5
$\begingroup$

How can I find the following integral:

$$\int^{1/2}_0 \int^{1-x}_x (x+y)^9(x-y)^9 \, dy \, dx $$

My thoughts:

Can we possibly convert this to spherical or use change of variables?

$\endgroup$

4 Answers 4

6
$\begingroup$

according to the shape of the area of integration and the shape of the function that is under integral, the easiest answer is to define variables $u=x+y$ and $v=x-y$then we have: $$\frac{1}{J}=\frac{\partial(u,v)}{\partial(x,y)}=\begin{vmatrix}1&1\\1&-1\end{vmatrix}\Rightarrow |J|=\frac{1}{2}$$ and the borders of the area of integration are $u=-v,u=1,v=0$ just draw the shape of the problem to understand this part.then:
$$\int^{1/2}_0 \int^{1-x}_x (x+y)^9(x-y)^9 \, dy \, dx=\int^{0}_{-1}\int^1_{-v} \frac{1}{2}u^9v^9\,du\,dv=\frac{-1}{400}$$


Edit: to answer the question asked in the comments:
why did I use $\frac{1}{|J|}$?
consider the analog situation for a 1-D integral where we have to use change of variables method:
$$\int\frac{x\,dx}{\sqrt{1-x^4}}$$
we have $u=x^2\Rightarrow du=2x\,dx$
meaning that we have $u$ as a function of $x$ so we can calculate $du=\frac{du}{dx}dx$ but in order to substitute $dx$ in the original integral we write $du=2x\,dx\Rightarrow dx=\frac{1}{2x}du$ which means that we need $dx=\frac{dx}{du}du$ to substitute in the original integral so
In a 1-D integral we have $u$ as a function of $x$ so we can calculate $\frac{du}{dx}$ but to substitute in the original integral we need $\frac{dx}{du}$
here is the same situation:
to substitute in the original integral, we need:
$$dx\,dy=\frac{\partial(x,y)}{\partial(u,v)}du\,dv=|J|du\,dv$$
But most of the time we have $u$ and $v$ as a function of $x$ and $y$:
$$u=f(x,y),v=g(x,y)$$
so we can compute $$\frac{1}{|J|}=\frac{\partial(u,v)}{\partial(x,y)}$$

$\endgroup$
7
  • $\begingroup$ Does the definition of your new variables have something to do with an affine transformation? $\endgroup$
    – Khallil
    Jul 27, 2015 at 20:37
  • 1
    $\begingroup$ @Khallil as you know we can use the method change of variables in 1-D integrals to calculate them easier just like that we can use change of variables in a 2-D or 3-D integral. and the jacobi above defines the relationship between old variables and new ones. change of variables is just mapping from a space to another space just like the affine transformation is. $\endgroup$ Jul 27, 2015 at 21:06
  • $\begingroup$ May I ask why you chose to define your variables in such a way? $\endgroup$
    – Khallil
    Jul 27, 2015 at 21:14
  • 1
    $\begingroup$ @Khallil because I want to find a way to calculate the above integral easier. if you draw the borders of the original integral in the question you will see that it's a triangle sides $x+y=1,x-y=0,x=0$ also you have the terms $x-y,x+y$ in the function so you will deduce that choosing the variables this way will make both the function and the borders easier. I should say that choosing the new variables is just a trick you will learn as much as you solve problem.practice makes perfect $\endgroup$ Jul 27, 2015 at 21:27
  • $\begingroup$ Thanks for the help, @sepideh! $\endgroup$
    – Khallil
    Jul 27, 2015 at 21:28
2
$\begingroup$

Hint: $$(x+y)^9(x-y)^9=((x+y)(x-y))^9=(x^2-y^2)^9$$

$\endgroup$
4
  • $\begingroup$ Okay, so we can introduce an 'r here by saying that $(-(x^2+y^2))^9=(-r^2)^9=-r^{18}$. Right? $\endgroup$ Jul 27, 2015 at 19:53
  • $\begingroup$ Okay I got that much. Now what do I do next? $\endgroup$ Jul 27, 2015 at 20:03
  • $\begingroup$ Actually, the expression is $(x^2-y^2)^9$. Now you can use binomial expansion of $(x^2-y^2)^9$ so that all the terms are separated in powers of $x$ & $y$ then integrating all, first w.r.t. $y$ then w.r.t. $x$ It can be done but the procedure/expansion will be lengthy. $\endgroup$ Jul 27, 2015 at 20:17
  • $\begingroup$ It doesn't seem efficient or intuitive. Just a hammering of algebra. $\endgroup$
    – Khallil
    Jul 27, 2015 at 20:39
1
$\begingroup$

How about a change of variable like $u=x+y, v=x-y$?

The Jacobian is $-\frac 12$, and the area of integration is the triangle bounded by the lines $x=y, x+y=1, x=0$

This translates as: $v$ varies from $v=0$ to $ v=u$ for the inner integral, and $u=0$ to $u=1$ for the outer integral.

Therefore we evaluate $$\int_{u=0}^{u=1}\int_{v=0}^{v=u}u^9v^9(-\frac 12)dvdu$$

The final answer is $-\frac{1}{400}$

$\endgroup$
1
  • $\begingroup$ Yes. Actually I was wanting something in this direction. Can you please show me this step-by-step if you don't mind? $\endgroup$ Jul 27, 2015 at 19:56
0
$\begingroup$

\begin{align} u & = x+y \\ v & = x-y \end{align} $$ du\,dv = \left|\frac{\partial (u,v)}{\partial(x,y)}\right|\,dx\,dy = 2\,dx\,dy $$

\begin{align} & \int^{1/2}_0 \int^{1-x}_x (x+y)^9(x-y)^9 \, dy \, dx \\[10pt] = {} & \int_0^1 \left( \int_{u-1}^0 u^9 v^9 2\,dv \right) \,du \\[10pt] = {} & \int_0^1 u^9 \frac{(-(u-1)^{10})} 5 \, du \end{align}

$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .