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$\int \int_B$ $\frac{\sqrt[3]{y-x}}{1+y+x}dx.dy$ where $ B $ is a triangle of vertices $ (0,0), (1,0) $ and $ (0,1) $

I tried to replace it, so it was $\sqrt [3] {\frac {u} {v}}$

But I couldn't find the limits of integration

I also used the "jacobi correction" there was this replacement there times $\frac {1} {2}$

Jean Marie
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trombho
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2 Answers2

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Hint:

Note that $(x,y)\in B\iff (y,x)\in B$ and $$f(x,y)=-f(y,x)$$

$f$ is the integrand, of course.

ajotatxe
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As @ajotatxe has pointed out, symmetry considerations show that the integral over $B$ must be zero.

However, to demonstrate the use of transformations, let $u=y-x,v=y+x$, then $\frac{\partial (u,v)}{\partial (x,y)}=\begin{vmatrix}-1&1 \\1&1\\\end{vmatrix}=-2.$

On $B, 0\le v\le 1$ and $-v\le u\le v.$ The integral then becomes

$$\int_0^1 \int_{-v}^v -\frac{1}{2}\frac{\sqrt[3]u}{1+v}dudv.$$

Now $\int_{-v}^v \sqrt[3]udu=[\frac{3}{4}u^\frac{4}{3}]_{-v}^v=0$ and so the entire integral is $0$.

  • I have upvoted your answer. Nevertheless I think very ambiguous to consider $\sqrt[3]{y-x}$ for negative values of $y-x$ in the question (and therefore in your solution $u^{4/3}$ for negative values of $u$). – Jean Marie Nov 17 '19 at 00:54
  • Thanks for the upvote! I am interested to understand your concern about cube roots of negative values since they have a unique real value. –  Nov 17 '19 at 01:00
  • Doing that, you run into very strange things : for example I just asked to Matlab $(-1)^{1/3} $ and the answer wasn't $-1$, it was the complex number $1/2+\sqrt{3}/2i$ ; another example about the same cubic root : I start from $-1$ assumed to be $(-1)^{1/3}=(-1)^{2/6}=((-1)^2)^{1/6}=1^{1/6}=1$. Therefore $-1=1$, etc... – Jean Marie Nov 17 '19 at 01:09
  • See the strange "rules" in https://math.stackexchange.com/q/952663. The "rule of life" should be that one should use fractional exponents only for positive numbers, and write for example $sign(u) * |u|^{1/3}$ instead of $u^{1/3}$ to cover both cases $u \geq 0$ and $u<0$. – Jean Marie Nov 17 '19 at 01:16
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    Thanks. Just like we all adopt the convention that $\sqrt{4}$ is $2$ I have always assumed the cube root notation to be unambiguous. I was pushing it using the fractional power though. With your interesting 'paradoxical' proof I think the issue is with sixths rather than thirds because there are two real sixth roots. –  Nov 17 '19 at 01:17
  • This issue has been raised several times on Maths SE : here is a nuanced answer (though I don't fully agree with it) : https://math.stackexchange.com/q/317546 – Jean Marie Nov 17 '19 at 01:45