How to find the area of Transformation on a Set D

Question

i have question that i asked here before but i didn't get clear answer , its not a Spam but i cant understand this question and i need help thank you.

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 $T(D)$.

i know that the new area can be anything i dont know why should i use the coordinates (0,0,1) , (0,1,0) i know that they are on the circule but how could i find the area when i dont know the shape of it on R3

The Questino is most likely about Vector Product / dot product

Reference : Question taken from Technion institute of Technology

Course : Calculas 2t for electrical Enginners.

heres what i got after thinking a bit :

the trasnformation gets only two points (1,0,0) , (0,1,0) Based on conditions from Set D .

thats all every other T(point) is a liner combination of the Image T(1,0,0) T(0,1,0).

so its a plane of the two vectors and each combination of them will still be on the plane so the area is the plane given by these vectors

so we use cross product and we get the area .

is this true ?.

Answer 1

Since a transformation is a vector function: $$T(x,y,y)=(2x-y+z)\vec i+(x-y+10z)\vec j+(3x-7y+2z)\vec k$$

Where, the bounds of integration are $z=0$ and $x^2+y^2=1$, given this one can use integration to integrate the function $T$ on $D$, and it will give you the area since if you notice $z=0$. Which means the Domain of integration is a disk or $0\leq r\leq 1, z=0$, and $0\leq \theta\leq 2\pi$. these will come handy later.

Let, $$x=cos(\theta), y=sin(\theta),z=0$$$$dx=-sin(\theta)d\theta,dy=cos(\theta)d\theta,dz=0$$

The integral, then becomes: $$\int\int_V2dxdy-2dzdx-17dydz$$

where V is the region given by the set $D$.

However, the integrands with the term $dz$ cancel out yielding: $$\int\int_V2dxdy$$

After putting in the values the integral ends up yielding 0. Which is not hard to see, Since $sin(2x)$ is periodic.

Answer 2

First of all, I presume there must $\leq$ sign instead of $=$ in the expression $x^2 + y^2 = 1$ in the definition of $D$. Here you indeed don't need to use double integral. I'll give you an intuitive solution to this problem.

Consider first not circle but square, what would happen to its area after this transformation? The area of the square can be found by two vectors that form its sides $u,v \in \mathbb{R}^3$ (e.g. using vector product $\|u \times v\|$), after applying transform these two vectors become $T(u), T(v) \in \mathbb{R}^3$. And the area is equal to $$\|T(u) \times T(v)\| = \|u \times v\| \cdot |\det T| \quad \text{(prove this)}$$ Intuitively, the area of a circle can be formed by infinetely many small squares, and hence it's possible to sum up all the areas and get the result: $$\text{Area}(D) \cdot |\det T|$$ Greetings from Technion!