8.2-bo'lim : Er yuzasi
5. \(y = \sin \left( {2x} \right)\) , \(\displaystyle 0 \le x \le \frac{\pi }{8}\) larni \(x\) o'qi atrofida aylantirish natijasida olingan ob'ektning sirt maydonini toping.
Yechim
E'tibor bering, biz ushbu muammoni qisman o'rnatdik (a) 2-muammo. Shunday qilib, biz bu erda muammoning o'rnatish qismining bosqichlarini qisqacha bayon qilamiz. Agar sizga barcha tafsilotlarni ko'rish kerak bo'lsa, iltimos, 2-muammodagi ishni ko'rib chiqing.
Mana bu muammo uchun \(ds\).
\[\frac{{dy}}{{dx}} = 2\cos \left( {2x} \right)\hspace{0.5in} \Rightarrow \hspace{0.5in} ds = \sqrt {1 + 4{{\cos }^2}\left( {2x} \right)} \,dx\]Sirt maydoni uchun integral,
\[SA = \int\limits_{0}^{{\frac{\pi }{8}}}{{2\pi \sin \left( {2x} \right)\,\sqrt {1 + 4{{\cos }^2}\left( {2x} \right)} \,dx}}\]2-qadam
Ushbu integralni baholash uchun bizga quyidagi trigni almashtirish kerak bo'ladi.
\[\begin{align*}\cos \left( {2x} \right) = \frac{1}{2}\tan \left( \theta \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \to \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 2\sin \left( {2x} \right)\,dx = \frac{1}{2}{\sec ^2}\left( \theta \right)\,d\theta \\ & \sqrt {1 + 4{{\cos }^2}\left( {2x} \right)} = \sqrt {1 + {{\tan }^2}\left( \theta \right)} = \sqrt {{{\sec }^2}\left( \theta \right)} = \left| {\sec \left( \theta \right)} \right|\end{align*}\]Mutlaq qiymatlar satrlari bilan ishlash uchun biz \(x\) chegaralarini \(\theta \) chegaralariga aylantirishimiz kerak. Mana bu ish.
\[\begin{align*}x & = 0:\,\,\,\cos \left( 0 \right) = 1 = \frac{1}{2}\tan \left( \theta \right)\,\,\,\,\,\,\,\, \to \,\,\,\,\,\theta = {\tan ^{ - 1}}\left( 2 \right) = 1.1071\\ x & = \frac{\pi }{8}:\,\,\,\cos \left( {\frac{\pi }{4}} \right) = \frac{{\sqrt 2 }}{2} = \frac{1}{2}\tan \left( \theta \right)\,\,\,\,\,\,\,\, \to \,\,\,\,\,\theta = {\tan ^{ - 1}}\left( {\sqrt 2 } \right) = 0.9553\end{align*}\]\(\theta \) mos keladigan diapazoni \(0.9553 \le \theta \le 1.1071\) hisoblanadi. Bu birinchi kvadrantda va u erda sekant ijobiy. Shuning uchun biz sekantga mutlaq qiymat satrlarini tushirishimiz mumkin.
3-qadam
Oldingi bosqichdagi barcha ishlarni birgalikda qo'yish,
\[SA = \int\limits_{0}^{{\frac{\pi }{8}}}{{2\pi \sin \left( {2x} \right)\,\sqrt {1 + 4{{\cos }^2}\left( {2x} \right)} \,dx}} = - \frac{\pi }{2}\int_{{1.1071}}^{{0.9553}}{{{{\sec }^3}\left( \theta \right)\,d\theta }}\]4-qadam
\({\sec ^3}\left( \theta \right)\) integrali uchun formuladan foydalanib, biz Trig funksiyalarni o'z ichiga olgan integrallar olamiz,
\[SA = - \frac{\pi }{2}\int_{{1.1071}}^{{0.9553}}{{{{\sec }^3}\left( \theta \right)\,d\theta }} = \left. { - \frac{\pi }{4}\left[ {\sec \left( \theta \right)\tan \left( \theta \right) + \ln \left| {\sec \left( \theta \right) + \tan \left( \theta \right)} \right|} \right]} \right|_{1.1071}^{0.9553} = \require{bbox} \bbox[2pt,border:1px solid black]{{1.8215}}\]E'tibor bering, siz ishlatgan o'nli kasrlar soniga qarab javobingiz bu erda berilganidan biroz farq qilishi mumkin. Kompyuter tomonidan olingan "aniq" javob 1,8222.