Question: I want to calculate to volume of the subset $\{ \begin{pmatrix} 2 & 33 & 444\\ 0 & 3 & 4 \\ 0 & 0 & 4 \end{pmatrix}\begin{pmatrix}x \\y \\ z \end{pmatrix}:x^2+y^2+z^2 \leq 1\}$ of $\mathbb{R}^3$. My thoughts are the following. I know that the determinant is a scaling factor, so the the linear transformation $T\begin{pmatrix}x \\y \\ z \end{pmatrix}=\begin{pmatrix} 2 & 33 & 444\\ 0 & 3 & 4 \\ 0 & 0 & 4 \end{pmatrix}\begin{pmatrix}x \\y \\ z \end{pmatrix}$ scales the unit sphere $x^2+y^2+z^2 \leq 1$ with a factor that is the determinant of matrix which I have calculated to 24. The volume of the unit sphere is $4\pi/3$ and so the volume of the subset is $24 \cdot 4\pi/3=32\pi$. I am not sure if this method works and I find it hard to see from the expression if it is a reasonable answer.

Answer: 
I like [Jack](https://math.stackexchange.com/users/9464/jack) answer. If you want to use a more general theorem of integration, you can use [substitution for multiple variables](https://en.m.wikipedia.org/wiki/Integration_by_substitution) theorem which states:

$$\int\_{T(U)} f(\mathbf y) \ d\mathbf{y}=\int\_U f(T(\mathbf x)) \vert \det T^\prime (\mathbf x) \vert \ d \mathbf x$$
Here $T$ is linear. So its differential at each point is itself. And as you want to compute a volume, you'll take for $f$ the constant map equal to $1$. And for $U$ the unit ball $B$. Hence, you get
$$\int\_{T(B)}\ d\mathbf{y}=\int\_B \vert \det T \vert \ d \mathbf x$$

Therefore
$$\text{Vol}(T(B)) = (\det T) \cdot \text{Vol}(B) =24\cdot \frac{4}{3} \pi=32\pi$$
You got the wrong determinant. 

In general, one has the following theorem in measure theory:

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> If $T$ is a linear transformation and $A$ is a measurable subset of $R^n$, then $T(A)$ is also Lebesgue measurable and has the measure ${\displaystyle |\det(T)|\,\lambda \,(A)}$, where $\lambda(A)$ is the Lebesgue measure of $A$.
> 
> 
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