Chapter 8.3, Problem 36E

To determine

To find : the volume of the solid.

Answer to Problem 36E

The volume of the solid is $\frac{5\pi }{6}$ .

Explanation of Solution

Given information : The curve about the y -axis $y=x^2,y=2-x,x=0$ for $x\ge 0$ .

Calculation :

  $\begin{array}{l}{y}_1=2-x\\ y_2=x^2\\ x=0\\ x\ge 0\end{array}$

Region specified rotate this around the y -axis.

  $V=2\pi {\displaystyle \underset{0}{\overset{1}{\int }}x\left(y_1-y_2\right)}dx$

Shell formula: $V=2\pi {\displaystyle \underset{0}{\overset{1}{\int }}rh}dx$

The volume of the solid:

  $\begin{array}{l}{x}^2=2-x\\ x^2+x-2=0\\ \left(x+2\right)\left(x-1\right)=0\\ x=1,-2\\ V=2\pi {\displaystyle \underset{0}{\overset{1}{\int }}x\left(2-x-x^2\right)}dx\\ V=2\pi {\displaystyle \underset{0}{\overset{2}{\int }}\left(2x-x^2-x^3\right)}dx\\ V=2\pi {\left[x^2-\frac{x^3}{3}-\frac{x^4}{4}\right]}_0^1\\ V=2\pi \left(1-\frac{1}{3}-\frac{1}{4}\right)\\ V=\frac{5\pi }{6}\end{array}$