The problem is asking to use the shell method to find the volume of a solid generated by revolving the region between the lines $y = 6 - x$, $y = 0$, and $x = 7$ around the x-axis.

Shell Method Formula:

For revolution about the x-axis, the shell method involves integrating with respect to $y$, and the formula is:

$V = 2\pi \int_{y_{\text{min}}}^{y_{\text{max}}} (\text{radius}) (\text{height}) \, dy$

Step 1: Express $x$ as a function of $y$

We are given the line equation $y = 6 - x$. To use the shell method, we solve for $x$ in terms of $y$:

$x = 6 - y$

Step 2: Define limits of integration

The region is bounded by $y = 0$ and the line $y = 6 - x$, which intersects the y-axis at $y = 6$. So, the limits of integration are from $y = 0$ to $y = 6$.

Step 3: Determine the radius and height for the shell

• Radius: Since the region is revolved around the x-axis, the radius is simply $y$.
• Height: The height is the distance from the y-axis to the curve, which is $7 - x$. Since $x = 6 - y$, the height becomes:

$\text{height} = 7 - (6 - y) = y + 1$

Step 4: Set up the integral

Now, we can set up the integral for the volume:

$V = 2\pi \int_{0}^{6} y(y + 1) \, dy$

Step 5: Simplify and integrate

First, expand the integrand:

$y(y + 1) = y^2 + y$

Now, integrate term by term: