17.2-bo'lim : Parametrik yuzalar
7. \({x^2} + {y^2} = 7\) silindr ichidagi \(2x + 3y + 6z = 9\) qismining sirt maydonini aniqlang.
Yechim
Avval sirtni parametrlashimiz kerak. Biz \(z\) o'qi markazida joylashgan silindr ichidagi qismni xohlayotganimiz sababli, avval \(z\) olish uchun tekislik tenglamasini echish mantiqan to'g'ri keladi.
\[z = \frac{3}{2} - \frac{1}{3}x - \frac{1}{2}y\]To'liq tekislik uchun parametrlash shundan iboratki,
\[\vec r\left( {x,y} \right) = \left\langle {x,y,\frac{3}{2} - \frac{1}{3}x - \frac{1}{2}y} \right\rangle \]Biz faqat muammo bayonotida keltirilgan silindr ichidagi qismni xohlaymiz, shuning uchun biz \(x\) va \(y\) ni \({x^2} + {y^2} \le 7\) diskidagilar bilan cheklashimiz kerak. Bu endi samolyotning faqat silindr ichidagi qismini beradi.
2-qadam
Keyinchalik, biz \({\vec r_x} \times {\vec r_y}\) ni hisoblashimiz kerak. Mana o'sha ish.
\[{\vec r_x} = \left\langle {1,0, - \frac{1}{3}} \right\rangle \hspace{0.5in}{\vec r_y} = \left\langle {0,1, - \frac{1}{2}} \right\rangle \] \[{\vec r_x} \times {\vec r_y} = \left| {\begin{array}{*{20}{c}}{\vec i}&{\vec j}&{\vec k}\\1&0&{ - \frac{1}{3}}\\0&1&{ - \frac{1}{2}}\end{array}} \right| = \frac{1}{3}\vec i + \frac{1}{2}\vec j + \vec k\]Endi bizga haqiqatan ham kerak bo'lgan narsa,
\[\left\| {{{\vec r}_x} \times {{\vec r}_y}} \right\| = \sqrt {{{\left( {\frac{1}{3}} \right)}^2} + {{\left( {\frac{1}{2}} \right)}^2} + 1} = \sqrt {\frac{{49}}{{36}}} = \frac{7}{6}\]3-qadam
U holda sirt maydoni uchun integral,
\[A = \iint\limits_{D}{{\frac{7}{6}\,dA}}\]Bu holda \(D\) biz 1-qadamda qayd etgan \(x\) va \(y\) limitlaridir. Demak, \(D\) bu shunchaki \({x^2} + {y^2} \le 7\) diskidir.
4-qadam
Bu holda integralni hisoblash juda oddiy. Biz qilishimiz kerak bo'lgan narsa shundan iboratki,
\[\iint\limits_{D}{{dA}} = {\mbox{Area of }}D\]Shunday qilib, sirt maydoni oddiy,
\[A = \iint\limits_{D}{{\frac{7}{6}\,dA}} = \frac{7}{6}\iint\limits_{D}{{dA}} = \frac{7}{6}\left[ {{\mbox{Area of }}D} \right] = \frac{7}{6}\left[ {\pi {{\left( {\sqrt 7 } \right)}^2}} \right] = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{49}}{6}\pi }}\]