consider the following regions.\nfind the volume of the solid generated by revolving the specified region…

Explanation:

Step1: Set up the disk - washer method formula

The region $R_1$ is bounded by $y = x$, $y = 1$, and $x = 0$. When revolving about the $y$-axis ($x = 0$), we use the disk - washer method. For a vertical axis of revolution, if we integrate with respect to $y$, the volume $V=\pi\int_{a}^{b}[R(y)]^{2}dy$. Here, for $y\in[0,1]$, the right - most curve is $x = y$.

Step2: Apply the integral formula

The volume formula becomes $V=\pi\int_{0}^{1}y^{2}dy$.

Step3: Integrate the function

Using the power rule for integration $\int y^{n}dy=\frac{y^{n + 1}}{n+1}+C$ ($n\neq - 1$), we have $\int_{0}^{1}y^{2}dy=\left[\frac{y^{3}}{3}\right]_{0}^{1}$.

Step4: Evaluate the definite integral

$\left[\frac{y^{3}}{3}\right]_{0}^{1}=\frac{1^{3}}{3}-\frac{0^{3}}{3}=\frac{1}{3}$. Then $V=\pi\times\frac{1}{3}=\frac{\pi}{3}$.

Answer:

$\frac{\pi}{3}$