Key Concepts and Formulas

• Area between curves: The area enclosed between two curves $y = f(x)$ and $y = g(x)$, where $f(x) \geq g(x)$ on the interval $[a, b]$, is given by $\int_a^b [f(x) - g(x)] \, dx$.
• Solving for intersection points: To find the points of intersection between two curves, set their equations equal to each other and solve for the variable.
• Integration properties: $\int x^n dx = \frac{x^{n+1}}{n+1} + C$, where $n \neq -1$.

Step-by-Step Solution

Step 1: Find the points of intersection of the two curves.

We are given the curves $y = kx^2$ and $x = ky^2$. To find their intersection points, we need to solve these equations simultaneously. Substituting $y = kx^2$ into the second equation, we get $x = k(kx^2)^2 = k^3x^4$. This simplifies to $k^3x^4 - x = 0$, which can be factored as $x(k^3x^3 - 1) = 0$. Thus, $x = 0$ or $k^3x^3 = 1$, implying $x^3 = \frac{1}{k^3}$, so $x = \frac{1}{k}$. When $x = 0$, $y = k(0)^2 = 0$. When $x = \frac{1}{k}$, $y = k\left(\frac{1}{k}\right)^2 = k\left(\frac{1}{k^2}\right) = \frac{1}{k}$. Therefore, the points of intersection are $(0, 0)$ and $\left(\frac{1}{k}, \frac{1}{k}\right)$.

Step 2: Express both curves as functions of x.

We have $y = kx^2$ and $x = ky^2$. We need to express the second equation as a function of $x$. From $x = ky^2$, we have $y^2 = \frac{x}{k}$, so $y = \sqrt{\frac{x}{k}}$. Since $k > 0$ and we are considering the area in the first quadrant, we take the positive square root.

Step 3: Set up the integral for the area between the curves.

The area enclosed between the curves is given by the integral of the difference between the two functions from $x = 0$ to $x = \frac{1}{k}$. Since $y = \sqrt{\frac{x}{k}}$ is above $y = kx^2$ in the region of interest, the area is: $\text{Area} = \int_0^{1/k} \left( \sqrt{\frac{x}{k}} - kx^2 \right) \, dx$

Step 4: Evaluate the integral.

$\text{Area} = \int_0^{1/k} \left( \frac{1}{\sqrt{k}} x^{1/2} - kx^2 \right) \, dx = \left[ \frac{1}{\sqrt{k}} \cdot \frac{x^{3/2}}{3/2} - k \cdot \frac{x^3}{3} \right]_0^{1/k}$ $= \left[ \frac{2}{3\sqrt{k}} x^{3/2} - \frac{k}{3} x^3 \right]_0^{1/k} = \left( \frac{2}{3\sqrt{k}} \left(\frac{1}{k}\right)^{3/2} - \frac{k}{3} \left(\frac{1}{k}\right)^3 \right) - (0 - 0)$ $= \frac{2}{3\sqrt{k}} \cdot \frac{1}{k^{3/2}} - \frac{k}{3} \cdot \frac{1}{k^3} = \frac{2}{3k^2} - \frac{1}{3k^2} = \frac{1}{3k^2}$

Step 5: Solve for k, given that the area is 1.

We are given that the area is 1 square unit. Therefore, $\frac{1}{3k^2} = 1$ $3k^2 = 1$ $k^2 = \frac{1}{3}$ Since $k > 0$, we have $k = \sqrt{\frac{1}{3}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$. However, our answer must be $\sqrt{3}$. Let us re-examine step 3.

Step 3 (Revised): Setting up the integral for area using y.

The area can also be calculated by integrating with respect to y. We have $x = \sqrt{y/k}$ and $x = ky^2$. The intersection points are $(0,0)$ and $(1/k, 1/k)$. Since $x = \sqrt{y/k}$ is to the right of $x=ky^2$ for $y \in [0, 1/k]$, we have:

$Area = \int_0^{1/k} (\sqrt{y/k} - ky^2) dy = \frac{1}{3k^2}$ as before.

Step 5 (Revised): Solving for k, given area is 1.

Thus, $\frac{1}{3k^2} = 1$, so $k^2 = \frac{1}{3}$ and $k = \frac{1}{\sqrt{3}}$.

Error Analysis:

It appears the correct answer given in the problem is incorrect. The derivation is correct, and the correct value of k is $\frac{1}{\sqrt{3}}$.

Revised Solution with Corrected Area Calculation and Adjustment to Given Answer

Let's re-examine the problem and assume the correct answer is indeed $\sqrt{3}$. Then, $\frac{1}{3k^2} = 1$ $3k^2 = 1$ $k^2 = \frac{1}{3}$ $k = \frac{1}{\sqrt{3}}$

If $k = \sqrt{3}$, then the area would be $\frac{1}{3 (\sqrt{3})^2} = \frac{1}{3(3)} = \frac{1}{9}$. Thus, the given area = 1 is incorrect. The problem statement contains an error.

Let's solve for what area would give $k = \sqrt{3}$. If $k = \sqrt{3}$, then Area $= \frac{1}{3 (\sqrt{3})^2} = \frac{1}{9}$.

Let's assume the area is 1/9. $\frac{1}{3k^2} = \frac{1}{9}$ $3k^2 = 9$ $k^2 = 3$ $k = \sqrt{3}$

Therefore, if the area was 1/9, then k would be $\sqrt{3}$. However, the problem states the area is 1.

Common Mistakes & Tips

• Incorrectly identifying the upper and lower curves: Always sketch the curves or test points to determine which curve is above the other in the region of interest.
• Forgetting the limits of integration: The limits of integration must correspond to the points of intersection of the curves.
• Sign errors: Be careful with signs when subtracting the functions and evaluating the integral.

Summary

We found the intersection points of the curves $y = kx^2$ and $x = ky^2$. We then set up the integral to find the area between the curves, expressing both equations in terms of x. Solving the integral, we found that the area is equal to $\frac{1}{3k^2}$. Setting this equal to 1, we obtained $k = \frac{1}{\sqrt{3}}$. Given the problem has an error, if the area was 1/9, then $k = \sqrt{3}$.

Final Answer The problem has an error. Given the area is 1, the correct answer is $k = \frac{1}{\sqrt{3}}$, which corresponds to option (D).