Computing double integral using linear algebra 
$$\iint_D (6x+2y) \, \mathrm d x \,\mathrm d y$$
where $D$ is the convex hull of $4$ given points,
$$D = \mbox{conv} \left\{ (0,0),(-2,6),(3,2),(1,8) \right\}$$
This is a parallelogram with "unit vectors" $(-2,6)$, $(3,2)$.

I wanted to give a try to solve the following problem with algebra instead of calculus. So, I thought about calculating the area of it and piling it up to get its volume. I got the cosine between both vectors with the dot product formula, which is $$\dfrac{6}{\sqrt{13\cdot40}}$$ and the sine with the Pythagoras identity which is $$\sqrt{\dfrac{484}{13\cdot40}}$$ I remember that $|a|\cdot |b|\cdot \sin\alpha$ gives the height, and $|a|\cdot |b|\cdot \cos\alpha$ gives the area. So I figured that maybe this would solve the integral problem for the area?
$$\dfrac{6}{\sqrt{13\cdot40}} \cdot \sqrt{13} \cdot \sqrt{40}\sqrt{\dfrac{484}{{13\cdot40}}}\cdot \sqrt{13} \cdot \sqrt{40}$$
That gives me $$6 \cdot 22 = 132$$ That's wrong but the right result is $$11 \cdot 22 = 242$$ so there might be something in that?
 A: The area of the domain is given by the cross product of two sides,
$$(2,-6)\times(3,2)=22.$$
Then the integral must be the area times the average value of the integrand, which by linearity is also the value of the function at the centroid, $\left(\dfrac12,4\right)$, hence
$$22\left(6\cdot\frac12+2\cdot4\right)=22\cdot11.$$

Your answer cannot be right, because you make no use of the coefficients $6$ and $2$ in the integrand. What you are evaluating is
$$|a|^2|b|^2\cos\alpha\sin\alpha$$
which has no justification.

Another solution is by noting that the volume is a truncated prism with bases $(0,0,0),$ $(-2,6,0),$ $(3,2,0),(1,8,0)$ and $(0,0,0),(-2,6,0),(3,2,22),(1,8,22)$, which is the half of a non-truncated prism of height $22$. Hence
$$22\cdot\frac{22}2.$$
A: Let
$$\begin{bmatrix} x\\ y\end{bmatrix} = \begin{bmatrix} 3 & -2\\ 2 & 6\end{bmatrix} \begin{bmatrix} u\\ v\end{bmatrix}$$
where the determinant of the matrix is $22$. Since 
$$6x + 2y = \begin{bmatrix} 6\\ 2\end{bmatrix}^\top \begin{bmatrix} x\\ y\end{bmatrix} = \begin{bmatrix} 6\\ 2\end{bmatrix}^\top \begin{bmatrix} 3 & -2\\ 2 & 6\end{bmatrix} \begin{bmatrix} u\\ v\end{bmatrix} = \begin{bmatrix} 22\\ 0\end{bmatrix}^\top \begin{bmatrix} u\\ v\end{bmatrix} = 22 u$$
we have
$$\iint_D (6x+2y) \, \mathrm d x \,\mathrm d y = \iint_{[0,1]^2 } (22 u)  (22 \, \mathrm d u \,\mathrm d v) = 22^2 \int_0^1 u \, \mathrm d u = \frac{22^2}{2} = 22 \cdot 11$$
