16.7-bo'lim : Grin teoremasi
5. \( \displaystyle \oint_{C}{{\left( {x{y^2} + {x^2}} \right)\,dx + \left( {4x - 1} \right)\,dy}}\) uchun Grin teoremasini tasdiqlang, bu erda \(C\) quyida ko'rsatilgan. (a) chiziqli integralni to'g'ridan-to'g'ri hisoblash va (b) chiziqli integralni hisoblash uchun Grin teoremasidan foydalanish.
Yechim
Shunday qilib, keling, egri chiziqlarni quyidagicha belgilash bilan muammoni boshlaylik:
Har bir egri chiziq uchun belgilangan yo'nalishdan keyin bu erda har bir egri chiziq uchun parametrlar mavjud.
\({C_1}:\,\,\vec r\left( t \right) = \left\langle {0,t} \right\rangle \hspace{0.25in}0 \le t \le 3\)\({C_2}:\,\,\vec r\left( t \right) = \left( {1 - t} \right)\left\langle {0,3} \right\rangle + t\left\langle { - 3,0} \right\rangle = \left\langle { - 3t,3 - 3t} \right\rangle \hspace{0.25in}0 \le t \le 1\)
\({C_3}:\,\,\vec r\left( t \right) = \left\langle {t,0} \right\rangle \hspace{0.25in} - 3 \le t \le 0\)
2-qadam
Mana bu egri chiziqlarning har biri bo'yicha baholangan chiziqli integrali.
\[\begin{align*}\oint_{{{C_1}}}{{\left( {x{y^2} + {x^2}} \right)\,dx + \left( {4x - 1} \right)\,dy}} & = \int_{0}^{3}{{\left[ {\left( 0 \right){{\left( t \right)}^2} + {{\left( 0 \right)}^2}} \right]\left( 0 \right)\,dt}} + \int_{0}^{3}{{\left[ {4\left( 0 \right) - 1} \right]\left( 1 \right)\,dt}}\\ & = \int_{0}^{3}{{ - 1\,dt}} = \left. { - t} \right|_0^3 = \underline { - 3} \end{align*}\] \[\begin{align*}\oint_{{{C_2}}}{{\left( {x{y^2} + {x^2}} \right)\,dx + \left( {4x - 1} \right)\,dy}} & = \int_{0}^{1}{{\left[ {\left( { - 3t} \right){{\left( {3 - 3t} \right)}^2} + {{\left( { - 3t} \right)}^2}} \right]\left( { - 3} \right)\,dt}} + \int_{0}^{1}{{\left[ {4\left( { - 3t} \right) - 1} \right]\left( { - 3} \right)\,dt}}\\ & = \int_{0}^{1}{{81{t^3} - 189{t^2} + 81t\,dt}} + \int_{0}^{1}{{36t + 3\,dt}}\\ & = \int_{0}^{1}{{81{t^3} - 189{t^2} + 117t + 3\,dt}}\\ & = \left. {\left( {\frac{{81}}{4}{t^4} - 63{t^3} + \frac{{117}}{2}{t^2} + 3t} \right)} \right|_0^1 = \underline {\frac{{75}}{4}} \end{align*}\] \[\begin{align*}\oint_{{{C_3}}}{{\left( {x{y^2} + {x^2}} \right)\,dx + \left( {4x - 1} \right)\,dy}} & = \int_{{ - 3}}^{0}{{\left[ {\left( t \right){{\left( 0 \right)}^2} + {{\left( t \right)}^2}} \right]\left( 1 \right)\,dt}} + \int_{{ - 3}}^{0}{{\left[ {4\left( t \right) - 1} \right]\left( 0 \right)\,dt}}\\ & = \int_{{ - 3}}^{0}{{{t^2}\,dt}} = \left. {\frac{1}{3}{t^3}} \right|_{ - 3}^0 = \underline 9 \end{align*}\]3-qadam
Endi biz to‘liq egri chiziq ustidagi chiziqli integralining qiymatini olish uchun oldingi bosqich natijalarini qo‘shishimiz kerak. Bu beradi,
\[\oint_{C}{{\left( {x{y^2} + {x^2}} \right)\,dx + \left( {4x - 1} \right)\,dy}} = \left( { - 3} \right) + \left( {\frac{{75}}{4}} \right) + \left( 9 \right) = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{99}}{4}}}\]b chiziqli integralni hisoblash uchun Grin teoremasidan foydalanish
Yechim
E'tibor bering, integraldagi aylana egri chiziq ijobiy yo'nalishda ekanligini ko'rsatadi va shuning uchun biz bu integralda Grin teoremasidan foydalanishimiz mumkin.
Bizda mavjud integraldan,
\[P = x{y^2} + {x^2}\hspace{0.25in}\hspace{0.25in}Q = 4x - 1\]Esda tutingki, \(P\) \(x\) ga, \(Q\) esa \(y\) ga ko'paytiriladi.
2-qadam
Green teoremasidan foydalanib, chiziqli integralga aylanadi,
\[\oint_{C}{{\left( {x{y^2} + {x^2}} \right)\,dx + \left( {4x - 1} \right)\,dy}} = \iint\limits_{D}{{4 - \left( {2xy} \right)\,dA}} = \iint\limits_{D}{{4 - 2xy\,dA}}\]\(D\) - egri chiziq bilan o'ralgan hudud.
3-qadam
Mintaqaning gipotenuzasi bo'ylab chiziq tenglamasi \(y = x + 3\) tomonidan berilganligini tekshirish uchun buni sizga qoldiramiz. Ushbu tenglamaga ega bo'lganimizdan so'ng, mintaqa uchun chegaralarni olish juda oson. Ular,
\[\begin{array}{c} - 3 \le x \le 0\\ 0 \le y \le x + 3\end{array}\]4-qadam
Endi biz ikki karrali integralni baholashimiz kerak. Bu erda baholash ishi.
\[\begin{align*}\oint_{C}{{\left( {x{y^2} + {x^2}} \right)\,dx + \left( {4x - 1} \right)\,dy}} & = \iint\limits_{D}{{4 - 2xy\,dA}}\\ & = \int_{{ - 3}}^{0}{{\int_{0}^{{x + 3}}{{4 - 2xy\,dy}}\,dx}}\\ & = \int_{{ - 3}}^{0}{{\left. {\left( {4y - x{y^2}} \right)} \right|_0^{x + 3}\,dx}}\\ & = \int_{{ - 3}}^{0}{{12 - 5x - 6{x^2} - {x^3}\,dx}}\\ & = \left. {\left( {12x - \frac{5}{2}{x^2} - 2{x^3} - \frac{1}{4}{x^4}} \right)} \right|_{ - 3}^0\\ & = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{99}}{4}}}\end{align*}\]Shunday qilib, biz chiziqli integralni to'g'ridan-to'g'ri integrallash orqali olingan kabi Grin teoremasini chiziqli integralga qo'llaganimizdan keyin xuddi shunday javob oldik.