a solid is built on this region with cross section perpendicular to the y - axis being semi - circles. find…

Explanation:

Step1: Express (x) in terms of (y)

Since (y = x^{3}), then (x=y^{\frac{1}{3}}).

Step2: Determine the diameter of the semi - circle

The cross - sections are perpendicular to the (y) - axis. The diameter (d) of each semi - circle is the (x) value at a given (y). So (d = y^{\frac{1}{3}}), and the radius (r=\frac{y^{\frac{1}{3}}}{2}).

Step3: Find the area formula for the semi - circle

The area of a semi - circle is (A=\frac{1}{2}\pi r^{2}). Substituting (r = \frac{y^{\frac{1}{3}}}{2}) into the formula, we get (A(y)=\frac{1}{2}\pi(\frac{y^{\frac{1}{3}}}{2})^{2}=\frac{\pi}{8}y^{\frac{2}{3}}).

Step4: Set up the integral for the volume

We integrate with respect to (y) from (y = 1) to (y = 8). The volume (V=\int_{a}^{b}A(y)dy), so (V=\int_{1}^{8}\frac{\pi}{8}y^{\frac{2}{3}}dy).

Step5: Integrate the function

Using the power rule (\int y^{n}dy=\frac{y^{n + 1}}{n+1}+C) ((n\neq - 1)), we have (\int\frac{\pi}{8}y^{\frac{2}{3}}dy=\frac{\pi}{8}\times\frac{y^{\frac{2}{3}+1}}{\frac{2}{3}+1}=\frac{\pi}{8}\times\frac{y^{\frac{5}{3}}}{\frac{5}{3}}=\frac{3\pi}{40}y^{\frac{5}{3}}).

Step6: Evaluate the definite integral

(V=\left[\frac{3\pi}{40}y^{\frac{5}{3}}\right]_{1}^{8}=\frac{3\pi}{40}(8^{\frac{5}{3}}-1^{\frac{5}{3}})=\frac{3\pi}{40}(32 - 1)=\frac{3\pi}{40}\times31=\frac{93\pi}{40}).

Answer:

(\frac{93\pi}{40})