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The problem asks for the mass of a metallic sheet \( L \), which is described in the Cartesian plane as:

\[
L = \{(x, y) \in \mathbb{R}^2 : 0 \leq x \leq 1, -x \leq y \leq x + 1\}
\]

The density of the sheet is given by the function \( \gamma(x, y) = x + y + 1 \).

The mass \( M \) of the lamina can be computed by integrating the density function over the region \( L \):

\[
M = \iint_{L} \gamma(x, y) \, dA
\]

Now, let's break down the region of integration. The bounds are:

- For \( x \), \( 0 \leq x \leq 1 \).
- For \( y \), \( -x \leq y \leq x + 1 \).

We can set up the double integral:

\[
M = \int_{0}^{1} \int_{-x}^{x+1} (x + y + 1) \, dy \, dx
\]

I will now calculate this double integral to find the mass of the lamina.The mass of the metallic sheet is \( \frac{25}{6} \).

Would you like more details on how the integral was computed, or do you have any further questions?

Here are some related questions:
1. What would change if the density function was \( \gamma(x, y) = x^2 + y^2 \)?
2. How would you find the center of mass of the lamina given this density function?
3. What are the physical implications of having a non-uniform density?
4. How does the shape of the region affect the computation of mass?
5. Can this type of problem be generalized to three dimensions?

**Tip:** In multivariable calculus, it's useful to sketch the region of integration to better understand the problem.

Uma lâmina metálica L descrita matematicamente no plano cartesiano como L={(x,y)∈R^2:0≤x≤1;-x≤y≤x+1} tem sua densidade modelada pela função γ(x,y)=x+y+1. A massa da lâmina é dada por: (calcular a massa).

Learn how to calculate the mass of a lamina with a non-uniform density described by γ(x, y) = x + y + 1. The problem is solved using double integration over a defined region in the Cartesian plane.

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