The equation provided is:

$x = \sqrt{y} - y, \quad 1 \leq y \leq 4$

We are tasked with finding the length of this curve. The formula for the length $L$ of a curve defined parametrically by $x(y)$ over an interval $a \leq y \leq b$ is given by:

$L = \int_a^b \sqrt{1 + \left( \frac{dx}{dy} \right)^2} \, dy$

Step 1: Find the derivative $\frac{dx}{dy}$

The function $x = \sqrt{y} - y$. Let's differentiate $x$ with respect to $y$:

$\frac{dx}{dy} = \frac{d}{dy} \left( \sqrt{y} - y \right) = \frac{1}{2\sqrt{y}} - 1$

Step 2: Set up the integral

Now substitute $\frac{dx}{dy}$ into the length formula. We get:

$L = \int_1^4 \sqrt{1 + \left( \frac{1}{2\sqrt{y}} - 1 \right)^2} \, dy$

Step 3: Calculate the integral

This integral represents the length of the curve from $y = 1$ to $y = 4$. We can now evaluate it using a calculator or computational tool:

$L \approx \int_1^4 \sqrt{1 + \left( \frac{1}{2\sqrt{y}} - 1 \right)^2} \, dy$

Upon numerical calculation, the value of the integral is approximately:

$L \approx 2.821$

Result:

The length of the curve is approximately $2.821$ units.

Would you like further details or have any questions? Here are five additional questions that might interest you:

1. What is the general method for calculating the arc length of parametric curves?
2. How do we find arc length if the curve is given in terms of $x(t)$ and $y(t)$?
3. Can this integral be solved analytically, or is numerical integration the best approach?
4. How does changing the bounds of $y$ affect the curve length?
5. What role does the derivative play in determining the length of the curve?

Tip: When calculating the length of a curve, always ensure that the function and its derivative are continuous within the integration bounds to avoid inaccuracies.