From x(x³ - 64) = 0, we get x=0 as one solution. For the second solution, we solve x³ = 64, which yields x=4 since 4³ = 64. These are the x -coordinates of the intersection points.

Substitute the x -values back into either equation (1) or (2) to find the corresponding y -coordinates. Using y = (x²)/(4) (from equation 2), we find that when x=0, y=0, and when x=4, y=4. Thus, the curves intersect at (0,0) and (4,4).

Substitute the x -values back into either equation (1) or (2) to find the corresponding y -coordinates. Using y = (x²)/(4) (from equation 2), we find that when x=0, y=0, and when x=4, y=4. Thus, the curves intersect at (0,0) and (4,4).

From y²=4x, we get y = ± 2√x. Since the region of interest is in the first quadrant (between x=0 and x=4), we consider the positive root, y_1 = 2√x. From x²=4y, we get y_2 = (x²)/(4). We observe that for x (0,4), 2√x > (x²)/(4). For example, at x=1, 2√1 = 2 and (1²)/(4) = 0.25. So, y_1 = 2√x is the upper curve and y_2 = (x²)/(4) is the lower curve.

From y²=4x, we get y = ± 2√x. Since the region of interest is in the first quadrant (between x=0 and x=4), we consider the positive root, y_1 = 2√x. From x²=4y, we get y_2 = (x²)/(4). We observe that for x (0,4), 2√x > (x²)/(4). For example, at x=1, 2√1 = 2 and (1²)/(4) = 0.25. So, y_1 = 2√x is the upper curve and y_2 = (x²)/(4) is the lower curve.

Integrate each term with respect to x. Recall that ∫ x^n dx = (x^(n+1))/(n+1). So, ∫ √x dx = ∫ x^(1/2) dx = x³/23/2 and ∫ x² dx = (x³)/(3).

Simplify the constants. 2 · (1)/(3/2) = 2 · (2)/(3) = (4)/(3) and (1)/(4) · (1)/(3) = (1)/(12).

Calculate the values: 4³/2 = (√4)³ = 2³ = 8. Simplify (64)/(12) by dividing both numerator and denominator by 4 to get (16)/(3). Subtract the terms.

Find the area of the region bounded by the curves $y^2 = 4x$ and $x^2 = 4y$ .

Answer: The area of the region bounded by the curves $y^2 = 4x$ and $x^2 = 4y$ is $\frac{16}{3}$ square units.

Step-by-step solution

Step 1: Identify the curves and find intersection points

We are given two parabolic equations. To find the area bounded by them, we first need to determine their points of intersection. We will solve these equations simultaneously to find the $(x, y)$ coordinates where they intersect.

Step 2: Substitute y from (2) into (1)

From equation (2), we can express $y$ as $y = \frac{x^2}{4}$ . Substitute this expression for $y$ into equation (1) to eliminate $y$ and obtain an equation solely in terms of $x$ .

Step 3: Solve for x-coordinates of intersection

Simplify the equation to solve for $x$ . Factor out $x$ from the equation $x^4 - 64x = 0$ . This gives us two possibilities for $x$ : $x=0$ or $x^3 - 64 = 0$ .

Step 4: Find the values of x

From $x(x^3 - 64) = 0$ , we get $x=0$ as one solution. For the second solution, we solve $x^3 = 64$ , which yields $x=4$ since $4^3 = 64$ . These are the $x$ -coordinates of the intersection points.

Step 5: Find the y-coordinates of intersection

Substitute the $x$ -values back into either equation (1) or (2) to find the corresponding $y$ -coordinates. Using $y = \frac{x^2}{4}$ (from equation 2), we find that when $x=0$ , $y=0$ , and when $x=4$ , $y=4$ . Thus, the curves intersect at $(0{,}0)$ and $(4{,}4)$ .

Step 6: Set up the integral for the area

To find the area enclosed by the two curves, we integrate the difference between the upper curve and the lower curve with respect to $x$ , from the first intersection point to the second. In this region, $y^2=4x$ means $y = 2\sqrt{x}$ (upper half) and $x^2=4y$ means $y = \frac{x^2}{4}$ (lower curve). We need to determine which curve is 'upper' and which is 'lower' in the interval $[0{,}4]$ .

Step 7: Determine upper and lower curves

From $y^2=4x$ , we get $y = \pm 2\sqrt{x}$ . Since the region of interest is in the first quadrant (between $x=0$ and $x=4$ ), we consider the positive root, $y_1 = 2\sqrt{x}$ . From $x^2=4y$ , we get $y_2 = \frac{x^2}{4}$ . We observe that for $x \in (0{,}4)$ , $2\sqrt{x} > \frac{x^2}{4}$ . For example, at $x=1$ , $2\sqrt{1} = 2$ and $\frac{1^2}{4} = 0.25$ . So, $y_1 = 2\sqrt{x}$ is the upper curve and $y_2 = \frac{x^2}{4}$ is the lower curve.

Step 8: Integrate to find the area

Now we set up the definite integral with the limits of integration from $x=0$ to $x=4$ and the integrand as the difference between the upper curve $y=2\sqrt{x}$ and the lower curve $y=\frac{x^2}{4}$ .

Step 9: Perform the integration

Integrate each term with respect to $x$ . Recall that $\int x^n dx = \frac{x^{n+1}}{n+1}$ . So, $\int \sqrt{x} dx = \int x^{1/2} dx = \frac{x^{3/2}}{3/2}$ and $\int x^2 dx = \frac{x^3}{3}$ .

Step 10: Simplify the integrated terms

Simplify the constants. $2 \cdot \frac{1}{3/2} = 2 \cdot \frac{2}{3} = \frac{4}{3}$ and $\frac{1}{4} \cdot \frac{1}{3} = \frac{1}{12}$ .

Step 11: Evaluate the definite integral

Apply the limits of integration by substituting the upper limit ( $x=4$ ) and the lower limit ( $x=0$ ) into the integrated expression and subtracting the result for the lower limit from the result for the upper limit.

Step 12: Calculate the numerical value

Calculate the values: $4^{3/2} = (\sqrt{4})^3 = 2^3 = 8$ . Simplify $\frac{64}{12}$ by dividing both numerator and denominator by 4 to get $\frac{16}{3}$ . Subtract the terms.

Step 13: Final calculation

Perform the final subtraction to get the area.

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