Find the volume of the solid that results when the region bounded by x=y2
 and x=y+12
 is revolved about the y
-axis.
Volume =
 

We are tasked with finding the volume of the solid formed when the region bounded by the curves \( x = y^2 \) and \( x = y + 12 \) is revolved around the **y-axis**. This problem requires the **method of cylindrical shells** or the **disk/washer method**. Since the region is being revolved around the y-axis, we’ll apply the **washer method**.

### Step 1: Find the points of intersection of the curves

We need to find the \( y \)-values where the curves intersect by setting \( x = y^2 \) equal to \( x = y + 12 \):

\[
y^2 = y + 12
\]

Rearranging this equation:

\[
y^2 - y - 12 = 0
\]

Factoring the quadratic:

\[
(y - 4)(y + 3) = 0
\]

Thus, the solutions are \( y = 4 \) and \( y = -3 \). These are the limits of integration.

### Step 2: Set up the volume integral using the washer method

The volume of the solid generated by revolving the region around the y-axis is given by:

\[
V = \pi \int_{y_1}^{y_2} \left[ R_{\text{outer}}^2 - R_{\text{inner}}^2 \right] \, dy
\]

Where:
- \( R_{\text{outer}} \) is the radius of the outer curve (i.e., \( x = y + 12 \)),
- \( R_{\text{inner}} \) is the radius of the inner curve (i.e., \( x = y^2 \)).

Thus, the volume integral is:

\[
V = \pi \int_{-3}^{4} \left[ (y + 12)^2 - (y^2)^2 \right] \, dy
\]

### Step 3: Expand and simplify the integrand

First, expand \( (y + 12)^2 \):

\[
(y + 12)^2 = y^2 + 24y + 144
\]

Now subtract \( (y^2)^2 = y^4 \):

\[
(y + 12)^2 - y^4 = y^2 + 24y + 144 - y^4
\]

Thus, the volume integral becomes:

\[
V = \pi \int_{-3}^{4} (-y^4 + y^2 + 24y + 144) \, dy
\]

### Step 4: Integrate

Now, we integrate each term:

\[
\int (-y^4 + y^2 + 24y + 144) \, dy = -\frac{y^5}{5} + \frac{y^3}{3} + 12y^2 + 144y
\]

### Step 5: Evaluate the definite integral

Now, we evaluate the expression at the limits \( y = -3 \) and \( y = 4 \):

1. At \( y = 4 \):

\[
-\frac{4^5}{5} + \frac{4^3}{3} + 12(4^2) + 144(4) = -\frac{1024}{5} + \frac{64}{3} + 12(16) + 144(4)
\]

2. At \( y = -3 \):

\[
-\frac{(-3)^5}{5} + \frac{(-3)^3}{3} + 12((-3)^2) + 144(-3) = -\frac{-243}{5} + \frac{-27}{3} + 12(9) + 144(-3)
\]

We’ll calculate these values now.The volume of the solid generated by revolving the region bounded by \( x = y^2 \) and \( x = y + 12 \) around the y-axis is approximately **868.93 cubic units**.

Would you like further details on any part of this process?

Here are 5 related questions for deeper exploration:
1. How would the method change if the region was revolved around the x-axis?
2. What is the geometric interpretation of the washer method in this context?
3. Can the same volume be found using the cylindrical shell method?
4. How does the shape of the curves influence the solid's volume?
5. How would the volume change if different functions bounded the region?

**Tip:** When solving volume of revolution problems, always check for symmetry—it can often simplify your calculations.

Find the volume of the solid that results when the region bounded by x = y^2 and x = y + 12 is revolved about the y-axis.

Learn how to calculate the volume of a solid formed by revolving the region bounded by x = y^2 and x = y + 12 around the y-axis. This detailed solution covers integration using the washer method, solving quadratic equations, and applying calculus concepts. Suitable for Grades 11-12 or early college.

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