Using an appropriate change of variables, calculate

Using an appropriate change of variables, calculate the area bounded by the curves: $$ \sqrt{\frac xa}+\sqrt{\frac yb}=1; \qquad \sqrt{\frac xa}+\sqrt{\frac yb}=2; \qquad \frac xa=\frac yb; \qquad 4\frac xa=\frac yb, $$ where $a>0,\,b>0$.

Step 1 · Substitute and compute the Jacobian

Let $u=\sqrt{x/a},\,v=\sqrt{y/b}$ (so $x=au^2,\,y=bv^2$, $u,v\geqslant0$). Then:

$$ \frac{\partial(x,y)}{\partial(u,v)} = \begin{vmatrix}2au&0\\0&2bv\end{vmatrix}=4abuv, $$

so $dx\,dy=4ab\,uv\,du\,dv$.

Step 2 · Transform the boundary curves

The four curves become: $u+v=1$, $u+v=2$, $v=u$ (from $x/a=y/b$), and $v=2u$ (from $4x/a=y/b$). The region in the $(u,v)$ plane is the curvilinear (here, straight-sided) quadrilateral between the two parallel lines $u+v=1,\,2$ and the two rays $v=u,\,v=2u$, with vertices at $\left(\frac12,\frac12\right)$, $\left(\frac13,\frac23\right)$, $(1,1)$, and $\left(\frac23,\frac43\right)$.

Step 3 · Set up the area integral

For a fixed value of $s=u+v\in[1,2]$, $v$ runs from $u$ to $2u$, i.e. $u$ runs from $\frac s3$ (where $u+2u=s$) to $\frac s2$ (where $u+u=s$), with $v=s-u$:

$$ A = \int_1^2\!\int_{s/3}^{s/2}4ab\,u(s-u)\,du\,ds. $$

Step 4 · Evaluate the integral

$$ \int_{s/3}^{s/2}(su-u^2)\,du = \left[\frac{su^2}2-\frac{u^3}3\right]_{s/3}^{s/2} = \left(\frac{s^3}8-\frac{s^3}{24}\right)-\left(\frac{s^3}{18}-\frac{s^3}{81}\right) = s^3\left(\frac18-\frac1{24}-\frac1{18}+\frac1{81}\right). $$

With common denominator $648$: $\frac18=\frac{81}{648},\,\frac1{24}=\frac{27}{648},\,\frac1{18}=\frac{36}{648},\,\frac1{81}=\frac8{648}$, so the bracket is $\dfrac{81-27-36+8}{648}=\dfrac{26}{648}=\dfrac{13}{324}$. Then:

$$ A = 4ab\cdot\frac{13}{324}\int_1^2 s^3\,ds = \frac{52ab}{324}\left[\frac{s^4}4\right]_1^2 = \frac{13ab}{81}\cdot\frac{16-1}4 = \frac{13ab}{81}\cdot\frac{15}4 = \frac{195ab}{324} = \frac{65ab}{108}. $$

Additional Notes

• This derivation matches the source exactly.
• Verified numerically for $a=b=1$: direct quadrature of $\int_1^2\int_{s/3}^{s/2}4u(s-u)\,du\,ds$ gives $\approx0.601852$, matching $\frac{65}{108}\approx0.601852$.