Chapter 7.3, Problem 41E

a.

To determine

To find: The volume of the equilateral triangle.

Answer to Problem 41E

The volume of the equilateral triangle is $V_T=2\sqrt{3}$ .

Explanation of Solution

Given information:

The base of a solid is the region between the curve is $y=2\sqrt{\sin x}$ .

Calculation:

The base of the triangle is given by,

  $y=2\sqrt{\sin x}$

Let’s use the formula,

  $V_T={\displaystyle {\int }_a^b\frac{b\left(x\right)h\left(x\right)}{2}dx}$

To use the PythagoreanTheorem, which states that $a^2+b^2=c^2$ , the equilateral triangle is cut into half to create a right triangle. Now that the right triangle is equilateral, its base will be $\frac{b\left(x\right)}{2}$ and its hypotenuse will be the same as $b\left(x\right)$ .

Solving the value of $h\left(x\right)$

  $\begin{array}{c}{\left[h\left(x\right)\right]}^2+{\left(\frac{2\sqrt{\sin x}}{2}\right)}^2={\left(2\sqrt{\sin x}\right)}^2\\ {\left[h\left(x\right)\right]}^2=4\sin x-\sin x\\ {\left[h\left(x\right)\right]}^2=3\sin x\\ h\left(x\right)=\sqrt{3\sin x}\end{array}$

Solving the above equation having an interval $\left[0,\pi \right]$ .

  $\begin{array}{c}{V}_T={\displaystyle {\int }_a^b\frac{b\left(x\right)h\left(x\right)}{2}dx}\\ ={\displaystyle {\int }_0^{\pi }\frac{\left(2\sqrt{\sin x}\right)\left(\sqrt{3\sin x}\right)}{2}}dx\\ ={\displaystyle {\int }_0^{\pi }\sqrt{3}\sin xdx}\\ ={\left.-\sqrt{3}\cos x\right|}_0^{\pi }\\ =-\sqrt{3}\left(\cos \pi -\cos 0\right)\\ =-\sqrt{3}\left(-1-1\right)\\ =2\sqrt{3}\end{array}$

Therefore, the volume of the equilateral triangle is $V_T=2\sqrt{3}$ .

b.

To determine

To find: The volume of the square with bases running from the $x$ -axis to the curve.

Answer to Problem 41E

The volume of the square is $V_s=8$ .

Explanation of Solution

Given information:

The base of a solid is the region between the curve is $y=2\sqrt{\sin x}$ .

Calculation:

The formula used to find the volume of the square with a base bounded by the curve and the $x$ -axis.

  $V_s={\displaystyle {\int }_a^{b}A\left(x\right)dx}$

Where $A\left(x\right)$ is from the $x$ -axis to the curve $y=2\sqrt{\sin x}$ .

The area of the square is given by,

  $A\left(x\right)=x^2$

Where $x$ is the side length of the square.

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

To find the volume of the square,

  $\begin{array}{c}{V}_s={\displaystyle {\int }_a^{b}A\left(x\right)dx}\\ ={\displaystyle {\int }_0^{\pi }4\sin xdx}\\ ={\left.-4\cos x\right|}_0^{\pi }\\ =-4\left(\cos \pi -\cos 0\right)\\ =-4\left(-1-1\right)\\ V_s=8\end{array}$

Therefore, the volume of the square is $V_s=8$ .