Berilgan chiziqlar bilan chegaralangan shaklni absissa o‘qi atrofida aylantirishdan hosil bo'lgan jism hajmini toping. $$ \cdots y=2 x-x^{2}, y=0 $$

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Berilgan funksiya va chegaralarni hisobga olgan holda, hosil bo‘ladigan jism hajmini topish uchun disk metodi yoki po‘sta metodi qo‘llaniladi. Bu yerda, absissa o‘qi (x o‘qi) atrofida aylantirish amalga oshiriladi.

### Berilgan:
- $$ y = 2x - x^2 $$
- $$ y = 0 $$

### Qadamlar:

1. **Chegaralarni aniqlash:**
   
   $$ y = 2x - x^2 $$ va $$ y = 0 $$ tenglamalarining kesishuv nuqtalarini topamiz:
   $$
   2x - x^2 = 0 \
   x(2 - x) = 0 \
   $$
   Shunday qilib, $$ x = 0 $$ va $$ x = 2 $$.

2. **Hajm formulasini yozish:**
   
   Absissa o‘qi atrofida aylantirish uchun hajm quyidagi integral bilan ifodalanadi:
   $$
   V = \pi \int_{a}^{b} [f(x)]^2 \, dx
   $$
   Bu yerda $$ a = 0 $$, $$ b = 2 $$, va $$ f(x) = 2x - x^2 $$.

3. **Integralni hisoblash:**
   $$
   V = \pi \int_{0}^{2} (2x - x^2)^2 \, dx
   $$
   Avval kvadratni kengaytiramiz:
   $$
   (2x - x^2)^2 = 4x^2 - 4x^3 + x^4
   $$
   Demak, integral:
   $$
   V = \pi \int_{0}^{2} (4x^2 - 4x^3 + x^4) \, dx
   $$
   Bu integralni hisoblaymiz:
   $$
   \int (4x^2 - 4x^3 + x^4) \, dx = \frac{4}{3}x^3 - x^4 + \frac{1}{5}x^5 + C
   $$
   Chegaralarni qo‘llaymiz:
   $$
   V = \pi \left[ \left( \frac{4}{3}(2)^3 - (2)^4 + \frac{1}{5}(2)^5 \right) - \left( \frac{4}{3}(0)^3 - (0)^4 + \frac{1}{5}(0)^5 \right) \right]
   $$
   Hisoblash:
   $$
   V = \pi \left( \frac{32}{3} - 16 + \frac{32}{5} \right)
   $$
   Qo‘shish uchun umumiy kasr topamiz:
   $$
   \frac{32}{3} - 16 + \frac{32}{5} = \frac{32 \cdot 5}{15} - \frac{16 \cdot 15}{15} + \frac{32 \cdot 3}{15} = \frac{160 - 240 + 96}{15} = \frac{16}{15}
   $$
   Shunday qilib:
   $$
   V = \pi \cdot \frac{16}{15} = \frac{16\pi}{15}
   $$

### Javob:
Aylanma jismning hajmi $$\frac{16}{15}\pi$$ birlik kubga teng.
Aylanma jismining hajmi $$\frac{16}{15}\pi$$ birlik kubga teng.

Berilgan chiziqlar bilan chegaralangan shaklni absissa o‘qi atrofida aylantirishdan hosil bo'lgan jism hajmini toping. \[ \cdots y=2 x-x^{2}, y=0 \]

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