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I need to solve the following integral using change of variables: $$\int\int_D\frac{\sqrt[3]{y-x}}{1+x+y}dA$$ where D is the triangle with vertices $(0,0)$, $(1,0)$ and $(0,1)$.

I tried to change the variables to $u=y-x$ and $v=1+x+y$, but then I couldn't solve the integral. Any tips on how to solve it (or the full solution) would be highly appreciated! Thanks in advance!

EDIT: I should've been more detailed in what I did. I tried the change $u=y-x$ and $v=1+x+y$, found the new domain and calculated the Jacobian. The integral then is: $$\int_1^2\int_{1-v}^{v-1}\frac{\sqrt[3]{u}}{v}\frac{1}{2}dudv$$ That is what I can't solve. I can solve it in u, but I don't even know if it's right: $$\frac{3}{8}\int_1^2\frac{{(v-1)}^{4/3}-{(1-v)}^{4/3}}{v}dv$$

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  • $\begingroup$ You are on the right track. $\endgroup$ – Pocho la pantera Oct 9 '13 at 20:07
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    $\begingroup$ Sym-me-try, sweeeet sym-me-try. What's $\int_{-a}^a \sqrt[3]{u}\,du$? $\endgroup$ – Daniel Fischer Oct 9 '13 at 20:26
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Your triangle is bounded above by the line $y=1-x,$ meaning $y\le 1-x,$ which we can rewrite as $x+y\le1.$ This suggests using $v=x+y$ as your other variable, rather than your choice of $v$.

Since we put $$u=y-x\\v=x+y,$$ then we have $$y=\frac{u+v}2\\x=\frac{v-u}2.$$ It can be shown, then, that the Jacobian is $-\frac12,$ so that $$\iint_D\frac{\sqrt[3]{y-x}}{1+x+y}\,dA=-\frac12\iint_D\frac{\sqrt[3]{u}}{1+v}\,du\,dv.$$ It remains only to find the new limits of integration that describe $D$.

Since $x$ is bounded below by $0,$ then $$0\le\frac{v-u}2\\0\le v-u\\u\le v$$ Since $y$ is bounded below by $0,$ then $$0\le\frac{u+v}2\\0\le u+v\\-v\le u.$$ Hence, we have $-v\le u\le v$ and $v\le 1$ (by previous discussion). Also, we have $0\le v$ (why?), and so the integral becomes $$-\frac12\int_0^1\int_{-v}^v\frac{\sqrt[3]{u}}{1+v}\,du\,dv=-\frac12\int_0^1\frac1{1+v}\left(\int_{-v}^v\sqrt[3]{u}\,du\right)\,dv.$$

Can you take it from there?

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  • $\begingroup$ I can take it until there. I did that with both changes, the one I stated in the OP and the one you showed, but then I don't actually know how to solve the integral. $\endgroup$ – MSEguest Oct 9 '13 at 20:23
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    $\begingroup$ I've altered the last bit of my answer to hopefully make it simpler to see what comes next. (Hint: Symmetry.) $\endgroup$ – Cameron Buie Oct 9 '13 at 20:27
  • $\begingroup$ Is it zero? I got to zero and just thought I was doing something wrong... $\endgroup$ – MSEguest Oct 9 '13 at 20:31
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    $\begingroup$ Yes, indeed! The change in variables let us find out that the integral actually has a nice value, which we couldn't have simply found any other way (that I know of). $\endgroup$ – Cameron Buie Oct 9 '13 at 20:32
  • $\begingroup$ Thank you very much then, you were of great help! $\endgroup$ – MSEguest Oct 9 '13 at 20:36
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with the change of variables you need to

1) Transform the Domain $D$ by using the $(x,y)\rightarrow (u,v)$ i.e. $(0,0)\rightarrow (0-0,1+0+0)=(0,1)$. Do this for all points.

2) Compute the Jacobian $$ \mathrm{dA}=\left| \begin{array}{cc} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{array} \right|\partial u\partial v. $$

Then combine them all together and solve with the appropriate limits based on the new transformed boundary of the integral.

Hope it helps.

Rob

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Hint: $$ \int\int_D\frac{\sqrt[3]{y-x}}{1+x+y}dA=\int_{1}^{2}\int_{A}^{B} \frac{\sqrt[3]{u}}{v}\,\,J\,\,du\, dv $$ where $A=A(v)$ and $B=B(v)$ and $J$ is the jacobian.

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  • $\begingroup$ That's what I did, A=1-v, B=v-1 and J=1/2. But then I just can't solve it. $\endgroup$ – MSEguest Oct 9 '13 at 20:16
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    $\begingroup$ I understand. Do you try interchanging the order of integration? $\endgroup$ – Pocho la pantera Oct 9 '13 at 20:22
  • $\begingroup$ Was about to try when Cameron Buie's comment made me realize what I was missing. But thanks for the help! $\endgroup$ – MSEguest Oct 9 '13 at 20:35
  • $\begingroup$ Yes, Cameron gave the key observation. $\endgroup$ – Pocho la pantera Oct 9 '13 at 20:39

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