Chapter 8.3, Problem 39E

a.

To determine

To find: The find the volume of given solid.

Answer to Problem 39E

The area is $\sqrt{3}\sin x$ Unit² and the volume of the solid is $2\sqrt{3}$ Unit³

Explanation of Solution

Given information : The base of the solid curve $y=2\sqrt{\sin x}$ and the interval $\left[0,\pi \right]$ on the X-axis.

The cross section perpendicular to the X-axis are Equilateral triangles with bases running from the X-axis to the curve as shown in the figure.

  

Formula used : The volume of a solid of known integrable cross-section area $A\left(x\right)$ from $x=a$ to $x=b$ is the integral of $A$ from $a$ to $b$ ,

  $V={\displaystyle \underset{a}{\overset{b}{\int }}A\left(x\right)dx}$ .

The area of a equilateral triangle of side “ $a$ ” is $\sqrt{\frac{3}{4}}{a}^2$ Unit²

Calculation :

The cross-section perpendicular to X-axis is equilateral triangle.

So, area

  $\begin{array}{c}A\left(x\right)=\sqrt{\frac{3}{4}}{y}^2\\ =\sqrt{\frac{3}{4}}{\left(2\sqrt{\sin x}\right)}^2\\ =\sqrt{\frac{3}{4}}\times 4\times sinx\\ =\sqrt{3}\sin x\end{array}$

And volume of the solid is,

  $\begin{array}{l}{\displaystyle \underset{0}{\overset{\pi }{\int }}A\left(x\right)}dx\\ ={\displaystyle \underset{0}{\overset{\pi }{\int }}\sqrt{3}\sin x}dx\\ =\sqrt{3}{\left[-\cos x\right]}_0^{\pi }\\ =\sqrt{3}\left[-\left(-1\right)-\left(-1\right)\right]\\ =\sqrt{3}\left(1+1\right)\\ =2\sqrt{3}\end{array}$

b.

To determine

To find: The find the volume of given solid.

Answer to Problem 39E

The volume of the solid is $8$ Unit³.

Explanation of Solution

Given information : The base of the solid curve $y=2\sqrt{\sin x}$ and the interval $\left[0,\pi \right]$ on the X-axis.

The cross section perpendicular to the X-axis are Equilateral triangles with bases running from the X-axis to the curve as shown in the figure.

  

The cross-section perpendicular to the X-axis are squares with bases running from the X-axis to the curve.

Formula used : The volume of a solid of known integrable cross-section area $A\left(x\right)$ from $x=a$ to $x=b$ is the integral of $A$ from $a$ to $b$ ,

  $V={\displaystyle \underset{a}{\overset{b}{\int }}A\left(x\right)dx}$ .

The area of a equilateral triangle of side “ $a$ ” is $\sqrt{\frac{3}{4}}{a}^2$ Unit²

The area of the Square of side “ $a$ ” is $a^2$ Unit²

The cross section perpendicular to X-axis is a square.

So, the area

  $\begin{array}{c}A\left(x\right)=y^2\\ ={\left(2\sqrt{\sin x}\right)}^2\\ =4\sin x\end{array}$

Volume of the solid is,

  $\begin{array}{l}{\displaystyle \underset{0}{\overset{\pi }{\int }}A\left(x\right)}dx\\ ={\displaystyle \underset{0}{\overset{\pi }{\int }}4\sin x}dx\\ =4{\left[-\cos x\right]}_0^{\pi }\\ =4\left[-\left(-1\right)-\left(-1\right)\right]\\ =4\left(1+1\right)\\ =8\end{array}$

The volume of the solid is $8$ Unit³.