ketch the region enclosed by 
 x + y^2 = 20 and x + y = 20
.
Decide whether to integrate with respect to 
 x or y
, and then find the area of the region.

The area is:

To solve the problem of finding the area of the region enclosed by the curves \( x + y^2 = 20 \) and \( x + y = 20 \), we can follow these steps:

### 1. Sketch the Curves
First, let's rewrite the equations to better understand the curves:

1. **Equation 1**: \( x + y^2 = 20 \)
   - Solving for \( x \): \( x = 20 - y^2 \)
   - This represents a parabola opening to the left with vertex at \( (20, 0) \).

2. **Equation 2**: \( x + y = 20 \)
   - Solving for \( x \): \( x = 20 - y \)
   - This represents a straight line with a slope of \(-1\).

### 2. Find the Points of Intersection
To find the points of intersection between the parabola and the line, set the equations equal to each other:

\[
20 - y^2 = 20 - y
\]

Simplifying, we get:

\[
y^2 = y
\]

\[
y(y - 1) = 0
\]

So, \( y = 0 \) or \( y = 1 \).

Substituting these values of \( y \) back into one of the original equations to find \( x \):

- For \( y = 0 \): \( x = 20 - 0^2 = 20 \).
- For \( y = 1 \): \( x = 20 - 1 = 19 \).

Thus, the points of intersection are \( (20, 0) \) and \( (19, 1) \).

### 3. Decide the Method of Integration
Given the nature of the equations (especially the parabola opening horizontally), it is easier to integrate with respect to \( y \).

### 4. Set Up the Integral
The area \( A \) can be found by integrating the difference between the two functions from \( y = 0 \) to \( y = 1 \):

\[
A = \int_{0}^{1} \left[(20 - y) - (20 - y^2)\right] dy
\]

Simplifying the integrand:

\[
A = \int_{0}^{1} \left(-y + y^2\right) dy
\]

### 5. Evaluate the Integral
Now, integrate:

\[
A = \int_{0}^{1} \left(-y + y^2\right) dy = \left[-\frac{y^2}{2} + \frac{y^3}{3}\right]_{0}^{1}
\]

Substituting the limits:

\[
A = \left[-\frac{1^2}{2} + \frac{1^3}{3}\right] - \left[-\frac{0^2}{2} + \frac{0^3}{3}\right]
\]

\[
A = \left[-\frac{1}{2} + \frac{1}{3}\right] = -\frac{3}{6} + \frac{2}{6} = -\frac{1}{6}
\]

Since we are finding the area, we take the absolute value:

\[
A = \frac{1}{6}
\]

### Final Answer:
The area of the region enclosed by the curves is \( \boxed{\frac{1}{6}} \).

---

Would you like more details on any part of this process, or do you have any questions?

### Related Questions:
1. How would you sketch the graph of \( x + y^2 = 20 \)?
2. What happens if you integrate with respect to \( x \) instead?
3. How do you find the intersection points of other types of curves?
4. How does the direction in which a parabola opens affect the integration method?
5. Can you explain the significance of taking the absolute value when calculating area?

### Tip:
When deciding whether to integrate with respect to \( x \) or \( y \), choose the variable that simplifies the integral and makes the bounds easier to determine.

Discover how to find the area enclosed by the curves x + y^2 = 20 and x + y = 20 using integration. Follow a step-by-step solution including finding intersection points, deciding integration direction, and evaluating the integral. Suitable for advanced high school students.

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