Find the volume of the solid generated by revolving the region between the y-axis and the curve x = 2/y, 1 ≤ y ≤ 4, about the y-axis.

Steps to Solve

1. Identify the region and function

The curve is given by the equation $x = \frac{2}{y}$. We need to consider the region bounded by this curve and the y-axis for $1 \leq y \leq 4$.

2. Set up the volume integral

To find the volume of the solid formed by revolving this region around the y-axis, we use the formula:

$$ V = \pi \int_{a}^{b} [R(y)]^2 , dy $$

where $R(y)$ is the distance from the y-axis to the curve.

3. Determine the function for $R(y)$

From the equation $x = \frac{2}{y}$, the radius function $R(y)$ is equal to $\frac{2}{y}$. Thus, we have:

$$ R(y) = \frac{2}{y} $$

4. Calculate the volume integral

Substituting our limits ($a = 1$ and $b = 4$) and the function into the volume formula, the integral becomes:

$$ V = \pi \int_{1}^{4} \left(\frac{2}{y}\right)^2 , dy $$

5. Evaluate the integral

We need to simplify the square and evaluate:

$$ \left(\frac{2}{y}\right)^2 = \frac{4}{y^2} $$

Thus,

$$ V = \pi \int_{1}^{4} \frac{4}{y^2} , dy $$

Now, calculating the integral:

$$ V = \pi \left[ -\frac{4}{y} \right]_{1}^{4} $$

Substituting the limits:

$$ V = \pi \left( -\frac{4}{4} + \frac{4}{1} \right) = \pi \left( -1 + 4 \right) = 3\pi $$