ParametricSurfaces - 9-masala yechimi | MathModel.uz

17.2-bo'lim : Parametrik yuzalar

9. $z = 3 + 2y + \frac{1}{4}{x^4}$ ning $y = {x^5}$, $x = 1$ va $x$ o'qi bilan chegaralangan $xy$ tekisligidagi mintaqadan yuqori bo'lgan qismining sirt maydonini aniqlang.

Yechim

Ushbu sirtni parametrlash juda oddiy. Bizda $z = f\left( {x,y} \right)$ ko'rinishidagi sirt tenglamasi mavjud va shuning uchun sirtning parametrlanishi,

\[\vec r\left( {x,y} \right) = \left\langle {x,y,3 + 2y + \frac{1}{4}{x^4}} \right\rangle \]

Endi, bu to'liq sirtning parametrizatsiyasi va biz faqat keyingi mintaqada joylashgan qismni xohlaymiz.

$The region in the $xy$-plane bounded by $y = {x^5}$, $x = 1$ and the $x$-axis, shaded in green. The curve $y = {x^5}$ starts at the origin, stays close to the $x$-axis until about $x = 0.5$ and then rises steeply to (1,1). The vertical line $x = 1$ forms the right edge of the region from (1,0) to (1,1) and the $x$-axis forms the bottom edge.$

Shunday qilib, sirtning faqat bir qismini olish uchun biz $x$ va $y$ ni quyidagi diapazonlarga cheklashimiz kerak,

\[\begin{align*}0 \le x \le 1\\ & 0 \le y \le {x^5}\end{align*}\]

Yon eslatmada biz 1-o'rinda ekanligimizni ko'rishimiz mumkin^st kvadrant bu erda va shuning uchun biz $x \ge 0$ va $y \ge 0$ ekanligini bilamiz. Shunday qilib, biz 1-dagi sirtni ko'rishimiz mumkin^st kvadrant har doim $xy$ tekisligidan yuqori bo'ladi va shuning uchun muammo bayonotida taklif qilinganidek, har doim yuqoridagi mintaqadan yuqori bo'ladi.

2-qadam

Keyinchalik, biz ${\vec r_x} \times {\vec r_y}$ ni hisoblashimiz kerak. Mana o'sha ish.

\[{\vec r_x} = \left\langle {1,0,{x^3}} \right\rangle \hspace{0.5in}{\vec r_y} = \left\langle {0,1,2} \right\rangle \] \[{\vec r_x} \times {\vec r_y} = \left| {\begin{array}{*{20}{c}}{\vec i}&{\vec j}&{\vec k}\\1&0&{{x^3}}\\0&1&2\end{array}} \right| = - {x^3}\vec i - 2\vec j + \vec k\]

Endi bizga haqiqatan ham kerak bo'lgan narsa,

\[\left\| {{{\vec r}_x} \times {{\vec r}_y}} \right\| = \sqrt {{{\left( { - {x^3}} \right)}^2} + {{\left( { - 2} \right)}^2} + {{\left( 1 \right)}^2}} = \sqrt {{x^6} + 5} \]

3-qadam

U holda sirt maydoni uchun integral,

\[A = \iint\limits_{D}{{\sqrt {{x^6} + 5} \,dA}} = \int_{0}^{1}{{\int_{0}^{{{x^5}}}{{\sqrt {{x^6} + 5} \,dy}}\,dx}}\]

Yuqoridagi integralda ta'kidlanganidek, $D$ 1-bosqichda topilgan $x$ va $y$ diapazonlari xolos.

4-qadam

Endi biz sirt maydonini olish uchun integralni baholashimiz kerak.

\[\begin{align*}A & = \int_{0}^{1}{{\int_{0}^{{{x^5}}}{{\sqrt {{x^6} + 5} \,dy}}\,dx}} = \int_{0}^{1}{{\left. {y\sqrt {{x^6} + 5} } \right|_0^{{x^5}}\,dx}}\\ & = \int_{0}^{1}{{{x^5}\,\sqrt {{x^6} + 5} \,dx}} = \left. {\frac{1}{9}{{\left( {{x^6} + 5} \right)}^{\frac{3}{2}}}} \right|_0^1 = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{1}{9}\left( {{6^{\frac{3}{2}}} - {5^{\frac{3}{2}}}} \right) = 0.3907}}\end{align*}\]