Calculate the area, in square units, bounded above by x=sqrt of y +4 and x=-y+6 and bounded below by the x-axis. Give your answer as an improper fraction, if necessary, and do not include units.

To find the area bounded by the curves \( x = \sqrt{y} + 4 \), \( x = -y + 6 \), and the x-axis (i.e., \( y = 0 \)), we'll follow these steps:

### 1. Find the points of intersection
We need to determine where the curves \( x = \sqrt{y} + 4 \) and \( x = -y + 6 \) intersect.

Set the equations equal to each other:

\[
\sqrt{y} + 4 = -y + 6
\]

Solve for \( y \):

\[
\sqrt{y} = -y + 2
\]

Square both sides to eliminate the square root:

\[
y = (-y + 2)^2
\]

Expand the right side:

\[
y = y^2 - 4y + 4
\]

Rearrange the equation:

\[
y^2 - 5y + 4 = 0
\]

Now, solve the quadratic equation:

\[
y = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(4)}}{2(1)}
\]

\[
y = \frac{5 \pm \sqrt{25 - 16}}{2}
\]

\[
y = \frac{5 \pm \sqrt{9}}{2}
\]

\[
y = \frac{5 \pm 3}{2}
\]

So, \( y = 4 \) and \( y = 1 \).

These are the points of intersection for the curves. 

### 2. Set up the integral
We calculate the area by integrating the difference between the rightmost function \( x = \sqrt{y} + 4 \) and the leftmost function \( x = -y + 6 \) from \( y = 1 \) to \( y = 4 \).

\[
\text{Area} = \int_{1}^{4} \left[(\sqrt{y} + 4) - (-y + 6)\right] \, dy
\]

Simplify the integrand:

\[
\text{Area} = \int_{1}^{4} \left(\sqrt{y} + y - 2\right) \, dy
\]

### 3. Evaluate the integral
We break the integral into three parts:

\[
\text{Area} = \int_{1}^{4} \sqrt{y} \, dy + \int_{1}^{4} y \, dy - \int_{1}^{4} 2 \, dy
\]

Evaluate each integral:

1. For \( \int \sqrt{y} \, dy \):

\[
\int \sqrt{y} \, dy = \int y^{1/2} \, dy = \frac{2}{3} y^{3/2}
\]

2. For \( \int y \, dy \):

\[
\int y \, dy = \frac{y^2}{2}
\]

3. For \( \int 2 \, dy \):

\[
\int 2 \, dy = 2y
\]

Substituting back:

\[
\text{Area} = \left[\frac{2}{3} y^{3/2} + \frac{y^2}{2} - 2y\right]_{1}^{4}
\]

Now calculate:

At \( y = 4 \):

\[
\frac{2}{3} \cdot 4^{3/2} + \frac{4^2}{2} - 2(4) = \frac{2}{3} \cdot 8 + 8 - 8 = \frac{16}{3} + 8 - 8 = \frac{16}{3}
\]

At \( y = 1 \):

\[
\frac{2}{3} \cdot 1^{3/2} + \frac{1^2}{2} - 2(1) = \frac{2}{3} \cdot 1 + \frac{1}{2} - 2 = \frac{2}{3} + \frac{1}{2} - 2 = \frac{4}{6} + \frac{3}{6} - \frac{12}{6} = \frac{-5}{6}
\]

Subtract the lower limit from the upper limit:

\[
\text{Area} = \frac{16}{3} - \left(-\frac{5}{6}\right) = \frac{16}{3} + \frac{5}{6} = \frac{32}{6} + \frac{5}{6} = \frac{37}{6}
\]

### Final Answer:
The area is \( \frac{37}{6} \). 

Would you like further details or have any questions?

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Here are five questions that can further enhance your understanding:

1. How would you approach finding the area if the curves were reversed?
2. What if the region was bounded by the y-axis instead of the x-axis?
3. How does squaring both sides in the equation impact the solution?
4. Can you derive the integral limits directly from the intersection points?
5. How would this problem change if the equation was in terms of x instead of y?

**Tip:** When solving integrals, always ensure that the limits of integration correspond to the correct region of interest.

Explore how to calculate the area bounded by the curves x = sqrt(y) + 4 and x = -y + 6, along with the x-axis. This detailed solution involves finding the intersection points, setting up the integral, and evaluating it. Ideal for advanced students in Grades 11-12.

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