The image has been opened. Let me now interpret and solve problem 3 from the image.

Problem 3 appears to involve a double integral:

$\iint_T (x+y) \, dx \, dy$

The region $T$ is described as the triangle with vertices $(1,1)$, $(1,2)$, and $(2,1)$.

Step 1: Determine the Region $T$

The region $T$ is a right triangle bounded by:

1. $x = 1$ (vertical line),
2. $y = 1$ (horizontal line), and
3. the line $x + y = 3$ (derived from the points $(1,2)$ and $(2,1)$).

Step 2: Set Up the Double Integral

For the region, $T$, the limits of integration for $x$ range from $1$ to $2$. For each fixed $x$, the limits for $y$ go from $1$ to $3 - x$.

Thus, the double integral becomes:

$\int_{1}^{2} \int_{1}^{3 - x} (x + y) \, dy \, dx$

Step 3: Solve the Inner Integral

The inner integral is with respect to $y$:

$\int_1^{3 - x} (x + y) \, dy = \int_1^{3 - x} x \, dy + \int_1^{3 - x} y \, dy$

Now, let's compute both terms:

$\int_1^{3 - x} x \, dy = x \cdot \left[(3 - x) - 1\right] = x(2 - x)$

$\int_1^{3 - x} y \, dy = \left[\frac{y^2}{2}\right]_1^{3 - x} = \frac{(3 - x)^2}{2} - \frac{1}{2}$

Therefore, the result of the inner integral is:

$x(2 - x) + \left[\frac{(3 - x)^2}{2} - \frac{1}{2}\right]$

Step 4: Solve the Outer Integral

Now, we integrate the result with respect to $x$ from 1 to 2:

$\int_1^2 \left[ x(2 - x) + \left(\frac{(3 - x)^2}{2} - \frac{1}{2}\right) \right] dx$

I'll compute this integral next.The value of the double integral is:

$\iint_T (x + y) \, dx \, dy = \frac{4}{3}$

This is the solution to problem 3.

Would you like further details or explanations?

Here are five related questions to explore:

1. How do you change the order of integration in double integrals?
2. How do you set up double integrals for general triangular regions?
3. What is the geometric interpretation of double integrals?
4. How would the integral change if the region was bounded differently?
5. How do you compute double integrals using polar coordinates?

Tip: For triangular regions, it's often helpful to rewrite one of the variables as a function of the other to simplify integration limits.