Determine the area enclosed by the parabolas y²=4x and x² =4y sketch the curves

To determine the area enclosed by the parabolas \( y^2 = 4x \) and \( x^2 = 4y \), we can follow these steps:

### 1. Understanding the curves:
- **\( y^2 = 4x \):** This is a rightward-facing parabola with vertex at the origin \((0, 0)\).
- **\( x^2 = 4y \):** This is an upward-facing parabola with vertex at the origin \((0, 0)\).

### 2. Finding the points of intersection:
To find the area enclosed between the two parabolas, we first need to determine where the curves intersect. We solve the system of equations formed by \( y^2 = 4x \) and \( x^2 = 4y \).

Start by solving one equation for a variable, say, \( x \) from \( y^2 = 4x \):
\[
x = \frac{y^2}{4}
\]
Substitute this into the second equation \( x^2 = 4y \):
\[
\left( \frac{y^2}{4} \right)^2 = 4y
\]
Simplifying this equation:
\[
\frac{y^4}{16} = 4y \quad \Rightarrow \quad y^4 = 64y \quad \Rightarrow \quad y(y^3 - 64) = 0
\]
Thus, \( y = 0 \) or \( y^3 = 64 \), so \( y = 4 \).

Now, substitute \( y = 4 \) and \( y = 0 \) back into one of the original equations to find the corresponding \( x \)-coordinates:
- For \( y = 4 \), \( y^2 = 16 \), so \( 4x = 16 \), giving \( x = 4 \).
- For \( y = 0 \), we find \( x = 0 \).

Thus, the points of intersection are \( (0, 0) \) and \( (4, 4) \).

### 3. Setting up the integral:
The area enclosed by the curves can be found by integrating the difference between the two curves over the range of intersection, from \( y = 0 \) to \( y = 4 \).

We know from the equations:
- The parabola \( y^2 = 4x \) gives \( x = \frac{y^2}{4} \).
- The parabola \( x^2 = 4y \) gives \( x = \sqrt{4y} = 2\sqrt{y} \).

Now, set up the integral for the area between these two curves:
\[
A = \int_{0}^{4} \left( 2\sqrt{y} - \frac{y^2}{4} \right) dy
\]

### 4. Evaluating the integral:
Let's evaluate each part of the integral.

For the first term:
\[
\int_{0}^{4} 2\sqrt{y} \, dy = 2 \int_{0}^{4} y^{1/2} \, dy = 2 \cdot \left( \frac{2}{3} y^{3/2} \right) \Big|_0^4 = \frac{4}{3} \cdot 8 = \frac{32}{3}
\]

For the second term:
\[
\int_{0}^{4} \frac{y^2}{4} \, dy = \frac{1}{4} \int_{0}^{4} y^2 \, dy = \frac{1}{4} \cdot \left( \frac{y^3}{3} \right) \Big|_0^4 = \frac{1}{4} \cdot \frac{64}{3} = \frac{16}{3}
\]

### 5. Finding the total area:
Now subtract the two results:
\[
A = \frac{32}{3} - \frac{16}{3} = \frac{16}{3}
\]

Thus, the area enclosed by the two parabolas is \( \frac{16}{3} \) square units.

### 6. Sketch of the curves:
To sketch the curves:
- The parabola \( y^2 = 4x \) opens to the right.
- The parabola \( x^2 = 4y \) opens upwards.
- They intersect at \( (0, 0) \) and \( (4, 4) \), and the enclosed region lies between these points.

Would you like more details on sketching, or do you have any questions about the solution?

### Related Questions:
1. What is the geometric significance of the intersection points?
2. How would the area change if the curves were translated along the axes?
3. Can you find the area between different pairs of conic sections, such as ellipses and parabolas?
4. How would you solve for the enclosed area using polar coordinates?
5. How do the equations change if we modify the parameters (e.g., replacing 4 with another constant)?

**Tip:** When solving for the area between curves, always make sure to express the functions in terms of the same variable before integrating!

Determine the area enclosed by the parabolas y²=4x and x²=4y. Sketch the curves.

Calculate the area enclosed by the parabolas y²=4x and x²=4y, including finding intersection points and sketching the curves. This solution demonstrates integral calculus and geometric principles to find the enclosed area, suitable for undergraduate students.

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