Chapter 7, Problem 48RE

To determine

To find:The volume of the solid generated by resolving the region bounded by the curves.

Answer to Problem 48RE

The volume of the solid generated is $V=9.7716\text{ cu}\text{. units}$ .

Explanation of Solution

Given information:

  $\begin{array}{l}y=3^{1-x^2}\\ y=\frac{x^2-3}{10}\end{array}$

Formula used:

The formula for cylindrical shell is,

  $V={\displaystyle {\int }_a^{b}2\pi rhdx}$

Calculation:

Let's graph thegiven area and draw some conclusions based on it.

  

From the graph, it is noted that the point of intersection is $\left(1.8933,0\right)$ .

The distance from the $y$ -axis to the vertical strip is $r=x$ , which is also based on the sketch. Consequently, the graph’sdifference is represented by the height of the strip.

Thus, the height of the strip is the difference between the graphs.

  $h=3^{1-x^2}-\frac{x^2-3}{10}$

Substituting the value of $h$ in the volume formula to get,

  $V={\displaystyle {\int }_0^{1.8933}2\pi \left(x\right)\left(3^{1-x^2}-\frac{x^2-3}{10}\right)dx}$

The above integral is evaluated by dividing into two integral,

  $V=I_1+I_2$

Where,

  $\begin{array}{l}{I}_1={\displaystyle {\int }_0^{1.8933}2\pi \left(x\right)\left(3^{1-x^2}\right)dx}\\ I_2=-{\displaystyle {\int }_0^{1.8933}2\pi \left(x\right)\left(\frac{x^2-3}{10}\right)dx}\end{array}$

Let’s take,

  $\begin{array}{c}u=1-x^2\\ du=-2xdx\end{array}$

Solving the integrals by using the formula,

  ${\displaystyle \int a^{u}du=\frac{a^u}{\ln a}+C}$

Thus,

  $\begin{array}{c}{I}_1=-\pi {\displaystyle {\int }_0^{1.8933}{3}^{u}du}\\ =-\pi {\left.\left[\frac{3^u}{\ln 3}\right]\right|}_0^{1.8933}\\ =-\pi {\left.\left[\frac{3^{1-x^2}}{\ln 3}\right]\right|}_0^{1.8933}\\ =\frac{-\pi }{\ln 3}\left(3^{1-{\left(1.8933\right)}^2}-3\right)\\ I_1\approx 8.4116\end{array}$

Let’s take,

  $\begin{array}{c}u=x^2-3\\ du=2xdx\end{array}$

Thus,

  $\begin{array}{c}{I}_2=-\frac{\pi }{10}{\displaystyle {\int }_0^{1.8933}udu}\\ =-\frac{\pi }{10}{\left.\left[\frac{u^2}{2}\right]\right|}_0^{1.8933}\\ =-\frac{\pi }{10}{\left.\left[\frac{{\left(x^2-3\right)}^2}{2}\right]\right|}_0^{1.8933}\\ =-\frac{\pi }{20}\left({\left({1.8933}^2-3\right)}^2-3^2\right)\\ I_2\approx 1.360\end{array}$

Now, adding both the integrals to get the volume of the solid,

  $\begin{array}{c}V=I_1+I_2\\ =8.4116+1.36\\ =9.7716\end{array}$

Therefore, the volume of the solid generated is $V=9.7716\text{ cu}\text{. units}$ .