid. — Figure 12. (a) The region to the left of the function $g(y)=\sqrt{4-y}$ over the $y\text{-axis}$ interval $\left[0,4\right].$ (b) The solid of revolution formed by revolving the region about the $y\text{-axis}\text{.}$

To find the volume, we integrate with respect to $y.$ We obtain

\begin{array}{cc} V & = \int_{c}^{d} π {[g (y)]}^{2} d y \\ = \int_{0}^{4} π {[\sqrt{4 - y}]}^{2} d y = π \int_{0}^{4} (4 - y) d y \\ = {π [4 y - \frac{y^{2}}{2}] |}_{0}^{4} = 8 π . \end{array}

$\begin{array}{cc}\hfill V& ={\displaystyle\int }_{c}^{d}\pi {\left[g(y)\right]}^{2}dy\hfill \\ & ={\displaystyle\int }_{0}^{4}\pi {\left[\sqrt{4-y}\right]}^{2}dy=\pi {\displaystyle\int }_{0}^{4}(4-y)dy\hfill \\ & ={\pi \left[4y-\frac{{y}^{2}}{2}\right]|}_{0}^{4}=8\pi .\hfill \end{array}$

The volume is $8\pi$ units³.

Determining Volumes by Slicing: Learn It 3

The Disk Method

the disk method

the disk method for solids of revolution around the yyy-axis

the disk method for solids of revolution around the $y$ -axis