Biz qilishimiz kerak bo'lgan birinchi narsa - \(R\) konvertatsiyasini aniqlash.

Birinchidan, \(R\) eskizi.

Hududning har bir chegarasining tenglamalari eskizda berilgan. Biz har bir cho'qqining koordinatalarini bilganimizni hisobga olsak, uning oddiy algebrasi chiziqda ikkita nuqta berilgan chiziq tenglamasini aniqlash uchun tenglamalarni tekshirishni sizga qoldiramiz.

E'tibor bering, ushbu mintaqa bo'ylab integratsiya qilish uchun tartibdan qat'i nazar, uchta integral kerak bo'ladi. Har bir integratsiya tartibi uchta integralni to'g'ri talab qilishini ko'rishingiz mumkinmi?

Keling, har bir chegara egri chizig'ini o'zgartiraylik.

\(\displaystyle y = x + 1:\hspace{0.25in}\,\,\,\,\frac{1}{3}\left( {4v - u} \right) = \frac{1}{3}\left( {v - u} \right) + 1\,\,\,\,\,\,\,\,\,\, \to \,\,\,\,\,\,\,3v = 3\,\,\,\hspace{0.25in}\to \,\,\,\,\,\,\,\,\,v = 1\)

\(\displaystyle y = 4x - 17:\hspace{0.25in}\frac{1}{3}\left( {4v - u} \right) = \frac{4}{3}\left( {v - u} \right) - 17\,\,\,\,\,\,\, \to \,\,\,\,\,\,\,3u = - 51\,\,\,\,\,\,\,\, \to \,\,\,\,\,\,\,\,\,u = - 17\)

\(\displaystyle y = x - 2:\hspace{0.25in}\,\,\,\frac{1}{3}\left( {4v - u} \right) = \frac{1}{3}\left( {v - u} \right) - 2\,\,\,\,\,\,\,\,\,\, \to \,\,\,\,\,\,\,3v = - 6\,\,\,\,\,\,\,\,\,\, \to \,\,\,\,\,\,\,\,\,v = - 2\)

\(\displaystyle y = 4x - 8:\hspace{0.25in}\frac{1}{3}\left( {4v - u} \right) = \frac{4}{3}\left( {v - u} \right) - 8\hspace{0.25in}\to \,\,\,\,\,\,\,3u = - 24\,\,\,\,\,\,\, \to \,\,\,\,\,\,\,\,\,u = - 8\)

Mana o'zgartirilgan hududning eskizi va o'zgartirilgan hududni asl hududga qaraganda integratsiya qilish ancha oson bo'lishini ta'kidlang.

Shunday qilib, o'zgartirilgan mintaqa uchun chegaralar:

\[\begin{array}{c} - 17 \le u \le - 8\\ \, - 2 \le v \le 1\end{array}\]

Shuni ham yodda tutingki, bu asl mintaqaga qaraganda integratsiya qilish uchun juda yaxshi mintaqa bo'ladi.

2-qadam

Bizga bu o'zgarishning yakobiyi kerak bo'ladi.

\[\frac{{\partial \left( {x,y} \right)}}{{\partial \left( {u,v} \right)}} = \left| {\begin{array}{*{20}{c}}{\displaystyle \frac{{\partial x}}{{\partial u}}}&\displaystyle {\frac{{\partial x}}{{\partial v}}}\\\displaystyle {\frac{{\partial y}}{{\partial u}}}&\displaystyle {\frac{{\partial y}}{{\partial v}}}\end{array}} \right| = \left| {\begin{array}{*{20}{c}}{ \displaystyle - \frac{1}{3}}&{\displaystyle \frac{1}{3}}\\{\displaystyle - \frac{1}{3}}&{\displaystyle \frac{4}{3}}\end{array}} \right| = - \frac{4}{9} - \left( { - \frac{1}{9}} \right) = - \frac{1}{3}\]

3-qadam

Endi biz integralni “yangi” \(uv\) koordinatalar tizimi nuqtai nazaridan yozishimiz mumkin.

\[\begin{align*}\iint\limits_{R}{{6x - 3y\,dA}} & = \int_{{ - 17}}^{{ - 8}}{{\int_{{ - 2}}^{1}{{\left[ {6\left( {\frac{1}{3}} \right)\left( {v - u} \right) - 3\left( {\frac{1}{3}} \right)\left( {4v - u} \right)} \right]\left| { - \frac{1}{3}} \right|\,dv}}\,du}}\\ & = \int_{{ - 17}}^{{ - 8}}{{\int_{{ - 2}}^{1}{{ - \frac{1}{3}\left( {2v + u} \right)\,dv}}\,du}}\end{align*}\]

Jacobian-ni qo'shishni unutmang va unga mutlaq qiymat chiziqlari kerakligini unutmang. Bunday holda, biz faqat mutlaq qiymat satrlarini minus belgisini tashlab, tushirishimiz mumkin, chunki bizda mutlaq qiymatda raqam bor.

Bundan tashqari, soddalashtirilgan integral integratsiyaning boshqa tartibini taklif qilmadi va shuning uchun biz faqat bittasini tanladik. Boshqa buyurtma bilan ishlash juda oson bo'lar edi.

4-qadam

Nihoyat, integralni baholaylik.

\[\begin{align*}\iint\limits_{R}{{6x - 3y\,dA}} & = \int_{{ - 17}}^{{ - 8}}{{\int_{{ - 2}}^{1}{{ - \frac{1}{3}\left( {2v + u} \right)\,dv}}\,du}}\\ & = \int_{{ - 17}}^{{ - 8}}{{\left. {\left( { - \frac{1}{3}\left( {{v^2} + uv} \right)} \right)} \right|_{ - 2}^1\,du}}\\ & = \int_{{ - 17}}^{{ - 8}}{{1 - u\,du}}\\ & = \left. {\left( {u - \frac{1}{2}{u^2}} \right)} \right|_{ - 17}^{ - 8}\\ & = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{243}}{2}}}\end{align*}\]