Question

1. about the y-axis

$x = 3y/2$

Explanation:

Step1: Identify the method

We use the washer method (or cylindrical shells, but washer is easier here) for volume about the y - axis. The region is bounded by \(x = 0\) (y - axis), \(x=\frac{3y}{2}\), and \(y = 2\) (since the horizontal line is at \(y = 2\)). The limits for \(y\) are from \(y = 0\) to \(y=2\). The outer radius \(R(y)\) and inner radius \(r(y)\): Wait, actually, when rotating about the y - axis, for a horizontal slice, the radius is \(x\). But in this case, the region is between \(x = 0\) and \(x=\frac{3y}{2}\) from \(y = 0\) to \(y = 2\)? Wait, no, the figure shows a region with a horizontal line at \(y = 2\), the y - axis, and the line \(x=\frac{3y}{2}\). Wait, maybe it's a region where for \(y\) from 0 to 2, the right boundary is \(x=\frac{3y}{2}\) and the left boundary is \(x = 0\)? Wait, no, the shaded region: when rotating about the y - axis, we can use the disk method (since there is no inner radius, because the left boundary is the y - axis, \(x = 0\)). Wait, the formula for the volume using the disk method (rotating about y - axis, horizontal slices) is \(V=\pi\int_{a}^{b}[R(y)]^{2}dy\), where \(R(y)\) is the radius (the x - value) as a function of \(y\).

From the line \(x=\frac{3y}{2}\), when \(y = 2\), \(x=\frac{3\times2}{2}=3\), which matches the x - axis mark at 3. The limits for \(y\) are from \(y = 0\) to \(y = 2\).

Step2: Set up the integral

So \(R(y)=\frac{3y}{2}\), \(a = 0\), \(b = 2\). Then the volume \(V=\pi\int_{0}^{2}(\frac{3y}{2})^{2}dy\)

Step3: Evaluate the integral

First, expand \((\frac{3y}{2})^{2}=\frac{9y^{2}}{4}\)

Then \(V=\pi\int_{0}^{2}\frac{9y^{2}}{4}dy=\frac{9\pi}{4}\int_{0}^{2}y^{2}dy\)

The integral of \(y^{2}\) is \(\frac{y^{3}}{3}\), so:

\(\frac{9\pi}{4}[\frac{y^{3}}{3}]_{0}^{2}=\frac{9\pi}{4}\times\frac{8}{3}-0=\frac{9\pi\times8}{12}=\frac{72\pi}{12} = 6\pi\)

Wait, that seems too simple. Wait, maybe I misidentified the region. Wait, maybe the region is bounded by \(y = 2\), the y - axis, and the line \(x=\frac{3y}{2}\), but also, is there another boundary? Wait, the left - hand side of the shaded region: maybe the line \(x = 0\) (y - axis) and the line \(x=\frac{3y}{2}\), and the top boundary \(y = 2\), bottom boundary \(y = 0\). So the volume when rotating about the y - axis:

Wait, let's re - check. The disk method: when rotating a region under \(x = f(y)\) from \(y=a\) to \(y = b\) about the y - axis, the volume is \(V=\pi\int_{a}^{b}[f(y)]^{2}dy\). Here, \(f(y)=\frac{3y}{2}\), \(a = 0\), \(b = 2\).

So \(V=\pi\int_{0}^{2}(\frac{3y}{2})^{2}dy=\pi\int_{0}^{2}\frac{9y^{2}}{4}dy=\frac{9\pi}{4}\times[\frac{y^{3}}{3}]_{0}^{2}=\frac{9\pi}{4}\times\frac{8}{3}=6\pi\)

Wait, but maybe the region is different. Wait, the other part of the figure has a line \(x + 2y=2\), but the circled problem is 8, about the y - axis, with the right figure. So in the right figure, the shaded region is between \(x = 0\), \(y = 2\), and \(x=\frac{3y}{2}\). So the volume calculation as above.

Answer:

The volume is \(6\pi\) (or if we made a mistake in the region, but based on the given figure and the line \(x=\frac{3y}{2}\), the volume is \(6\pi\))