Chapter 6.2, Problem 2E

To determine

Find the volume of the solid obtained by rotating the region bounded by the given curves.

Answer to Problem 2E

The volume of the solid obtained by rotating the region bounded by the given curves is $\frac{8}{15}\pi$

Explanation of Solution

Given information: Given curves $y=1-x^2,y=0$

Calculation:

Draw the graph of $y=1-x^2,y=0$ rotate around $x-\text{axis}$ .

  

Area of typical disk:

  $\begin{array}{l}A=\pi r^2\\ A=\pi {\left(1-x^2\right)}^2\text{ Since }r=1-x^2\end{array}$

Volume $={\displaystyle \underset{a}{\overset{b}{\int }}A\left(x\right)}dx,$ where $A\left(x\right)=$ area of typical disk.

Here $X$ values for this graph range from $0\text{ to }1$ thus, $a=0$ , $b=1$ .

  $A\left(x\right)=\pi {\displaystyle \underset{0}{\overset{1}{\int }}{\left(1-x^2\right)}^2}$

  $A\left(x\right)=\pi {\displaystyle \underset{0}{\overset{1}{\int }}\left(1-2x^2+x^4\right)}dx$

Integrate each term using power rule, ${\displaystyle \underset{a}{\overset{b}{\int }}{x}^{n}dx=\frac{x^{n+1}}{n+1}}+c$

  $A\left(x\right)=\pi {\left[x-\frac{2}{3}{x}^3+\frac{1}{5}{x}^5\right]}_0^1$

Use fundamental theorem, ${\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)dx=F\left(b\right)}-F\left(a\right)\text{ where }F\left(X\right)$ is the Ant derivative of $f\left(x\right)$

  $=\pi \left\{\left[1-\frac{2}{3}{\left(1\right)}^3+\frac{1}{5}{\left(1\right)}^5\right]-\left(0\right)\right\}$

  $=\frac{8\pi }{15}$

Thus, volume of the solid obtained by rotating the region bounded by the given curves is $\frac{8}{15}\pi$