The value of the integral
\[
\iint_D 3(x^2 + y^2)\, dx\, dy,
\]
where $D$ is the shaded triangular region, is __________ (rounded off to the nearest integer).
Show Hint
For symmetric triangular regions, convert the double integral into $\int_{x}\int_{y}$ using the linear boundaries.
The triangular region $D$ has vertices:
\[
(0,0),\quad (4,4),\quad (4,-4)
\]
The left boundary is the $y$-axis, the right boundary is $x=4$,
and the two slanted boundaries are:
\[
y = x,\quad y = -x
\]
Thus, $x$ varies from 0 to 4.
For a fixed $x$, $y$ varies from $-x$ to $x$.
So the integral becomes:
\[
\iint_D 3(x^2 + y^2)\, dx\, dy
=
\int_{0}^{4} \int_{-x}^{x} 3(x^2 + y^2)\, dy\, dx
\]
First integrate with respect to $y$:
\[
\int_{-x}^{x} 3(x^2 + y^2)\, dy
= 3\left[ x^2 y + \frac{y^3}{3} \right]_{-x}^{x}
\]
Evaluate:
\[
= 3\left( x^3 + \frac{x^3}{3} -(-x^3 - \frac{x^3}{3})\right)
\]
\[
= 3\left( \frac{4x^3}{3} + \frac{4x^3}{3}\right)
= 3\left( \frac{8x^3}{3} \right)
= 8x^3
\]
Now integrate with respect to $x$:
\[
\int_{0}^{4} 8x^3\, dx
= 8 \left[ \frac{x^4}{4} \right]_{0}^{4}
= 2 (4^4)
= 2 \times 256
= 512
\]
Thus the value of the integral is:
\[
\boxed{512}
\]
