Q. 64 - Calculus

Question

The shell and disc methods for estimating the volume of if both solids of revolution are computable, they always provide the same result. Show that this holds true for any solid of rotation defined by rotating the area between the x-axis on $[a, b]$ and a linear function $y = mx + c,$ where $m, c, a,$ and $b$ are positive.

Step-by-Step Solution

Answer

The volume obtained by disc method and shell method yields the same result.

Step 1: Given information

The linear function $y=mx+c$

Step 2: Calculation

The major goal is to demonstrate that, given a solid in revolution, both the disc approach and the shell method provide the same volume.

Take into account the linear function$y=mx+c$, where $m$ and $c$ are positive numbers. Consider the solid formed by rotating the area enclosed by the line about the x-axis on$[a,b],$ where $a$ and $b$ are positive, to be the solid of revolution.

The scenario as shown in the accompanying diagram

The region created by the line $y=mx+c$ on $[a,b]$ is seen in the diagram.

Think about a solid of revolution created by rotating the area about an axis that is horizontal to $[a,b].$

Think of the solid's cross sections as discs with radii defined by the continuous function $r\left(x\right).$

The solid's volume is then given by the next definite integral.

$\left(\begin{array}{l}\text{ Volume of solid }\\ \text{ with disc cross sections }\end{array}\right)=\pi {\int }_a^b\left(r\right(x){)}^{2}dx$

Consider discs with a radius of $r\left(x\right)=m x+c$ in this case for the region defined by the linear function $y=m x+c.$

To calculate the solid's revolution volume, use the formula.

$$\begin{gathered} \pi {\int }_a^b\left(r\right(x){)}^{2}dx=\pi {\int }_a^b(mx+c{)}^{2}dx \\ =\pi {\left[\frac{(mx+c{)}^3}{3m}\right]}_a^b \text{ [Integrate] } \\ =\frac{\pi }{3m}\left((mb+c{)}^3-(ma+c{)}^3\right)\text{ [Apply limits] } \end{gathered}$$

The solid of rotation has a volume of due to the disc approach is

$\frac{\pi }{3m}\left((mb+c{)}^3-(ma+c{)}^3\right).$

Step 3: Further Calculation

Think about a solid of revolution created by rotating a region about an axis that is horizontal to $[a, b].$

Use nested shells to describe the solid, with the continuous functions $r\left(y\right)$ and $h\left(y\right)$ standing in for their average radii and heights on$[a, b].$

The solid's volume is then determined by the definite integral shown below:

 $\left(\begin{array}{l}\text{ Volume of solid }\\ \text{ with nested shells }\end{array}\right)=2\pi {\int }_a^{b}r\left(y\right)h\left(y\right)dy$

The limit of integration in this case is between $0$ and $mb+c$ for the region defined by the linear function $y=mx+c$. Do the integral calculation by adding the integrals from $0$ to $ma+c$ and $ma+c$ to $mb+c$

Think about shells whose radii are $r\left(y\right)=y.$

Each shell's height for the first integral is $h\left(y\right)=b-a.$

Solve for x using the equation $y=mx+c$ to get the height of the shells in the second integral.

$y=mx+c$

$y-c=mx \left[Subtractconbothsides\right]$

$\frac{y-c}{m}=x \left[Dividemonbothsides\right]$

To calculate the height, use the value of $x$ as follows.

$$\begin{gathered} h\left(y\right)=b-x \\ =b-\frac{y-c}{m} \end{gathered}$$

To calculate the solid's revolution volume, use the formula.

$$\begin{gathered} 2\pi {\int }_a^{b}r\left(y\right)h\left(y\right)dy=2\pi {\int }_0^{ma+c}y(b-a)dy+2\pi {\int }_{ma+c}^{mb+c}y\left(b-\frac{y-c}{m}\right)dy \\ =2\pi {\left[\frac{y^2}{2}(b-a)\right]}_0^{ma+c}+2\pi {\int }_{ma+c}^{mb+c}\left(by-\frac{y^2}{m}+\frac{cy}{m}\right)dy [\text{ Simplify]} \\ =\frac{2\pi }{2}(b-a\left)\right(ma+c{)}^2+2\pi {\left[\frac{b{y}^2}{2}-\frac{y^3}{3m}+\frac{c{y}^2}{2m}\right]}_{ma+c}^{mb+c} [\text{ Simplify]} \\ =\pi (b-a\left)\right(ma+c{)}^2+2\pi {\left[\frac{mb+c}{2m}{y}^2-\frac{y^3}{3m}\right]}_{ma+c}^{mb+c} \left[\begin{array}{l}\text{ Group coefficients }\\ \text{ of like-terms }\end{array}\right] \end{gathered}$$

Apply the integration limit and further simplify as follows:

$$\begin{gathered} 2\pi {\int }_a^{b}r\left(y\right)h\left(y\right)dy=\pi (b-a)(ma+c{)}^2+2\pi {\left[\frac{mb+c}{2m}{y}^2-\frac{y^3}{3m}\right]}_{ma+c}^{mb+c} \\ =\left\{\begin{array}{c}\pi (b-a)(ma+c{)}^2+2\pi \left(\begin{array}{l}\left(\frac{mb+c}{2m}\right)(mb+c{)}^2-\frac{(mb+c{)}^3}{3m}\\ -\left(\frac{mb+c}{}\right)(ma+c{)}^2+\frac{(ma+c{)}^3}{}\end{array}\right)\end{array}\right. \end{gathered}$$

$=\left\{\pi (b-a)(ma+c{)}^2+\frac{2\pi }{6m}\left(\begin{array}{l}3(mb+c{)}^3-2(mb+c{)}^3\\ -3(mb+c)(ma+c{)}^2+2(ma+c{)}^3\end{array}\right)\right\}\left[\mathrm{LCM}\right]$

Add $3m$ to the first term on the right side, multiply by $3m$, and then further simplify as follows:

$$\begin{gathered} 2\pi {\int }_a^{b}r\left(y\right)h\left(y\right)dy=\left\{\frac{\pi }{3m}3m(b-a)(ma+c{)}^2+\frac{\pi }{3m}\left(\begin{array}{l}3(mb+c{)}^3-2(mb+c{)}^3\\ -3(mb+c)(ma+c{)}^2+2(ma+c{)}^3\end{array}\right)\right\}=\frac{\pi }{3m}\left\{\left(\begin{array}{l}3mb(ma+c{)}^2-3ma(ma+c{)}^2+3(mb+c{)}^3-2(mb+c{)}^3\\ -3mb(ma+c{)}^2-3c(ma+c{)}^2+2(ma+c{)}^3\end{array}\right)\right\} \\ =\frac{\pi }{3m}\left(-3ma(ma+c{)}^2+(mb+c{)}^3-3c(ma+c{)}^2+2(ma+c{)}^3\right) \\ =\frac{\pi }{3m}\left((mb+c{)}^3+2(ma+c{)}^3-3(ma+c{)}^2(ma+c)\right) \end{gathered}$$

Simplify further as follows:

$$\begin{gathered} 2\pi {\int }_a^{b}r\left(y\right)h\left(y\right)dy=\frac{\pi }{3m}\left((mb+c{)}^3+2(ma+c{)}^3-3(ma+c{)}^2(ma+c)\right) \\ =\frac{\pi }{3m}\left((mb+c{)}^3+2(ma+c{)}^3-3(ma+c{)}^3\right) \\ =\frac{\pi }{3m}\left((mb+c{)}^3-(ma+c{)}^3\right) \end{gathered}$$

Consequently, the solid of evolution using the shell approach has a value of

$=\frac{\pi }{3m}\left((mb+c{)}^3-(ma+c{)}^3\right)$

To determine whether the outcome is the same, compare this to the volume discovered using the disc approach.

As a result, both the disc approach and the shell method produce the same volume for a solid in revolution.