9.5-bo'lim : Parametrik tenglamalar bilan sirt maydoni
3. 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.
Yechim
Bu erda bizga kerak bo'lgan birinchi narsa quyidagi ikkita hosiladir.
\[\frac{{dx}}{{dt}} = - 3\pi \sin \left( {\pi t} \right)\hspace{0.25in}\hspace{0.25in}\frac{{dy}}{{dt}} = 5\]2-qadam
Ushbu muammo uchun bizga \(ds\) kerak bo'ladi.
\[ds = \sqrt {{{\left[ { - 3\pi \sin \left( {\pi t} \right)} \right]}^2} + {{\left[ 5 \right]}^2}} \,dt = \sqrt {9{\pi ^2}{{\sin }^2}\left( {\pi t} \right) + 25} \,dt\]3-qadam
U holda sirt maydoni uchun integral,
\[\begin{align*}SA & = \int_{{}}^{{}}{{2\pi x\,ds}} = \int_{0}^{{\frac{1}{2}}}{{2\pi \left( {3\cos \left( {\pi t} \right)} \right)\sqrt {9{\pi ^2}{{\sin }^2}\left( {\pi t} \right) + 25} \,dt}}\\ & \hspace{0.25in}\hspace{0.25in}\,\,\,\,\, = 6\pi \int_{0}^{{\frac{1}{2}}}{{\cos \left( {\pi t} \right)\sqrt {9{\pi ^2}{{\sin }^2}\left( {\pi t} \right) + 25} \,dt}}\end{align*}\]Sirt maydoni uchun formula bilan ehtiyot bo'lishni unutmang! Amaldagi formula biz aylanayotgan o'qga bog'liq.
4-qadam
Xo'sh, bu juda yoqimsiz ko'rinishdagi integral, lekin biz ba'zida bunday integrallar bilan shug'ullanishimiz kerak. Agar birinchi navbatda quyidagi almashtirishdan foydalansak, biz biroz soddalashtiramiz.
\[u = \sin \left( {\pi t} \right)\,\,\,\,\, \to \,\,\,\,\,\,{\sin ^2}\left( {\pi t} \right) = {u^2}\hspace{0.25in}\,\,\,\,\,du = \pi \cos \left( {\pi t} \right)\] \[t = 0:\,\,\,\,\,u = \sin \left( 0 \right) = 0\hspace{0.25in}\hspace{0.25in}\hspace{0.25in}\hspace{0.25in}t = \frac{1}{2}:\,\,\,\,\,\,u = \sin \left( {\frac{1}{2}\pi } \right) = 1\]Ushbu almashtirish bilan integral,
\[SA = 6\int_{0}^{1}{{\sqrt {9{\pi ^2}{u^2} + 25} \,du}}\]5-qadam
Ushbu integralni quyidagi (biroz tartibsiz ...) triglarni almashtirish bilan baholash mumkin.
\[t = \frac{5}{{3\pi }}\tan \theta \hspace{0.25in}dt = \frac{5}{{3\pi }}{\sec ^2}\theta \,d\theta \] \[\sqrt {9{\pi ^2}{u^2} + 25} = \sqrt {25{{\tan }^2}\theta + 25} = 5\sqrt {{{\tan }^2}\theta + 1} = 5\sqrt {{{\sec }^2}\theta } = 5\left| {\sec \theta } \right|\]Sekantdagi mutlaq qiymatdan xalos bo'lish uchun chegaralarni \(\theta \) chegaralariga aylantirish kerak bo'ladi.
\[\begin{align*}u & = 0:\hspace{0.25in}\hspace{0.25in}0 = \frac{5}{{3\pi }}\tan \theta \hspace{0.25in} \to \hspace{0.25in}\tan \theta = 0\hspace{0.25in}\,\, \to \hspace{0.25in}\theta = 0\\ u & = 1:\hspace{0.25in}\hspace{0.25in}\,1 = \frac{5}{{3\pi }}\tan \theta \hspace{0.25in} \to \hspace{0.25in}\tan \theta = \frac{{3\pi }}{5}\hspace{0.25in}\, \to \hspace{0.25in}\theta = {\tan ^{ - 1}}\left( {\frac{{3\pi }}{5}} \right) = 1.0830\end{align*}\]OK, bu muammo uchun \(\theta \) ning mos diapazoni \(0 \le \theta \le 1.0830\) (birinchi kvadrant) va bu diapazonda sekant ijobiy ekanligini bilamiz. Shunday qilib, ildiz bo'ladi,
\[\sqrt {9{\pi ^2}{u^2} + 25} = 5\sec \theta \]Keyin sirt maydoni,
\[\begin{align*}SA & = \int_{0}^{{\frac{1}{2}}}{{2\pi \left( {3\cos \left( {\pi t} \right)} \right)\sqrt {9{\pi ^2}{{\sin }^2}\left( {\pi t} \right) + 25} \,dt}}\\ & = 6\int_{0}^{1}{{\sqrt {9{\pi ^2}{u^2} + 25} \,du}}\\ & = 6\int_{0}^{{1.0830}}{{\left( {5\sec \theta } \right)\left( {\frac{5}{{3\pi }}{{\sec }^2}\theta } \right)\,d \theta}}\\ & = 6\int_{0}^{{1.0830}}{{\frac{{25}}{{3\pi }}{{\sec }^3}\theta \,d\theta}}\\ & = \left. {\frac{{25}}{\pi }\left( {\sec \theta tan\theta + ln\left| {\sec \theta + \tan \theta } \right|} \right)} \right|_0^{1.0830} = \require{bbox} \bbox[2pt,border:1px solid black]{{43.0705}}\end{align*}\]Bu muammo biroz chalkash edi, lekin bu sizni bunday muammolarni hal qila olmasligingizga qaror qilishiga yo'l qo'ymang! Ular bajarilishi mumkin va ko'pincha ba'zi nisbatan oddiy almashtirishlar bilan soddalashtirilishi mumkin.