Chapter A5.1, Problem 38E

(a)

To determine

To show: the generated volumes have the ratio 4:1.

Explanation of Solution

Given:

The parabola $y^2=kx,$$k>0,$

  

Proof:

The equation of solid formed by revolving A around the $y$ -axis,

  $\begin{array}{l}{V}_A={\displaystyle {\int }_0^{\sqrt{kx}}\pi }{\left(\frac{y^2}{k}\right)}^{2}dy\\ =\frac{\pi }{k^2}{\displaystyle {\int }_0^{\sqrt{kx}}{y}^4}dy\\ =\frac{\pi x^2\sqrt{kx}}{5}\end{array}$

And, the volume of the right circular cylinder formed by revolving the rectangle about the $y$ -axis is, $V_{\text{Total}}=\pi x^2\sqrt{kx}$ .

The volume of the solid formed by revolving $B$ around the $y$ -axis is

  $\begin{array}{l}{V}_B=V_{\text{Total }}-V_A\\ =\pi x^2\sqrt{kx}-\frac{\pi x^2\sqrt{kx}}{5}\\ =\frac{4\pi x^2\sqrt{kx}}{5}\end{array}$

Thus, the ratio of $V_B$ to $V_A$ is 4: 1.

Conclusion:

Therefore, the ratio of $V_B$ to $V_A$ is 4: 1.

(b)

To determine

To find: the ratio of the volume of the solids.

Answer to Problem 38E

The ratio of $V_B$ to $V_A$ is 1: 1.

Explanation of Solution

Calculation:

  $\begin{array}{}$ V_B={\displaystyle {\int }_0^x\pi }{(\sqrt{kt})}^{2}dt$$ =\pi k{\displaystyle {\int }_0^{x}t}dt$$ =\frac{\pi k{x}^2}{2}$\end{array}$

And, the volume of the right circular cylinder formed by revolving the rectangle about the $x$ -axis is.

  $\begin{array}{}$ V_{\text{Tout }}=\pi {(\sqrt{bx})}^{2}x$$ =\pi k{x}^2$\end{array}$

The volume of the solid formed by revolving $B$ around the $x$ -axis is

  $\begin{array}{l}{V}_B=V_{Total}-V_A\\ =\pi k{x}^2-\frac{\pi k{x}^2}{2}\\ =\frac{\pi k{x}^2}{2}\end{array}$

Thus, the ratio of $V_B$ to $V_A$ is 1: 1.

Conclusion:

Therefore, the ratio of $V_B$ to $V_A$ is 1: 1.