To find the constant $c$ in the joint probability density function $f(x,y) = c \cdot x^2 \cdot (1+y)$ over the region $0 \leq x \leq 3$ and $0 \leq y \leq 3$, we use the fact that the total probability must equal 1. This can be expressed mathematically as:

$\int_{0}^{3} \int_{0}^{3} c \cdot x^2 \cdot (1+y) \, dx \, dy = 1$

Step 1: Compute the double integral

1. Integrate with respect to $x$:

$\int_{0}^{3} \int_{0}^{3} c \cdot x^2 \cdot (1+y) \, dx \, dy = c \int_{0}^{3} \left[\int_{0}^{3} x^2 \cdot (1+y) \, dx \right] dy$

First, calculate the inner integral with respect to $x$:

$\int_{0}^{3} x^2 \cdot (1+y) \, dx = (1+y) \int_{0}^{3} x^2 \, dx = (1+y) \left[\frac{x^3}{3}\right]_{0}^{3} = (1+y) \cdot \frac{27}{3} = 9(1+y)$

Step 2: Integrate with respect to $y$

Now, calculate the outer integral with respect to $y$:

$c \int_{0}^{3} 9(1+y) \, dy = 9c \int_{0}^{3} (1+y) \, dy = 9c \left[ y + \frac{y^2}{2} \right]_{0}^{3}$

Evaluating this integral:

$9c \left[ 3 + \frac{9}{2} \right] = 9c \cdot \frac{15}{2} = \frac{135c}{2}$

Step 3: Set the total probability to 1

Set the integral equal to 1:

$\frac{135c}{2} = 1$

Solve for $c$:

$c = \frac{2}{135}$

Final Answer:

The constant $c$ is $\boxed{\frac{2}{135}}$.

Would you like more details on any step or further questions?

Here are five related questions:

1. How do you verify if a function is a valid joint probability density function?
2. What is the significance of the constant $c$ in a probability density function?
3. Can the marginal distributions be calculated for this joint PDF?
4. How do you find the probability of a specific region using a joint PDF?
5. What are the properties of a joint probability density function?

Tip: When dealing with joint probability density functions, always check the limits of integration to ensure they cover the entire region of interest.