Find the volume of the solid obtained by rotating the region bounded by y=7x+56, y=0, and x=0 about the y-axis. (Hint: The region is a triangle to the left of the y-axis.)Find the volume of the solid obtained by rotating the region bounded by y=7x+56, y=0, and x=0 about the y-axis. (Hint: The region is a triangle to the left of the y-axis.)

To find the volume of the solid obtained by rotating the region bounded by \( y = 7x + 56 \), \( y = 0 \), and \( x = 0 \) about the \( y \)-axis, we'll use the method of cylindrical shells.

### Step 1: Understand the Region

The given region is bounded by the line \( y = 7x + 56 \), the \( y \)-axis (\( x = 0 \)), and the line \( y = 0 \). The line \( y = 7x + 56 \) intersects the \( y \)-axis at \( y = 56 \) (since at \( x = 0 \), \( y = 56 \)).

The line intersects the \( x \)-axis where \( y = 0 \):
\[
0 = 7x + 56 \implies x = -\frac{56}{7} = -8
\]
So the region is bounded by \( x = -8 \), \( x = 0 \), and \( y = 0 \).

### Step 2: Set Up the Integral (Using Shell Method)

Using the method of cylindrical shells, the volume \( V \) is given by:
\[
V = 2\pi \int_{a}^{b} x \cdot f(x) \, dx
\]
Here, \( f(x) = 7x + 56 \), and the limits of integration are from \( x = -8 \) to \( x = 0 \).

So the integral becomes:
\[
V = 2\pi \int_{-8}^{0} x(7x + 56) \, dx
\]

### Step 3: Solve the Integral

First, expand the integrand:
\[
V = 2\pi \int_{-8}^{0} (7x^2 + 56x) \, dx
\]

Now, integrate term by term:
\[
V = 2\pi \left[ \frac{7x^3}{3} + \frac{56x^2}{2} \right]_{-8}^{0}
\]

Evaluate the integral at the bounds \( x = 0 \) and \( x = -8 \):
- At \( x = 0 \):
\[
\frac{7(0)^3}{3} + \frac{56(0)^2}{2} = 0
\]
- At \( x = -8 \):
\[
\frac{7(-8)^3}{3} + \frac{56(-8)^2}{2} = \frac{7(-512)}{3} + \frac{56(64)}{2}
\]
\[
= -\frac{3584}{3} + 1792
\]

Now, calculate:
\[
-\frac{3584}{3} + \frac{5376}{3} = \frac{1792}{3}
\]

Thus, the volume \( V \) is:
\[
V = 2\pi \times \frac{1792}{3} = \frac{3584\pi}{3} \text{ cubic units}
\]

### Final Answer:
The volume of the solid obtained by rotating the region about the \( y \)-axis is \( \frac{3584\pi}{3} \) cubic units.

---

Would you like further details on any step or a different method? Here are some related questions:

1. What is the volume when rotating about the \( x \)-axis instead?
2. How does the setup change if using the disk method?
3. What would be the volume if the boundaries change to \( y = 7x + 28 \)?
4. Can this volume be calculated using polar coordinates?
5. How does the cylindrical shell method apply to different regions?

**Tip:** Always visualize the region and axis of rotation before choosing a method for volume calculation.

Learn how to find the volume of the solid obtained by rotating the region bounded by y = 7x + 56, y = 0, and x = 0 about the y-axis using the cylindrical shells method. Detailed step-by-step solution provided. Suitable for college-level calculus students.

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