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I'm studying to a Calculus test and I couldn't solve the following exercise:

Calculate $\iint_D \cos \left( \frac{x-y}{x+y} \right)\,dA$ where $D$ is the region of the plane $xy$ limited by $x+y=1$, $x=0$ and $y=0$.

After a little of thinking I've got the following:

$$ \int_0^1\int_0^{1-y}\cos\left(\frac{x-y}{x+y}\right)\,dx\,dy $$

I have no idea how to solve it. Wolfram Alpha couldn't solve it too.

jimjim
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Rodrigo Siqueira
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  • There is some theorem, (basically the Stokes thm for $2$ dimension) which leads this integral back to a line integral over the boundary of $D$, which now consists of $3$ segments, and the fraction seems easier on all of them. I forgot the details, a starting point could be http://en.wikipedia.org/wiki/Stokes_theorem. – Berci Mar 02 '13 at 22:50
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    @Berci you're thinking of Green's Theorem (fyi) – apnorton Mar 02 '13 at 22:59

2 Answers2

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Put $u=(x+y)/2$ and $v=(x-y)/2$. Reversing, $x=u+v,\ y=u-v$.

Then the border $x+y=1$ becomes $u=1/2$, while $x=0$ becomes $v=-u$ and $y=0$ becomes $v=u$. There will be a factor of 2 or 1/2 (I'll leave that to you to look up) because of the jacobian of the transformation, but the new setup of the iterated integral as $$\int_0^{1/2} \int_{-u}^u cos(v/u)\ dv\ du$$ becomes doable and if i did it right comes out $\sin(1)/4$.

The topic to check is change of variables in a double integral, look for the "jacobian" of the transformation. Here the jacobian is definitely a constant, either 2 or 1/2, I just forgot which one to multiply the transformed integral by. However I think this is the way needed to get the integral.

ADDED: I just integrated your form numerically on maple, and comparing things found it agrees with the value $\sin(1)/2$. So looks like the integral jacobian factor multiplier is 2, rather than 1/2. This makes sense since the area of the new triangle iterated over is only half the original integrating triangle.

coffeemath
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So, I've done the following:

$$ \iint_D \cos \left( \frac{x-y}{x+y} \right) \, dA, \,\,\,\, D: \{x+y=1, \, x=0, \, y=0\} $$

Substitution: $$ \alpha = x-y \therefore x = \frac{\alpha+\beta}{2} \\ \beta = x+y \therefore y = \frac{\beta-\alpha}{2} \\ J = \begin{vmatrix}\alpha_x & \alpha_y\\\beta_x & \beta_y\end{vmatrix} = \begin{vmatrix}1 & -1 \\ 1 & 1\end{vmatrix} = 2 \\ $$ So, $x+y=1$ becomes $\beta = 1$, $x = 0$ becomes $\alpha = -\beta$ and $y = 0$ becomes $\alpha=\beta$. Let's setup the integral: $$ \int_0^1 \int_{-\beta}^\beta \cos \left( \frac{\alpha}{\beta} \right) J \, d\alpha \, d\beta \\ = 2 \int_0^1 \int_{-\beta}^\beta \cos \left( \frac{\alpha}{\beta} \right) \, d\alpha \, d\beta $$

Variable substitution: $$ u = \frac{\alpha}{\beta} \\ du = \frac{d\alpha}{\beta} $$

The integral becomes: $$ 2 \int_0^1 \int_{-1}^1 \cos u \, \beta \, du \, d\beta $$

Solving: $$ \begin{align*} 2 \int_0^1 \int_{-1}^1 \cos u \; \beta \; du \; d\beta & = 2 \int_0^1 \left[ \sin u \, \beta \right]_{-1}^1 \, d\beta \\ &= 2 \int_0^1 2\beta\sin 1 \, d\beta \\ &= 2 \left[2 \sin 1 \frac{\beta^2}{2} \right]_0^1 \\ &= 2 \sin 1 \end{align*} $$

Rodrigo Siqueira
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    The value should be $(1/2)\sin 1$. I think you have to compute the jacobian of the inverse map, which is $1/2$ since that of the direct map is $2$. Also the numeric evaluation of the original untransformed integral is $(1/2)\sin 1$ according to when I ran it numerically on maple. – coffeemath Mar 04 '13 at 17:25
  • I seriously didn't understand why I should have computed the inverse of the jacobian. – Rodrigo Siqueira Mar 04 '13 at 23:42