Simple double integral - I can't get it right after substitution

Question

Calculate the area bound between the following lines:

1. $x=y$

2. $x=2y$

3. $x+y=a$

4. $x+y=3a$

$a>0$ Now, there's an obvious substitution:

$$\frac{x}{y}=u$$

and

$$x+y=v$$

which gives us $u \in [1,2]$, and $v \in [a, 3a]$

Calculating the jacobian, I get that $$J = \frac{v}{(u+1)^2}$$

This is because $y=\frac{v}{u+1}$ and $x = \frac{vu}{u+1}$. I got this by plugging in $x=yu$ in the second substitution and expressed both $x$ and $y$ in terms of $v$ and $u$.

Now my integral is:

$$\int_a^{3a}\int_1^2 \frac{v}{(u+1)^2}dudv$$ which evaluates to exactly $$\frac{2}{3}a^2$$

However, my workbook solution is $$\frac{7}{120}a^2$$

Can anyone tell me where I went wrong?

Answer 1

Let's check the result using coordinate geometry. We note that the intersection of $x = y$ and $x+y = 3a$ is simply $(x,y) = (\frac{3}{2} a, \frac{3}{2} a)$, and similarly, the intersection of $x = 2y$ and $x+y = 3a$ is $(x,y) = (2a,a)$. So the triangle with vertices $(0,0), (\frac{3}{2} a, \frac{3}{2} a), (2a, a)$ is a right triangle with legs of length $$\frac{3}{\sqrt{2}}a, \quad \frac{a}{\sqrt{2}}$$ and area $$\frac{3}{4}a^2.$$ Then we must subtract out the area of the smaller triangle that is bounded by $x = y$, $x = 2y$, and $x + y = a$, which of course has $(1/3)^2 = 1/9$ the area of the original triangle; i.e., the desired area is $$\frac{8}{9} \frac{3}{4} a^2 = \frac{2}{3} a^2.$$ Your answer is correct.

Answer 2

$$ J= \begin{bmatrix} \partial_ux & \partial_vx \\ \partial_uy & \partial_vy \end{bmatrix} $$ so if: $$ \begin{matrix} x=\frac{uv}{u+1} & \text{and} & y=\frac{v}{u+1} \end{matrix} $$ then: $$ J= \begin{bmatrix} \frac{v}{(u+1)^2} & \frac{u}{u+1} \\ -\frac{v}{(u+1)^2} & \frac{1}{u+1} \end{bmatrix} $$ so: $$|J|=\frac{v}{(u+1)^3}+\frac{uv}{(u+1)^3}=\frac{v}{(u+1)^2}$$ so this step is definitely correct :)