Find the area of the region enclosed by the graphs of the given equations by partitioning (integrating with respect to) the y-axis. x = y^2+2, 6y=18-x

To find the area of the region enclosed by the given equations by integrating with respect to the y-axis, we follow these steps:

### Step 1: Express the equations in terms of \( y \)
Given the equations:
1. \( x = y^2 + 2 \)
2. \( 6y = 18 - x \)

First, rewrite the second equation to express \( x \) in terms of \( y \):
\[
6y = 18 - x \implies x = 18 - 6y
\]

Now we have the equations:
\[
x_1(y) = y^2 + 2 \quad \text{and} \quad x_2(y) = 18 - 6y
\]

### Step 2: Find the points of intersection
To find the points of intersection, set \( x_1(y) = x_2(y) \):
\[
y^2 + 2 = 18 - 6y
\]

Rearrange the equation:
\[
y^2 + 6y - 16 = 0
\]

This is a quadratic equation. Solve it using the quadratic formula:
\[
y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \( a = 1 \), \( b = 6 \), and \( c = -16 \):
\[
y = \frac{-6 \pm \sqrt{6^2 + 4 \times 1 \times 16}}{2 \times 1} = \frac{-6 \pm \sqrt{36 + 64}}{2} = \frac{-6 \pm \sqrt{100}}{2} = \frac{-6 \pm 10}{2}
\]

This gives two solutions:
\[
y = \frac{-6 + 10}{2} = 2 \quad \text{and} \quad y = \frac{-6 - 10}{2} = -8
\]

So the points of intersection are at \( y = 2 \) and \( y = -8 \).

### Step 3: Set up the integral
The area of the region enclosed by the curves is given by the integral:
\[
\text{Area} = \int_{y_1}^{y_2} \left[ x_2(y) - x_1(y) \right] \, dy
\]
where \( y_1 = -8 \) and \( y_2 = 2 \).

Substitute the functions:
\[
\text{Area} = \int_{-8}^{2} \left[(18 - 6y) - (y^2 + 2)\right] \, dy
\]

Simplify the integrand:
\[
\text{Area} = \int_{-8}^{2} \left(16 - 6y - y^2 \right) \, dy
\]

### Step 4: Evaluate the integral
Now, evaluate the integral:
\[
\int_{-8}^{2} \left(16 - 6y - y^2 \right) \, dy
\]

Integrate each term separately:
\[
\int_{-8}^{2} 16 \, dy = 16y \Big|_{-8}^{2} = 16(2) - 16(-8) = 32 + 128 = 160
\]

\[
\int_{-8}^{2} -6y \, dy = -6\frac{y^2}{2} \Big|_{-8}^{2} = -3y^2 \Big|_{-8}^{2} = -3(4 - 64) = -12 + 192 = 180
\]

\[
\int_{-8}^{2} -y^2 \, dy = -\frac{y^3}{3} \Big|_{-8}^{2} = -\frac{8}{3} + \frac{512}{3} = -\frac{8}{3} + \frac{512}{3} = \frac{-8 + 512}{3} = \frac{504}{3} = 168
\]

Summing these, the total area is:
\[
\text{Area} = 160 + 180 - 168 = 172 \, \text{square units}
\]

### Final Answer
The area of the region enclosed by the curves is **172 square units**.

---

Would you like more details or have any questions?

Here are some related questions to explore:
1. How would the area change if we integrated with respect to the x-axis instead?
2. How does shifting the graph of one equation vertically affect the enclosed area?
3. What is the geometric interpretation of the area found?
4. Can you find the centroid of the region bounded by these curves?
5. What is the volume of the solid generated when this region is revolved about the y-axis?
6. How would the problem change if the curves intersect at different points?
7. What happens if one of the curves is modified, say \( x = y^2 + 3 \)?
8. How does the area compare if we consider another region, like between the x-axis and one of the curves?

**Tip:** Always check the points of intersection carefully to ensure accurate limits for your integral.

Learn how to calculate the area enclosed by the curves x = y^2 + 2 and 6y = 18 - x using integration with respect to the y-axis. Detailed step-by-step solution provided. Suitable for advanced high school students and college-level calculus learners.

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