So I have this exercise:
A is defined as $A := \{ (x,y) \in \mathbb{R}^2: 0<y<x \text{ and } 0<x+y<1\}$
We must show that $φ : A → \{(s, t) ∈ \mathbb{R}^2 : 0 < s < t < 1\}$ $φ((x, y)) = (x − y, x + y)$ is a $C^1$ Diffeomorphism.
We must calculate $\int_{A}(x + y)^{1/3} (x − y)^{1/2} dλ_2(x, y)$
I have already proven 1. but with 2. I dont know If I have done everything right
So we the assumptions of 1. we can use the integration by substitution with $s=x-y$ and $t=x+y$
We have $1/2 \int_{A}(x + y)^{1/3} (x − y)^{1/2} dλ_2(x, y)=\\ 1/2\int_{φ(A)}(t)^{1/3} (s)^{1/2} dλ_2(t, s)=\\ 1/2\int_{0}^{1}(t)^{1/3}(\int_{0}^{t}(s)^{1/2}ds)dt=\\ 1/2\int_{0}^{1}(t)^{1/3}[2/3s^{3/2}]_{0}^{t}=\\ 1/2 \cdot 2/3 \int_{0}^{1} t^{11/6}dt=\\ 1/2 \cdot 2/3\cdot 6/17[t^{17/6}]_{0}^{1}=1/2 \cdot 2/3\cdot 6/17=2/17$
Are all steps correct?
You have to adjust by Jacobian when you make a substitution. The general formula is $$d\lambda_2(x,y)={\LARGE\mid}\frac{\partial(x,y)}{\partial(t,s)}{\LARGE\mid} d\lambda_2(t,s),$$ where $\frac{\partial(x,y)}{\partial(t,s)}$ denotes the Jacobian determinant $$\frac{\partial(x,y)}{\partial(t,s)}=\frac{\partial x}{\partial t}\frac{\partial y}{\partial s}-\frac{\partial x}{\partial s}\frac{\partial y}{\partial t}.$$ Therefore, you need to multiply $1/2$ at the substitution. There seems to be a careless error after the integral on $s$, too: the last two lines will be $$\begin{split}& =\frac{1}{2}\int^1_0 \frac{2}{3} t^{\frac{11}{6}}dt\\ &=\frac{1}{3}\bigl[\frac{6}{17}t^{\frac{17}{6}}\bigr]^1_0\\ &=\frac{2}{17}. \end{split}$$