Let $D \subset \mathbb{R}^3$ be the set $D = \{(x, y, z) \mid x^2 + y^2 = 1~\text{and}~z = 0\}$. Let $T : \mathbb{R}^3 \to \mathbb{R}^3$ be the linear transformation defined by $T(x, y, z) = (2x − y + z, x − y + 10z, 3x − 7y + 2z)$. Find the area of $T(D)$.
understanding the body to calculate the area : $$z = 0 \implies T(D) = T(x,y,0) = (2x-y,x-y,3x-7y)$$
Now let's name: \begin{align*} x' & = 2x-y\\ y' & = x-y\\ z' & = 3x-7y \end{align*}
Solving the equations we get $(x'-y')^2 + (x'-2y')^2 = 1$. This is the body of several parabolas (I think). How can I calculate the area?
This is an easier question than you are making it.
What does the transformation do the area square that encloses the circle?
Does it transform the area of all shapes in the plane equally? (yes it does)
What does it to the area of the square with vertices $(0,0,0),(1,0,0), (0,1,0),(1,1,0)$?
This square is easy to work with. And we know the area is 1.
$T(1,0,0) = (2,1,3)\\ T(0,1,0) = (-1,-1,-7)$
As for the other vertexes $T\mathbf 0 = \mathbf 0$
$T(1,1,0) = T(1,0,0)+T(0,1,0)$
But we don't really need to worry about the other vertices as
$\|(2,1,3)\times(-1,-1,-7)\|$
Tells us the area of the parallelogram.
This tells you how much areas in the plane will scale.
What was the area of D before the transformation?