17.2-bo'lim : Parametrik yuzalar
10. \({u^2} + {v^2} = 4\) tsilindrining ichida yotgan quyidagi parametrik tenglama bilan berilgan sirt qismining sirt maydonini aniqlang.
\[\vec r\left( {u,v} \right) = \left\langle {2u,vu,1 - 2v} \right\rangle \]Yechim
Muammo bayonotida bizga sirt parametrlari allaqachon berilgan, shuning uchun bu muammo uchun bu haqda tashvishlanishga hojat yo'q. Hali qilishimiz kerak bo'lgan narsa - \(u\) va \(v\)ni \({u^2} + {v^2} \le 4\) diskiga cheklashimiz kerakligini tan olishdir.
2-qadam
Keyinchalik, biz \({\vec r_u} \times {\vec r_v}\) ni hisoblashimiz kerak. Mana o'sha ish.
\[{\vec r_u} = \left\langle {2,v,0} \right\rangle \hspace{0.5in}{\vec r_v} = \left\langle {0,u, - 2} \right\rangle \] \[{\vec r_u} \times {\vec r_v} = \left| {\begin{array}{*{20}{c}}{\vec i}&{\vec j}&{\vec k}\\2&v&0\\0&u&{ - 2}\end{array}} \right| = - 2v\vec i + 4\vec j + 2u\vec k\]Endi bizga haqiqatan ham kerak bo'lgan narsa,
\[\left\| {{{\vec r}_u} \times {{\vec r}_v}} \right\| = \sqrt {{{\left( { - 2v} \right)}^2} + {{\left( 4 \right)}^2} + {{\left( {2u} \right)}^2}} = \sqrt {4{u^2} + 4{v^2} + 16} = 2\sqrt {{u^2} + {v^2} + 4} \]3-qadam
U holda sirt maydoni uchun integral,
\[A = \iint\limits_{D}{{2\sqrt {{u^2} + {v^2} + 4} \,dA}}\]Bu erda \(D\) \({u^2} + {v^2} \le 4\) diskidir.
4-qadam
\(D\) disk bo'lgani uchun bu integral uchun eng yaxshi variant qutb koordinatalarining quyidagi "versiyasi" dan foydalanishdir.
\[u = r\cos \theta \hspace{0.25in} v = r\sin \theta \hspace{0.25in} {u^2} + {v^2} = {r^2} \hspace{0.25in} dA = r\,dr\,d\theta \]Ushbu \(D\) uchun qutb koordinatalari chegaralari:
\[\begin{array}{c}0 \le \theta \le 2\pi \\ 0 \le r \le 2\end{array}\]Shunday qilib, qutb koordinatalariga o'tkazishning integralini beradi,
\[A = \iint\limits_{D}{{2\sqrt {{u^2} + {v^2} + 4} \,dA}} = \int_{0}^{{2\pi }}{{\int_{0}^{2}{{2r\sqrt {{r^2} + 4} \,dr}}\,d\theta }}\]5-qadam
Endi biz sirt maydonini olish uchun integralni baholashimiz kerak.
\[\begin{align*}A & = \int_{0}^{{2\pi }}{{\int_{0}^{2}{{2r\sqrt {{r^2} + 4} \,dr}}\,d\theta }} = \int_{0}^{{2\pi }}{{\left. {\frac{2}{3}{{\left( {{r^2} + 4} \right)}^{\frac{3}{2}}}} \right|_0^2\,d\theta }}\\ & = \int_{0}^{{2\pi }}{{\frac{2}{3}\left( {{8^{\frac{3}{2}}} - {4^{\frac{3}{2}}}} \right)\,d\theta }} = \left. {\frac{2}{3}\left( {{8^{\frac{3}{2}}} - 8} \right)\theta } \right|_0^{2\pi } = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{32}}{3}\pi\left( {\sqrt 8 - 1} \right) = 61.2712}}\end{align*}\]