7. if the region enclosed by the y - axis, the line y = 2, and the curve y = ∛x is revolved about the y…

Explanation:

Step1: Solve the curve for x

Given $y = \sqrt[3]{x}$, then $x=y^{3}$.

Step2: Determine the limits of integration

The region is bounded by $y = 0$ (since it's enclosed by the $y$-axis) and $y = 2$.

Step3: Use the disk - method formula

The formula for the volume of a solid of revolution about the $y$-axis using the disk method is $V=\pi\int_{a}^{b}[R(y)]^{2}dy$. Here, $R(y)=y^{3}$, $a = 0$, and $b = 2$. So $V=\pi\int_{0}^{2}(y^{3})^{2}dy=\pi\int_{0}^{2}y^{6}dy$.

Step4: Integrate

Using the power - rule for integration $\int y^{n}dy=\frac{y^{n + 1}}{n+1}+C$ ($n\neq - 1$), we have $\int_{0}^{2}y^{6}dy=\left[\frac{y^{7}}{7}\right]_{0}^{2}=\frac{2^{7}}{7}-\frac{0^{7}}{7}=\frac{128}{7}$.

Step5: Find the volume

$V=\pi\times\frac{128}{7}=\frac{128\pi}{7}$.

Answer:

$\frac{128\pi}{7}$