If $x^{3/2}+y^{3/2}=r^{3/2}$, then $u=r$ and $v=r-x$.

[mathmari]

Hey!

Let $D$ be the space in the first quadrant of the $xy$-plane that is defined by the inequality $x^{\frac{3}{2}}+y^{\frac{3}{2}} \leq \alpha^{\frac{3}{2}}$ with $\alpha>0$. I want to transform $\iint_D f(x,y) dx dy$ to an integral on the triangle $E$ of the $uv$-plane that is defined by the inequalities $0 \leq u \leq \alpha$ and $0 \leq v \leq \alpha-u$.
Which new variables do we define here? (Wondering) Do we maybe define the following variables?
$$u=\left (x^{\frac{3}{2}}+y^{\frac{3}{2}}\right )^{\frac{2}{3}}$$ Then we would get $0\leq \left (x^{\frac{3}{2}}+y^{\frac{3}{2}}\right )^{\frac{2}{3}}\leq \left (a^{\frac{3}{2}}\right )^{\frac{2}{3}} \Rightarrow 0\leq u\leq a$.

But what about $v$ ? (Wondering)

 

[I like Serena]

Hi mathmari! (Smile)

What does the region look like?
Can we draw it?
Perhaps we can see then how best to transform it into a triangle. (Thinking)

You current idea for u corresponds more or less with a polar length, which would suggest that v could be an angle. That may not be the most natural fit though, and seems unlikely to lead to an upper bound of $\alpha -u$. (Nerd)

 

[mathmari]

What does the region look like?
Can we draw it?
Perhaps we can see then how best to transform it into a triangle. (Thinking)

The region is in the following form:

[DESMOS=-5,5,-5,5]x^{\frac{3}{2}}+y^{\frac{3}{2}}\le a^{\frac{3}{2}};a=2[/DESMOS] Do we not have to change only the curve that goes through the points $(0,2)$ and $(2,0)$ to get a triangle? (Wondering)

 

[I like Serena]

The region is in the following form:

Do we not have to change only the curve that goes through the points $(0,2)$ and $(2,0)$ to get a triangle?

Indeed. Hmmm, could we for instance define $u=x$, and try to find a $v$ such that the curve becomes a straight line? (Wondering)

 

[mathmari]

Indeed. Hmmm, could we for instance define $u=x$, and try to find a $v$ such that the curve becomes a straight line? (Wondering)

Does the curve has to become a straight line that passes through the points $(2,0)$ and $(0,2)$ ? (Wondering)

 

[I like Serena]

Does the curve has to become a straight line that passes through the points $(2,0)$ and $(0,2)$?

That is what I would propose we try.
But there is no 'have to' to it.

 

[mathmari]

That is what I would propose we try.
But there is no 'have to' to it.

So, does it holds that $v=-u+2$ ? (Wondering)

 

[I like Serena]

So, does it holds that $v=-u+2$ ? (Wondering)

That only holds on 1 of the 3 bounding curves (assuming we have $\alpha=2$).
But we can start with that bounding curve, and then try to generalize. (Thinking)

 

[mathmari]

That only holds on 1 of the 3 bounding curves (assuming we have $\alpha=2$).
But we can start with that bounding curve, and then try to generalize. (Thinking)

Will the triangle be bounded by the lines $v=-x+a$, $u=0$, $v=0$ ? (Wondering)

 

[I like Serena]

Will the triangle be bounded by the lines $v=-x+a$, $u=0$, $v=0$ ? (Wondering)

Yes.
Now suppose we have a point (x,y) inside D with, say, an $r$ such that $x^{3/2}+y^{3/2}=r^{3/2}$ and $0\le r \le \alpha$,
What would $u$ and $v$ be then in terms of $x$ and $y$? (Wondering)