To find the area bounded by the curves $y = x^2$, $x - y = 3$, $x = -3$, and $x = 2$ using a double integral, we follow these steps:

1. Identify the region:

   • The curve $y = x^2$ is a parabola opening upwards.
   • The line $x - y = 3$ can be rewritten as $y = x - 3$.
   • We need to find the intersection points of $y = x^2$ and $y = x - 3$.

2. Find intersection points: Set $x^2 = x - 3$: $x^2 - x + 3 = 0.$ The discriminant is: $D = (-1)^2 - 4 \cdot 1 \cdot 3 = 1 - 12 = -11.$ Since $D < 0$, there are no real intersections.

   We will instead check the boundaries set by $x = -3$ and $x = 2$.

3. Determine the bounds of integration:

   • For $x = -3$, $y = (-3)^2 = 9$.
   • For $x = 2$, $y = 2^2 = 4$.

4. Set up the double integral: The area $A$ can be found using: $A = \int_{-3}^{2} \left( (x - 3) - x^2 \right) \, dx,$ where $(x - 3)$ is the upper boundary and $x^2$ is the lower boundary.

5. Evaluate the integral: $A = \int_{-3}^{2} (x - 3 - x^2) \, dx = \int_{-3}^{2} (-x^2 + x - 3) \, dx.$

   Now, we compute the integral: $\int (-x^2 + x - 3) \, dx = -\frac{x^3}{3} + \frac{x^2}{2} - 3x.$ Evaluating from $-3$ to $2$: [ A = \left[-\frac{(2)^3}{3} + \frac{(2)^2}{2} - 3(2)\right] - \left[-\frac{(-3)^3}{3} + \frac{(-3)^2}{2} - 3(-3)\right]. ]

   Calculating the values: [ = \left[-\frac{8}{3} + 2 - 6\right] - \left[\frac{27}{3} + \frac{9}{2} + 9\right]. ] [ = \left[-\frac{8}{3} - 4\right] - \left[9 + 4.5 + 9\right]. ] [ = \left[-\frac{8}{3} - \frac{12}{3}\right] - [22.5]. ] $= -\frac{20}{3} - 22.5 = -\frac{20}{3} - \frac{67.5}{3} = -\frac{87.5}{3}.$ Thus, the area $A$ is: $A = \frac{87.5}{3} \text{ square units}.$

Summary

The area bounded by the curves is $\frac{87.5}{3}$ square units.

Do you need any details or have questions? Here are some related questions for further exploration:

1. How would the area change if the bounds were altered?
2. What is the significance of the intersection points in determining the area?
3. Can this method be applied to other bounded areas defined by different functions?
4. How does changing the shape of the curves affect the integration process?
5. What are alternative methods for calculating the area between curves?

Tip: Always sketch the curves to visualize the bounded area before setting up the integral.