The image contains a calculus problem that reads:

"Find the volume V of the solid obtained by rotating the region bounded by the curves y = x^3 and y = 5x about the line x = 3."

Solution:

To find the volume $V$ of the solid formed by rotating the region bounded by $y = x^3$ and $y = 5x$ about the line $x = 3$, we can use the method of cylindrical shells.

Step 1: Set up the integral for the volume using the cylindrical shells method.

The volume $V$ using the cylindrical shells method is given by:

$V = 2\pi \int_{a}^{b} (r_{\text{shell}})(h_{\text{shell}}) \, dx$

Where:

• $r_{\text{shell}} = \text{distance from the axis of rotation} = 3 - x$
• $h_{\text{shell}} = \text{height of the shell} = 5x - x^3$ (difference between the functions)

Step 2: Determine the bounds of integration.

We need to find the intersection points of the curves $y = x^3$ and $y = 5x$:

$x^3 = 5x$

$x(x^2 - 5) = 0$

This gives:

$x = 0 \quad \text{or} \quad x = \sqrt{5} \quad \text{or} \quad x = -\sqrt{5}$

However, since we're dealing with the region bounded by $x = 0$ and $x = \sqrt{5}$, we will integrate from 0 to $\sqrt{5}$.

Step 3: Set up the integral.

$V = 2\pi \int_{0}^{\sqrt{5}} (3 - x)(5x - x^3) \, dx$

Step 4: Expand and integrate.

Expanding the integrand:

$(3 - x)(5x - x^3) = 15x - 3x^3 - 5x^2 + x^4$

The integral becomes:

[ V = 2\