Using an appropriate change of variables, calculate
Using an appropriate change of variables, calculate the area bounded by the curve: $$ \left(\frac xa+\frac yb\right)^2 = \frac xa-\frac yb, \qquad y>0. $$
Explanation of Method
The curve mixes $\frac xa+\frac yb$ and $\frac xa-\frac yb$, so the natural substitution is $u=\frac xa+\frac yb,\ v=\frac xa-\frac yb$, which turns the curve into the simple parabola $u^2=v$ and rectifies the region into the area between a parabola and a line in the $(u,v)$ plane.
- Substitute $u=\frac xa+\frac yb,\ v=\frac xa-\frac yb$ and find the Jacobian.
- Transform the boundary curve and the constraint $y>0$ into $(u,v)$ coordinates.
- Integrate over the resulting region.
Step 1 · Substitute and invert
Let $u=\dfrac xa+\dfrac yb,\ v=\dfrac xa-\dfrac yb$. The curve becomes $u^2=v$. Solving for $x,y$: adding and subtracting the two definitions gives $\dfrac xa=\dfrac{u+v}2,\ \dfrac yb=\dfrac{u-v}2$, so:
$$ x=\frac{a(u+v)}2, \qquad y=\frac{b(u-v)}2. $$
Step 2 · Compute the Jacobian
$$ \frac{\partial(x,y)}{\partial(u,v)} = \begin{vmatrix} \frac a2 & \frac a2 \\ \frac b2 & -\frac b2 \end{vmatrix} = -\frac{ab}4-\frac{ab}4 = -\frac{ab}2, \qquad \left|\frac{\partial(x,y)}{\partial(u,v)}\right|=\frac{ab}2. $$
So $dx\,dy=\dfrac{ab}2\,du\,dv$.
Step 3 · Transform the region
The constraint $y>0$ becomes $\dfrac{b(u-v)}2>0\Rightarrow u>v$. Combined with the curve $v=u^2$, the condition $u>v=u^2$ gives $u(1-u)>0$, i.e. $0<u<1$. So the region in the $(u,v)$ plane is bounded above by the line $v=u$ and below by the parabola $v=u^2$, for $0<u<1$.
Step 4 · Set up and evaluate the area integral
$$ A = \int_0^1\!\int_{u^2}^{u}\frac{ab}2\,dv\,du = \frac{ab}2\int_0^1(u-u^2)\,du = \frac{ab}2\left[\frac{u^2}2-\frac{u^3}3\right]_0^1 = \frac{ab}2\cdot\frac16 = \frac{ab}{12}. $$
Additional Notes
- This derivation matches the source exactly.
- Verified numerically: direct quadrature of $\frac{ab}2\int_0^1(u-u^2)\,du$ gives exactly $\frac1{12}ab$, matching to full precision.
Result
The area is: