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 alredy proven 1. but with 3. I dont know If I have done everzthing right
To compute the integral $\int_{A}(x + y)^{1/3} (x − y)^{1/2} dλ_2(x, y)$, I used the change of variables formula:
$$\int_A f(x,y) d\lambda_2(x,y) = \int_{\phi(A)} f(\phi^{-1}(s,t)) |J_{\phi^{-1}}(s,t)| d\lambda_2(s,t)$$
where $|J_{\phi^{-1}}(s,t)|$ is the determinant of the Jacobian matrix of the inverse function $\phi^{-1}(s,t)$.
To calculate $\int_{A}(x + y)^{1/3} (x − y)^{1/2} dλ_2(x, y)$, we can use the change of variables formula. Let $s = x-y$ and $t = x+y$, so that $x = \frac{1}{2}(s+t)$ and $y = \frac{1}{2}(t-s)$. Then the region $A$ is transformed into the region ${(s, t) \in \mathbb{R}^2 : 0 < s < t < 1}$. The Jacobian of this transformation is $$J(s,t) = \begin{pmatrix} \frac{\partial x}{\partial s} & \frac{\partial x}{\partial t} \\ \frac{\partial y}{\partial s} & \frac{\partial y}{\partial t} \end{pmatrix} = \begin{pmatrix} \frac{1}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{1}{2} \end{pmatrix}$$ so we have $|J(s,t)| = \frac{1}{2}$.
We can now use the transformation rule to compute the integral over $A$ in terms of the integral over ${(s,t) \in \mathbb{R}^2 : 0 < s < t < 1}$. Let $f(s,t) = (t-s)^{1/2}(t+s)^{1/3}$ and $g(x,y) = (x+y)^{1/3}(x-y)^{1/2}$. Then we have:
\begin{align*} \int_A g(x,y) d\lambda_2(x,y) &= &= \int_{0}^{1}\int_{s}^{1}f(s,t)|J_\varphi(\varphi^{-1}(s,t))|dt ds \ &= \int_{0}^{1}\int_{s}^{1}(s)^{1/2}(t)^{1/3}1/2 dt ds \ &= 1/2 \int_{0}^{1} (s)^{1/2}[3/4t^{4/3}]_{s}^{1}\ &=...=1/6 \end{align*}
Therefore, $\int_{A}(x + y)^{1/3} (x − y)^{1/2} d\lambda_2(x, y) = \frac{1}{6}$.
I want to check if I have done some errors since it is my first time using the integration by substitution. Can someone help me?
Perform the following change of coordiantes: $$\begin{cases} s=x-y\\ t=x+y \end{cases}$$ Now, you have to see what is the new domain of integration with this coordinates: $$\begin{cases} 0<y<x\\ 0<x+y<1 \end{cases} \to \begin{cases} 0<(t-s)/2<(t+s)/2\\ 0<t<1 \end{cases}\to \begin{cases} 0<t-s<t+s\\ 0<t<1 \end{cases} $$ To visualize the first inequality, you can draw the graph (it's a triangle); now you're ready to integrate: $$I=\int_0^1\int_0^t \frac{1}{2}t^{1/3}s^{1/2}dsdt=\frac{2}{17} $$