Find the area of the region bounded by: sqrt(frac xa)+
Find the area of the region bounded by: $$ \sqrt{\frac xa}+\sqrt[4]{\frac yb}=1, \qquad x\geqslant0,\ y\geqslant0. $$
Explanation of Method
Although this curve can be forced into the generalized-polar-coordinates framework (with exponent $\alpha=4$), doing so leads to a messy quadratic-in-$u$ intermediate step. It is far simpler to solve the boundary curve directly for $y$ as a function of $x$ using the single substitution $t=\sqrt{x/a}$, which rationalizes both roots at once.
- Substitute $t=\sqrt{x/a}$ to solve for $y(x)$ explicitly.
- Set up $A=\int_0^a y\,dx$ in terms of $t$.
- Evaluate using the Beta-function formula for $\int_0^1 t^{m}(1-t)^{n}\,dt$.
Step 1 · Solve for $y(x)$
Let $t=\sqrt{x/a}$, so $x=at^2$, $dx=2at\,dt$, and $t$ ranges over $[0,1]$ as $x$ ranges over $[0,a]$. The boundary equation $\sqrt{x/a}+\sqrt[4]{y/b}=1$ becomes $\sqrt[4]{y/b}=1-t$, so:
$$ y = b(1-t)^4. $$
Step 2 · Set up the area integral
$$ A = \int_0^a y\,dx = \int_0^1 b(1-t)^4\cdot2at\,dt = 2ab\int_0^1 t(1-t)^4\,dt. $$
Step 3 · Evaluate via the Beta function
Recognizing $\displaystyle\int_0^1 t(1-t)^4\,dt=B(2,5)=\frac{\Gamma(2)\Gamma(5)}{\Gamma(7)}=\frac{1!\cdot4!}{6!}=\frac{24}{720}=\frac1{30}$:
$$ A = 2ab\cdot\frac1{30} = \frac{ab}{15}. $$
Additional Notes
- A note on the source computation: the source attempted this problem via generalized polar coordinates $x=ar\cos^4\varphi,\,y=br\sin^4\varphi$, which turns the boundary curve into a quadratic equation in $\sqrt r$; the derivation became increasingly complicated at this stage and was never carried through to a resolved closed form, ending in an unreliable numeric guess ($\approx0.191\,ab$). The direct substitution $t=\sqrt{x/a}$ above sidesteps this difficulty entirely and gives a clean closed form; note also that $\frac1{15}\approx0.0667$, not $0.191$, so the source's final numeric guess was itself substantially in error.
- Verified numerically: direct quadrature of $2\int_0^1 t(1-t)^4\,dt$ gives exactly $0.0666\overline6$, matching $\frac1{15}$ to full machine precision (so $A=\frac{ab}{15}$ for general $a,b$).
Result
The area is: