Find the area enclosed by the following curve: (x²+y²)^5
Find the area enclosed by the following curve: $$ (x^2+y^2)^5 = x^2y^2. $$
Explanation of Method
Converting to polar coordinates reduces the curve to $r^6=\frac14\sin^22\theta$, a four-petaled rose-like curve. As in Problem 123, symmetry reduces the computation to one petal times four; the resulting integral $\int_0^{\pi/2}(\sin2\theta)^{2/3}\,d\theta$ is not elementary but is a standard Beta-function value.
- Convert to polar coordinates, obtaining $r=(\sin^22\theta)^{1/6}/2^{1/3}$.
- Set up the area as $4\times$ one petal, and reduce the resulting integral to a Beta-function integral in $u=2\theta$.
- Evaluate using the Beta/Gamma-function formula for $\int_0^{\pi/2}\sin^p u\,du$.
Step 1 · Convert to polar coordinates
With $x=r\cos\theta$, $y=r\sin\theta$: $x^2+y^2=r^2$ and $x^2y^2=r^4\cos^2\theta\sin^2\theta$, so:
$$ r^{10}=r^4\cos^2\theta\sin^2\theta \quad\Longrightarrow\quad r^6=\cos^2\theta\sin^2\theta=\frac{\sin^22\theta}4 \quad(r\neq0). $$
So $r^2=\left(\dfrac{\sin^22\theta}4\right)^{1/3}=\dfrac{|\sin2\theta|^{2/3}}{4^{1/3}}$, and as in Problem 123 this traces a four-petaled shape.
Step 2 · Set up the area as four petals
$$ A = 4\cdot\frac12\int_0^{\pi/2} r^2\,d\theta = \frac2{4^{1/3}}\int_0^{\pi/2}(\sin2\theta)^{2/3}\,d\theta. $$
Substituting $u=2\theta$, $du=2\,d\theta$, and using the symmetry of $\sin u$ about $u=\frac\pi2$ over $[0,\pi]$:
$$ \int_0^{\pi/2}(\sin2\theta)^{2/3}\,d\theta = \frac12\int_0^\pi(\sin u)^{2/3}\,du = \int_0^{\pi/2}(\sin u)^{2/3}\,du. $$
So, since $\dfrac2{4^{1/3}}=2^{1/3}$:
$$ A = \frac2{4^{1/3}}\int_0^{\pi/2}\sin^{2/3}u\,du = 2^{1/3}\int_0^{\pi/2}\sin^{2/3}u\,du. $$
Step 3 · Evaluate via the Beta-function formula
Using $\displaystyle\int_0^{\pi/2}\sin^pu\,du=\frac{\sqrt\pi}2\cdot\frac{\Gamma\left(\frac{p+1}2\right)}{\Gamma\left(\frac{p+2}2\right)}$ with $p=\frac23$:
$$ \int_0^{\pi/2}\sin^{2/3}u\,du = \frac{\sqrt\pi}2\cdot\frac{\Gamma\left(\frac56\right)}{\Gamma\left(\frac43\right)}. $$
So:
$$ A = 2^{1/3}\cdot\frac{\sqrt\pi}2\cdot\frac{\Gamma\left(\frac56\right)}{\Gamma\left(\frac43\right)} = \frac{\sqrt\pi\,\Gamma\left(\frac56\right)}{2^{2/3}\,\Gamma\left(\frac43\right)}. $$
Additional Notes
- Numeric evaluation: $\Gamma\left(\frac56\right)\approx1.12879$, $\Gamma\left(\frac43\right)\approx0.89298$, $\sqrt\pi\approx1.77245$, $2^{2/3}\approx1.58740$, giving $A\approx1.41143$.
- Verified numerically by direct quadrature of the original polar integral $4\cdot\frac12\int_0^{\pi/2}r^2\,d\theta$: gives $\approx1.41143$, matching the closed form to five decimal places.
- Using the reflection and duplication formulas for the Gamma function, this can be further rewritten as $A=\dfrac{2\sqrt3\,\pi^2}{2^{1/3}\,\Gamma\left(\frac13\right)^3}$, a fully equivalent closed form (also verified numerically to match, $\approx1.41143$) that avoids reference to $\Gamma\left(\frac56\right)$ and $\Gamma\left(\frac43\right)$.
Result
The area is: