this question is designed to be answered without a calculator. the region bounded by y = x² and y = √x is…

Explanation:

Step1: Find intersection points

Set $x^{2}=\sqrt{x}$, then $x^{4}-x = 0$, $x(x^{3}-1)=0$. So $x = 0$ and $x = 1$ are the intersection - points.

Step2: Use the shell - method formula

The volume $V$ of the solid of revolution about the $y$-axis using the shell method is $V=2\pi\int_{a}^{b}x\left(f(x)-g(x)\right)dx$, where $a = 0$, $b = 1$, $f(x)=\sqrt{x}$, and $g(x)=x^{2}$. So $V = 2\pi\int_{0}^{1}x(\sqrt{x}-x^{2})dx=2\pi\int_{0}^{1}(x^{\frac{3}{2}}-x^{3})dx$.

Step3: Integrate term - by - term

$\int(x^{\frac{3}{2}}-x^{3})dx=\frac{2}{5}x^{\frac{5}{2}}-\frac{1}{4}x^{4}+C$.

Step4: Evaluate the definite integral

$V = 2\pi\left[\frac{2}{5}x^{\frac{5}{2}}-\frac{1}{4}x^{4}\right]_{0}^{1}=2\pi\left(\frac{2}{5}-\frac{1}{4}\right)$. $V = 2\pi\times\frac{8 - 5}{20}=2\pi\times\frac{3}{20}=\frac{3\pi}{10}$.

Answer:

$\frac{3\pi}{10}$