سوال ۲۷ کنکور ارشد مهندسی کامپیوتر ۱۴۰۴

The curve $(x^{2}+y^{2})^{2}=a^{2}(x^{2}-y^{2})$ is the lemniscate of Bernoulli. Convert to polar coordinates with $x^{2}+y^{2}=r^{2}$ and $x^{2}-y^{2}=r^{2}\cos 2\theta$:

$(r^{2})^{2}=a^{2}\,r^{2}\cos 2\theta \quad\Longrightarrow\quad r^{2}=a^{2}\cos 2\theta.$

The curve exists only where $\cos 2\theta\ge 0$, i.e. one loop spans $-\tfrac{\pi}{4}\le\theta\le\tfrac{\pi}{4}$, and a second symmetric loop spans $\tfrac{3\pi}{4}\le\theta\le\tfrac{5\pi}{4}$.

Area of one loop (using $A=\tfrac12\int r^{2}\,d\theta$):

$A_{\text{loop}}=\frac{1}{2}\int_{-\pi/4}^{\pi/4} a^{2}\cos 2\theta\,d\theta =\frac{a^{2}}{2}\left[\frac{\sin 2\theta}{2}\right]_{-\pi/4}^{\pi/4} =\frac{a^{2}}{2}\cdot\frac{1-(-1)}{2}=\frac{a^{2}}{2}.$

Total enclosed area (two congruent loops):

$A=2\cdot\frac{a^{2}}{2}=a^{2}.$

So the answer is $a^{2}$ (option 2).

Exam trick: Recognize $(x^{2}+y^{2})^{2}=a^{2}(x^{2}-y^{2})$ as the lemniscate $r^{2}=a^{2}\cos 2\theta$; its total area is exactly $a^{2}$. Don't forget the curve has two loops — integrating one half-period gives only $\tfrac{a^{2}}{2}$ (the trap in option 1).