I want to check if my solution are right. Can someone help me?
I have this problem:
Let $D:=\{ x \in \mathbb{R}^3 : x_1+x_2+x_3=1\}$
Show if $\int_{D}x_1x_2x_3dS_2$ exists. If so, calculate it
So we have this formula: $\int_{φ(U)} f dS_k = \int_{U} (f ◦ φ) (det(Dφ)^T Dφ)^{1/2} dλ_k$
Where $U ⊆ \mathbb{R}^k$ open, $φ ∈ C^1 (U; \mathbb{R}^n )$ such that for all $x ∈ U$ we have rank $Dφ(x) = k$, and such that $φ : U → φ(U)$ is a homoomorphism.
So what I practically need to prove is that:
$rank(Dφ(x)) = k$
$φ$ is Homoomorphism: $φ$ should be bijective and $φ$ $φ^{-1}$ should be continous
So for 1. I have written for $φ(x_1,x_2)=(x_1,x_2,1-x_1-x_2)$ and therefore $Dφ=\begin{pmatrix} 1 & 0 \\ 0 & 1\\ -1& -1 \end{pmatrix}$ and therefore as rank $2$
- we can say that $φ$ is continous since $x_1,x_2,1-x_1-x_2$ are continous functions. $φ$ is bijective and it is trivial that $φ^{-1}$ is also continous.
So we can calculate $\int_{D}x_1x_2x_3dS_2$
To do it we first need to calculate $(det(Dφ)^T Dφ)^{1/2}$
It follows: $(det(Dφ)^T Dφ)=det(\begin{pmatrix} 1 & 0 \\ 0 & 1\\ -1& -1 \end{pmatrix}^T \cdot\begin{pmatrix} 1 & 0 \\ 0 & 1\\ -1& -1 \end{pmatrix})=det(\left(\begin{matrix} 2 & 1 \\ 1 & 2 \end{matrix}\right))=3$
So we have $\int_{D}x_1x_2x_3dS_2=\sqrt{3}\int_{0}^{1-x_1}\int_{0}^{1} x_1x_2(1-x_1-x_2)dx_1dx_2$
so far everything right? I am not sure about the limits of my integral and the function $φ$. I think that the function may be $φ(x_1,x_2)=(x_1,x_2,x_1-x_2)$ and not $φ(x_1,x_2)=(x_1,x_2,1-x_1-x_2)$.