Find the volume of the solid of revolution formed by rotating the shaded area in Quadrant 1 between $$ x=1 $$ and $$ x+2 y=7 $$ about the $$ y $$-axis. (graph not to scale)

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To find the volume of the solid of revolution formed by rotating the shaded area in Quadrant 1 between $$ x = 1 $$ and the line $$ x + 2y = 7 $$ about the $$ y $$-axis, we first need to express the line in terms of $$ y $$.

Starting with the equation of the line:

$$
x + 2y = 7
$$

we can solve for $$ y $$:

$$
2y = 7 - x \quad \Rightarrow \quad y = \frac{7 - x}{2}
$$

Next, we need to determine the limits of integration. The line intersects the $$ y $$-axis when $$ x = 0 $$:

$$
0 + 2y = 7 \quad \Rightarrow \quad 2y = 7 \quad \Rightarrow \quad y = \frac{7}{2} = 3.5
$$

Thus, the line intersects the $$ y $$-axis at $$ (0, 3.5) $$.

Now, we need to find the volume of the solid formed by rotating the area between $$ x = 1 $$ and the line $$ x + 2y = 7 $$ about the $$ y $$-axis. We will use the method of cylindrical shells.

The volume $$ V $$ of the solid of revolution can be calculated using the formula:

$$
V = 2\pi \int_{a}^{b} x \cdot f(x) \, dx
$$

where $$ f(x) $$ is the height of the shell at position $$ x $$, and $$ a $$ and $$ b $$ are the limits of integration.

In our case, the height of the shell is given by $$ y = \frac{7 - x}{2} $$, and the limits of integration are from $$ x = 1 $$ to $$ x = 7 $$ (the $$ x $$-coordinate where the line intersects the $$ y $$-axis).

Thus, we have:

$$
V = 2\pi \int_{1}^{5} x \cdot \frac{7 - x}{2} \, dx
$$

This simplifies to:

$$
V = \pi \int_{1}^{5} x(7 - x) \, dx
$$

Now, we can expand the integrand:

$$
x(7 - x) = 7x - x^2
$$

So the volume integral becomes:

$$
V = \pi \int_{1}^{5} (7x - x^2) \, dx
$$

Now we can compute the integral:

$$
\int (7x - x^2) \, dx = \frac{7x^2}{2} - \frac{x^3}{3}
$$

Evaluating this from $$ 1 $$ to $$ 5 $$:

$$
V = \pi \left[ \left( \frac{7(5^2)}{2} - \frac{5^3}{3} \right) - \left( \frac{7(1^2)}{2} - \frac{1^3}{3} \right) \right]
$$

Calculating the first part:

$$
\frac{7(25)}{2} - \frac{125}{3} = \frac{175}{2} - \frac{125}{3}
$$

Finding a common denominator (which is 6):

$$
\frac{175 \cdot 3}{6} - \frac{125 \cdot 2}{6} = \frac{525 - 250}{6} = \frac{275}{6}
$$

Now for the second part:

$$
\frac{7(1)}{2} - \frac{1}{3} = \frac{7}{2} - \frac{1}{3}
$$

Finding a common denominator (which is 6):

$$
\frac{7 \cdot 3}{6} - \frac{1 \cdot 2}{6} = \frac{21 - 2}{6} = \frac{19}{6}
$$

Now we can combine both parts:

$$
V = \pi \left( \frac{275}{6} - \frac{19}{6} \right) = \pi \left( \frac{256}{6} \right) = \frac{128\pi}{3}
$$

Thus, the volume of the solid of revolution is:

$$
\boxed{\frac{128\pi}{3}}
$$
The volume of the solid of revolution is $$ \frac{128\pi}{3} $$.

Find the volume of the solid of revolution formed by rotating the shaded area in Quadrant 1 between \( x=1 \) and \( x+2 y=7 \) about the \( y \)-axis. (graph not to scale)

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