Consider the following.
y = x2
y = 6 − x,  Find the area of the region by integrating with respect to y.

To find the area of the region enclosed by the curves \( y = x^2 \) and \( y = 6 - x \), we need to set up an integral with respect to \( y \). Let's follow these steps:

### 1. Solve for \( x \) in terms of \( y \):

For the first curve, \( y = x^2 \), solve for \( x \):
\[
x = \pm \sqrt{y}
\]

For the second curve, \( y = 6 - x \), solve for \( x \):
\[
x = 6 - y
\]

### 2. Find the points of intersection:

To find the limits of integration, set the two equations equal to each other:
\[
x^2 = 6 - x
\]
Rearrange this into a standard quadratic form:
\[
x^2 + x - 6 = 0
\]
Solve using the quadratic formula:
\[
x = \frac{-1 \pm \sqrt{1^2 - 4(1)(-6)}}{2(1)} = \frac{-1 \pm \sqrt{1 + 24}}{2} = \frac{-1 \pm 5}{2}
\]
Thus, \( x = 2 \) and \( x = -3 \).

Now substitute these \( x \)-values back into either equation for \( y \) to get the \( y \)-limits of integration:

For \( x = 2 \):
\[
y = 2^2 = 4
\]
For \( x = -3 \):
\[
y = (-3)^2 = 9
\]

So, the region we are interested in is between \( y = 4 \) and \( y = 9 \).

### 3. Set up the integral:

The area can be computed by integrating the difference between the rightmost curve (\( x = 6 - y \)) and the leftmost curve (\( x = -\sqrt{y} \)) with respect to \( y \), from \( y = 4 \) to \( y = 9 \).

Thus, the area \( A \) is given by:
\[
A = \int_4^9 \left( (6 - y) - (-\sqrt{y}) \right) \, dy
\]
\[
A = \int_4^9 \left( 6 - y + \sqrt{y} \right) \, dy
\]

### 4. Compute the integral:

Now, let's compute each part:

\[
A = \int_4^9 \left( 6 \right) \, dy - \int_4^9 \left( y \right) \, dy + \int_4^9 \left( \sqrt{y} \right) \, dy
\]

Evaluate each integral:

1. \( \int_4^9 6 \, dy = 6[y]_4^9 = 6(9 - 4) = 6 \times 5 = 30 \)

2. \( \int_4^9 y \, dy = \left[ \frac{y^2}{2} \right]_4^9 = \frac{9^2}{2} - \frac{4^2}{2} = \frac{81}{2} - \frac{16}{2} = \frac{65}{2} \)

3. \( \int_4^9 \sqrt{y} \, dy = \int_4^9 y^{1/2} \, dy = \left[ \frac{2}{3} y^{3/2} \right]_4^9 = \frac{2}{3} (9^{3/2} - 4^{3/2}) = \frac{2}{3} (27 - 8) = \frac{2}{3} \times 19 = \frac{38}{3} \)

### 5. Sum the results:

Now sum up all the terms:
\[
A = 30 - \frac{65}{2} + \frac{38}{3}
\]

To combine these, first find a common denominator. The least common denominator of 2 and 3 is 6:

\[
A = 30 - \frac{195}{6} + \frac{76}{6}
\]
\[
A = 30 - \frac{195 - 76}{6} = 30 - \frac{119}{6}
\]
\[
A = 30 - 19.83 = 10.17 \, \text{square units}.
\]

Thus, the area of the region is approximately \( 10.17 \) square units.

Would you like more details or have any questions?

Here are some related questions to explore further:

1. How does changing the curves affect the area between them?
2. What happens if we change the method and integrate with respect to \( x \)?
3. Can we generalize the procedure for curves with more complex intersections?
4. How do we handle regions where the curves cross each other?
5. What is the significance of symmetry in problems like this?

**Tip:** When calculating areas between curves, always visualize and identify which curve is "above" or "below" before setting up the integral.

Consider the following.
y = x^2
y = 6 − x,  Find the area of the region by integrating with respect to y.

Learn how to find the area of the region enclosed by the curves y = x^2 and y = 6 - x by integrating with respect to y. This detailed solution walks you through solving for x, finding points of intersection, and calculating the integral. Suitable for high school and college-level calculus students.

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