1. Masalani bayon qilamiz: $x=3\cos t$, $y=8\sin t$ parametrik tenglamalar bilan berilgan chiziq va $y \geq 4\sqrt{3}$ chiziq bilan chegaralangan shakl yuzini topish kerak. 2. Bu parametrik tenglamalar elipsni ifodalaydi: $$\frac{x^2}{3^2} + \frac{y^2}{8^2} = 1.$$ Ya'ni, $$\frac{x^2}{9} + \frac{y^2}{64} = 1.$$ 3. Chegara chizig'i $y = 4\sqrt{3}$ bo'lib, bu elipsni yuqoridan kesib o'tadi. Shakl elipsning yuqori qismi va $y=4\sqrt{3}$ chizig'i orasidagi maydon. 4. Yuzani topish uchun integral formulasi: $$S = \int_{x_1}^{x_2} (y_{elips} - y_{chiziq}) \, dx,$$ bu yerda $y_{elips} = 8\sqrt{1 - \frac{x^2}{9}}$, $y_{chiziq} = 4\sqrt{3}$. 5. $y=4\sqrt{3}$ chiziq elipsni kesish nuqtalarini topamiz: $$4\sqrt{3} = 8\sin t \Rightarrow \sin t = \frac{4\sqrt{3}}{8} = \frac{\sqrt{3}}{2}.$$ Bu $t = \frac{\pi}{3}$ va $t = \frac{2\pi}{3}$ nuqtalar. 6. $x$ koordinatalarini topamiz: $$x = 3\cos t,$$ $$x_1 = 3\cos \frac{2\pi}{3} = 3 \times (-\frac{1}{2}) = -\frac{3}{2},$$ $$x_2 = 3\cos \frac{\pi}{3} = 3 \times \frac{1}{2} = \frac{3}{2}.$$ 7. Integralni hisoblaymiz: $$S = \int_{-\frac{3}{2}}^{\frac{3}{2}} \left(8\sqrt{1 - \frac{x^2}{9}} - 4\sqrt{3}\right) dx = 2 \int_0^{\frac{3}{2}} \left(8\sqrt{1 - \frac{x^2}{9}} - 4\sqrt{3}\right) dx,$$ chunki funksiya simmetrik. 8. O'zgaruvchi almashtirish: $$x = 3\sin \theta, \quad dx = 3\cos \theta d\theta,$$ chegaralar: $$x=0 \Rightarrow \theta=0,$$ $$x=\frac{3}{2} \Rightarrow \sin \theta = \frac{1}{2} \Rightarrow \theta = \frac{\pi}{6}.$$ 9. Integral: $$S = 2 \int_0^{\frac{\pi}{6}} \left(8\sqrt{1 - \sin^2 \theta} - 4\sqrt{3}\right) 3\cos \theta d\theta = 2 \int_0^{\frac{\pi}{6}} (8\cos \theta - 4\sqrt{3}) 3\cos \theta d\theta.$$ 10. Soddalashtiramiz: $$S = 6 \int_0^{\frac{\pi}{6}} (8\cos^2 \theta - 4\sqrt{3} \cos \theta) d\theta = 6 \left(8 \int_0^{\frac{\pi}{6}} \cos^2 \theta d\theta - 4\sqrt{3} \int_0^{\frac{\pi}{6}} \cos \theta d\theta \right).$$ 11. Integral qiymatlari: $$\int_0^{\frac{\pi}{6}} \cos^2 \theta d\theta = \int_0^{\frac{\pi}{6}} \frac{1+\cos 2\theta}{2} d\theta = \left. \frac{\theta}{2} + \frac{\sin 2\theta}{4} \right|_0^{\frac{\pi}{6}} = \frac{\pi}{12} + \frac{\sin \frac{\pi}{3}}{4} = \frac{\pi}{12} + \frac{\sqrt{3}}{8},$$ $$\int_0^{\frac{\pi}{6}} \cos \theta d\theta = \sin \theta \big|_0^{\frac{\pi}{6}} = \frac{1}{2}.$$ 12. Hisoblaymiz: $$S = 6 \left(8 \left(\frac{\pi}{12} + \frac{\sqrt{3}}{8}\right) - 4\sqrt{3} \times \frac{1}{2} \right) = 6 \left( \frac{8\pi}{12} + 8 \times \frac{\sqrt{3}}{8} - 2\sqrt{3} \right) = 6 \left( \frac{2\pi}{3} + \sqrt{3} - 2\sqrt{3} \right) = 6 \left( \frac{2\pi}{3} - \sqrt{3} \right).$$ 13. Yakuniy javob: $$\boxed{S = 6 \left( \frac{2\pi}{3} - \sqrt{3} \right)}.$$