find the volume of the solid whose base is the region bounded by the $x$-axis and the curve $y =…

Explanation:

Step1: Recall the formula for the volume of a solid with known cross - sections

The volume $V$ of a solid with cross - sectional area $A(x)$ from $x = a$ to $x = b$ is given by $V=\int_{a}^{b}A(x)dx$. Since the cross - sections are semi - circles perpendicular to the $x$ - axis, and the diameter of each semi - circle is $d = y=\sqrt{\cos(\frac{x}{5})}$. The radius of each semi - circle $r=\frac{1}{2}\sqrt{\cos(\frac{x}{5})}$.

Step2: Find the area formula for a semi - circle

The area of a semi - circle is $A=\frac{1}{2}\pi r^{2}$. Substituting $r = \frac{1}{2}\sqrt{\cos(\frac{x}{5})}$ into the formula, we get $A(x)=\frac{1}{2}\pi(\frac{1}{2}\sqrt{\cos(\frac{x}{5})})^{2}=\frac{\pi}{8}\cos(\frac{x}{5})$.

Step3: Set up the integral for the volume

We are integrating from $a =-\frac{5\pi}{2}$ to $b=\frac{5\pi}{2}$. So, $V=\int_{-\frac{5\pi}{2}}^{\frac{5\pi}{2}}\frac{\pi}{8}\cos(\frac{x}{5})dx$.

Step4: Evaluate the integral

Since $\int\cos(\frac{x}{5})dx = 5\sin(\frac{x}{5})+C$, then $\int_{-\frac{5\pi}{2}}^{\frac{5\pi}{2}}\frac{\pi}{8}\cos(\frac{x}{5})dx=\frac{\pi}{8}\times5\left[\sin(\frac{x}{5})\right]{-\frac{5\pi}{2}}^{\frac{5\pi}{2}}$. [ \begin{align*} \frac{5\pi}{8}\left[\sin(\frac{x}{5})\right]{-\frac{5\pi}{2}}^{\frac{5\pi}{2}}&=\frac{5\pi}{8}\left(\sin(\frac{\pi}{2})-\sin(-\frac{\pi}{2})\right)\ &=\frac{5\pi}{8}(1 + 1)\ &=\frac{5\pi}{4} \end{align*} ]

Answer:

$\frac{5\pi}{4}$