Find the area enclosed by the following curve (a>0)
Find the area enclosed by the following curve ($a>0$): $$ (x^2+y^2)^2 = 2ax^3. $$
Explanation of Method
Converting to polar coordinates collapses the quartic/cubic relation to the simple curve $r=2a\cos^3\theta$. Since $r\geqslant0$ requires $\cos\theta\geqslant0$, the whole loop is traced once as $\theta$ ranges over $\left[-\frac\pi2,\frac\pi2\right]$, and the area follows from the standard polar-area formula together with the classical reduction formula for $\int\cos^6\theta\,d\theta$.
- Convert to polar coordinates to get $r=2a\cos^3\theta$.
- Identify the valid range of $\theta$ and exploit symmetry.
- Evaluate $\int_0^{\pi/2}\cos^6\theta\,d\theta$ via the Wallis reduction formula.
Step 1 · Convert to polar coordinates
With $x=r\cos\theta$, $y=r\sin\theta$: $x^2+y^2=r^2$ and $x^3=r^3\cos^3\theta$, so:
$$ r^4 = 2ar^3\cos^3\theta \quad\Longrightarrow\quad r = 2a\cos^3\theta \quad(r\neq0). $$
Step 2 · Determine the range of $\theta$
Since $r\geqslant0$, we need $\cos\theta\geqslant0$, i.e. $-\frac\pi2\leqslant\theta\leqslant\frac\pi2$. The curve is symmetric about the $x$-axis (replacing $\theta\to-\theta$ leaves $r$ unchanged), so the full loop is traced as $\theta$ runs over this interval, and:
$$ A = \frac12\int_{-\pi/2}^{\pi/2} r^2\,d\theta = \int_0^{\pi/2} r^2\,d\theta = 4a^2\int_0^{\pi/2}\cos^6\theta\,d\theta. $$
Step 3 · Evaluate $\int_0^{\pi/2}\cos^6\theta\,d\theta$
Using the Wallis reduction $I_n=\frac{n-1}nI_{n-2}$ (valid for $\int_0^{\pi/2}\cos^n\theta\,d\theta$), with $I_0=\frac\pi2$:
$$ I_2=\frac12\cdot\frac\pi2=\frac\pi4, \qquad I_4=\frac34\cdot\frac\pi4=\frac{3\pi}{16}, \qquad I_6=\frac56\cdot\frac{3\pi}{16}=\frac{5\pi}{32}. $$
Step 4 · Combine
$$ A = 4a^2\cdot\frac{5\pi}{32} = \frac{5\pi a^2}{8}. $$
Additional Notes
- Verified numerically: direct quadrature of $\int_{-\pi/2}^{\pi/2}\frac12(2a\cos^3\theta)^2\,d\theta$ gives $\approx1.96350\,a^2$, matching $\frac{5\pi}8\approx1.96350$.
Result
The area is: