Questions: Find the volume of the solid obtained by revolving the region bounded by the graphs (y=frac3x), the (x)-axis, (x=1), and (x=5) about the (y)-axis. (Express numbers in exact form. Use symbolic notation and fractions where needed.)

$Find the volume of the solid obtained by revolving the region bounded by the graphs (y=frac3x), the (x)-axis, (x=1), and (x=5) about the (y)-axis. (Express numbers in exact form. Use symbolic notation and fractions where needed.)$

Transcript text: Find the volume of the solid obtained by revolving the region bounded by the graphs $y=\frac{3}{x}$, the $x$-axis, $x=1$, and $x=5$ about the $y$-axis. (Express numbers in exact form. Use symbolic notation and fractions where needed.) \[ V= \]

Solution

Solution Steps

Hint

To solve this problem, we use the method of cylindrical shells, integrating the volume of each infinitesimally thin shell, which is the product of the shell's circumference and its height, across the interval of interest. The height of each shell corresponds to the function value, and the radius is the distance from the shell to the axis of revolution.

Step 1: Define the Problem

We need to find the volume $ V $ of the solid obtained by revolving the region bounded by the graphs $ y = \frac{3}{x} $, the $ x $-axis, $ x = 1 $, and $ x = 5 $ about the $ y $-axis.

Step 2: Set Up the Integral

Using the method of cylindrical shells, the volume $ V $ can be expressed as: \[ V = \int_{a}^{b} 2 \pi x y \, dx \] where $ y = \frac{3}{x} $, $ a = 1 $, and $ b = 5 $. Thus, the integrand becomes: \[ V = \int_{1}^{5} 2 \pi x \left(\frac{3}{x}\right) \, dx = \int_{1}^{5} 6 \pi \, dx \]

Step 3: Evaluate the Integral

Now we evaluate the integral: \[ V = 6 \pi \int_{1}^{5} 1 \, dx = 6 \pi [x]_{1}^{5} = 6 \pi (5 - 1) = 6 \pi \cdot 4 = 24 \pi \]