The problem in the image asks to use the shell method to calculate the volume of the solid formed by revolving the region bounded by the curves $x = 0$, $y = 0$, and $y = -y^2 + 6y + 7$ around the x-axis, between $y = 0$ and $y = 7$.

Step-by-Step Approach

We will use the shell method for this problem. The shell method formula for volume is given by:

$V = 2\pi \int_{a}^{b} ( \text{radius} )( \text{height} ) \, dy$

Here:

• The radius is the distance from the x-axis, which is simply $y$ for each shell.
• The height is the horizontal distance between the curves, which is $x = 35 - y^2$ (since the other boundary is $x = 0$).

We revolve the region around the x-axis, so the limits of integration are from $y = 0$ to $y = 7$.

The volume integral becomes:

$V = 2\pi \int_{0}^{7} y (35 - y^2) \, dy$

Solving the Integral

Let's expand and integrate:

$V = 2\pi \int_{0}^{7} (35y - y^3) \, dy$

Now, integrate term by term:

$\int 35y \, dy = \frac{35y^2}{2}$ $\int y^3 \, dy = \frac{y^4}{4}$

Now, substitute these into the integral:

$V = 2\pi \left[ \frac{35y^2}{2} - \frac{y^4}{4} \right]_{0}^{7}$

Evaluating at the limits:

At $y = 7$: