The image shows two curves and an enclosed region in the Cartesian plane. The curves are:

• $y = x - 4$
• $x = y^2 - 2$

The objective is to calculate the area of the region enclosed by these curves using integration. The given formula for the area is represented as:

$\int_c^d m(y) \, dy$

where $c$ and $d$ are the limits of integration, and $m(y)$ is a function describing the curve in terms of $y$.

Steps to solve:

1. Find the limits of integration $c$ and $d$: These correspond to the $y$-values where the two curves intersect.

   To find the points of intersection, solve the system of equations:

   • $y = x - 4$
   • $x = y^2 - 2$

   Substitute $x = y^2 - 2$ into $y = x - 4$:

   $y = (y^2 - 2) - 4$ $y = y^2 - 6$ $y^2 - y - 6 = 0$

   Solve the quadratic equation $y^2 - y - 6 = 0$ using the quadratic formula:

   $y = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-6)}}{2(1)}$ $y = \frac{1 \pm \sqrt{1 + 24}}{2}$ $y = \frac{1 \pm \sqrt{25}}{2}$ $y = \frac{1 \pm 5}{2}$ Thus, $y = 3$ and $y = -2$ are the limits of integration.

   Therefore, $c = -2$ and $d = 3$.

2. Find the function $m(y)$: This function represents the difference between the two curves at a given $y$-value. Since $x = y^2 - 2$ describes the rightmost curve and $y = x - 4$ describes the leftmost curve, the function is:

   $m(y) = (y^2 - 2) - (y + 4)$ Simplifying: $m(y) = y^2 - y - 6$

3. Set up the integral: The area is found by integrating $m(y)$ from $y = -2$ to $y = 3$:

   $A = \int_{-2}^{3} (y^2 - y - 6) \, dy$

4. Solve the integral: $\int (y^2 - y - 6) \, dy = \frac{y^3}{3} - \frac{y^2}{2} - 6y$ Evaluate from $y = -2$ to $y = 3$.

Now, let me calculate the area for you.The area of the region enclosed by the curves is $\frac{-125}{6}$, which corresponds to an absolute value of approximately 20.83 square units (since area is always positive).

Would you like further details on any part of the solution?

Here are five related questions that expand on this concept:

1. How do you determine which curve represents the upper and lower bounds in a region?
2. What is the significance of using definite integrals for area calculations?
3. How can you find the intersection points of more complex curves?
4. How does the orientation of the curves affect the area calculation?
5. Can this method be extended to three-dimensional surfaces?

Tip: When finding areas between curves, always ensure that you subtract the lower curve from the upper curve within the integration limits to avoid negative areas.