The region D above lies between the two red lines and the red parabola y=(1)/(25)x^(2) It can be describe in two ways. 1. If we visualize the region having "top" and "bottom" boundaries, express each as functions of x and provide the interval of x-values that covers the entire region. "top" boundary g_2(x)=? "bottom" boundary g_1(x)=? interval of x values that covers the region =? 2. If we visualize the region having "right" and "left" boundaries, express each as functions of yand provide the interval of y-values that covers the entire region. "top" boundary f_2(y)=? "bottom" boundary f_1(y)=? interval of y values that covers the region =?

Let's analyze the region \( D \) defined by the red lines and the red parabola \( y = \frac{1}{25}x^2 \).

**1. Describing the Region with Top and Bottom Boundaries:**

For the top and bottom boundaries, we express the boundaries as functions of \( x \).

Given:
- The red lines can be written as \( y = 4x \) and \( y = x^2 + 3 \).
- The red parabola is \( y = \frac{1}{25}x^2 \).

However, from your description, we only need the parabola and the boundaries. Let's confirm the actual setup:

### Step 1: Finding Intersection Points
Set \( y = 4x \) and \( y = \frac{1}{25}x^2 \):
\[ 4x = \frac{1}{25}x^2 \]
\[ 100x = x^2 \]
\[ x(x - 100) = 0 \]
Thus, the intersection points are at \( x = 0 \) and \( x = 100 \).

### Top and Bottom Boundaries
- **Bottom Boundary \( g_1(x) \)**: This is the parabola, so \( g_1(x) = \frac{1}{25}x^2 \).
- **Top Boundary \( g_2(x) \)**: This is the line, so \( g_2(x) = 4x \).
- **Interval of \( x \)-values**: From \( x = 0 \) to \( x = 100 \).

So, the region is described as:
\[
g_1(x) = \frac{1}{25}x^2
\]
\[
g_2(x) = 4x
\]
\[
\text{Interval of } x \text{ values: } 0 \leq x \leq 100
\]

**2. Describing the Region with Right and Left Boundaries:**

For the right and left boundaries, we express the boundaries as functions of \( y \).

### Step 2: Solving for \( x \) in Terms of \( y \)
- From \( y = \frac{1}{25}x^2 \), solve for \( x \):
\[ x = \pm 5\sqrt{y} \]
Since we're dealing with the left and right boundaries, we take the positive and negative solutions.
- From \( y = 4x \), solve for \( x \):
\[ x = \frac{y}{4} \]

### Right and Left Boundaries
- **Left Boundary \( f_1(y) \)**: This is the negative solution of the parabola, so \( f_1(y) = -5\sqrt{y} \).
- **Right Boundary \( f_2(y) \)**: This is the line, so \( f_2(y) = \frac{y}{4} \).

### Interval of \( y \)-values
- For the line \( y = 4x \), at \( x = 100 \), \( y = 4 \cdot 100 = 400 \).
So, the interval of \( y \)-values is from \( y = 0 \) to \( y = 400 \).

So, the region is described as:
\[
f_1(y) = -5\sqrt{y}
\]
\[
f_2(y) = \frac{y}{4}
\]
\[
\text{Interval of } y \text{ values: } 0 \leq y \leq 400
\]

Would you like more details or have any questions?

Here are 5 numbered questions you might ask next:
1. How do you set up a double integral to find the area of this region?
2. How do you find the volume of the solid obtained by rotating this region about the x-axis?
3. How do you find the volume of the solid obtained by rotating this region about the y-axis?
4. How do you find the centroid of this region?
5. What is the arc length of the parabola from \( x = 0 \) to \( x = 100 \)?

Tip: When visualizing regions bounded by curves, always determine the intersection points first as they define the limits of integration and the bounds for the functions involved.

Explore how to describe region D between two red lines and a parabola with functions of x and y-values. Understand the intervals and boundaries to calculate area and volume. Suitable for advanced mathematics students and enthusiasts.

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