Calculate the points of intersection of 2y = 8x − 2x2 and 2y = 2x. Make a neat sketch of the graphs and show the representative strip that you will use to calculate the area bounded by the two graphs.

The given equations are $xy = -5$ and $y = x - 6$.
From the second equation, express $x$ in terms of $y$: $x = y + 6$.
Substitute this into the first equation:

Factor the quadratic equation:

This gives $y = -1$ or $y = -5$.
Now find the corresponding $x$ values:
If $y = -1$, then $x = -1 + 6 = 5$. Intersection point: $(5, -1)$.
If $y = -5$, then $x = -5 + 6 = 1$. Intersection point: $(1, -5)$.
The limits of integration for $y$ are from $c = -5$ to $d = -1$.

From $xy = -5$, we get $x = -\frac{5}{y}$.
From $y = x - 6$, we get $x = y + 6$.
To determine which function is $x_R$ (right) and $x_L$ (left), pick a test value for $y$ between $-5$ and $-1$, for example $y = -3$:
For $x = -\frac{5}{y}$: $x = -\frac{5}{-3} = \frac{5}{3} \approx 1.67$.
For $x = y + 6$: $x = -3 + 6 = 3$.
Since $3 > \frac{5}{3}$, the curve $x = y+6$ is to the right ($x_R$) and $x = -\frac{5}{y}$ is to the left ($x_L$).
So, $x_R = y+6$ and $x_L = -\frac{5}{y}$.

Since the region is below the x-axis (where $y$ is negative), the radius of the cylindrical shell is $|y| = -y$.

Evaluate at the upper limit $y = -1$:

Evaluate at the lower limit $y = -5$:

Subtract the lower limit value from the upper limit value:

The volume generated is $\frac{64\pi}{3}$.