1. Muammo: Parametrik tenglamalar $x=3\cos t$, $y=8\sin t$ bilan berilgan egri va $y=4\sqrt{3}$ chiziq bilan chegaralangan va $y \geq 4\sqrt{3}$ sharti ostidagi shakl yuzasini topish. 2. Egri tenglamalari ellipsani ifodalaydi: $$\frac{x^2}{9} + \frac{y^2}{64} = 1.$$ Bu yerda $a=3$, $b=8$. 3. $y=4\sqrt{3}$ chiziq ellipsani kesib, yuqoridagi qismni ajratib oladi. $4\sqrt{3} \approx 6.928$. 4. Ellipsaning yuqori qismidagi $y \geq 6.928$ bo'lim yuzini topamiz. Ellipsaning yuqori yarim yuzasi uchun $y=8\sin t$, $t \in [0,\pi]$. 5. $y=4\sqrt{3}$ chiziq uchun $8\sin t = 4\sqrt{3} \Rightarrow \sin t = \frac{4\sqrt{3}}{8} = \frac{\sqrt{3}}{2}$. 6. $\sin t = \frac{\sqrt{3}}{2}$ bo'lganda $t=\frac{\pi}{3}$ (yuqori yarim ellipsada). 7. Yuzani integral yordamida hisoblaymiz: $$S = \int_{t=\frac{\pi}{3}}^{\pi} y(t) x'(t) dt,$$ bu yerda $x'(t) = -3\sin t$. 8. Shaklning yuzasi: $$S = \int_{\frac{\pi}{3}}^{\pi} 8\sin t \cdot (-3\sin t) dt = -24 \int_{\frac{\pi}{3}}^{\pi} \sin^2 t dt = 24 \int_{\frac{\pi}{3}}^{\pi} \sin^2 t dt,$$ chunki integral chegaralari o'zgarganda ishora ham o'zgaradi. 9. $\sin^2 t = \frac{1 - \cos 2t}{2}$ formulasini qo'llaymiz: $$S = 24 \int_{\frac{\pi}{3}}^{\pi} \frac{1 - \cos 2t}{2} dt = 12 \int_{\frac{\pi}{3}}^{\pi} (1 - \cos 2t) dt = 12 \left[ t - \frac{\sin 2t}{2} \right]_{\frac{\pi}{3}}^{\pi}.$$ 10. Hisoblaymiz: $$12 \left( \pi - \frac{\sin 2\pi}{2} - \frac{\pi}{3} + \frac{\sin \frac{2\pi}{3}}{2} \right) = 12 \left( \pi - 0 - \frac{\pi}{3} + \frac{\sqrt{3}/2}{2} \right) = 12 \left( \frac{2\pi}{3} + \frac{\sqrt{3}}{4} \right).$$ 11. Yakuniy javob: $$S = 12 \cdot \frac{2\pi}{3} + 12 \cdot \frac{\sqrt{3}}{4} = 8\pi + 3\sqrt{3}.$$