9.5-bo'lim : Parametrik tenglamalar bilan sirt maydoni
2. Parametrik egri chiziqni berilgan o‘q atrofida aylantirish natijasida olingan jismning sirt maydonini aniqlang. Egri chiziq \(t\) ning berilgan diapazoni uchun aynan bir marta chiqadi deb taxmin qilishingiz mumkin.
\(x = 9 + 2{t^2}\hspace{0.25in}y = 4t\hspace{0.25in}0 \le t \le 2\)-ni \(x\) o'qi atrofida aylantiring.
Yechim
Bu erda bizga kerak bo'lgan birinchi narsa quyidagi ikkita hosiladir.
\[\frac{{dx}}{{dt}} = 4t\hspace{0.25in}\hspace{0.25in}\frac{{dy}}{{dt}} = 4\]2-qadam
Ushbu muammo uchun bizga \(ds\) kerak bo'ladi.
\[ds = \sqrt {{{\left[ {4t} \right]}^2} + {{\left[ 4 \right]}^2}} \,dt = \sqrt {16{t^2} + 16} \,dt = 4\sqrt {{t^2} + 1} \,dt\]E'tibor bering, biz ishning qolgan qismini biroz soddalashtirish uchun ildizdan 16 ni ajratdik.
3-qadam
U holda sirt maydoni uchun integral,
\[SA = \int_{{}}^{{}}{{2\pi y\,ds}} = \int_{0}^{2}{{2\pi \left( {4t} \right)\left( {4\sqrt {{t^2} + 1} } \right)\,dt}} = 32\pi \int_{0}^{2}{{t\sqrt {{t^2} + 1} \,dt}}\]Sirt maydoni uchun formula bilan ehtiyot bo'lishni unutmang! Amaldagi formula biz aylanayotgan o'qga bog'liq.
4-qadam
Bu tez almashtirish bilan hisoblash uchun oddiy integraldir. Mana integral ish,
\[SA = 32\pi \int_{0}^{2}{{t\sqrt {{t^2} + 1} \,dt}} = \left. {\frac{{32}}{3}\pi {{\left( {{t^2} + 1} \right)}^{\frac{3}{2}}}} \right|_0^2 = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{32}}{3}\pi \left( {{5^{\,\frac{3}{2}}} - 1} \right) = 341.1464}}\]