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

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, ikkita integral kerak bo'ladi. Har bir integratsiya tartibi ikkita integralni to'g'ri talab qilishini ko'rishingiz mumkinmi?

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

\(\displaystyle y = \frac{3}{2}x + 3:\hspace{0.25in}\,\,\,\,\frac{1}{4}\left( {3u + v + 12} \right) = \frac{3}{2}\left( {\frac{1}{2}} \right)\left( {u - v} \right) + 3\,\,\,\,\hspace{0.25in}\to \,\,\,\,\,\,\,v = 0\)

\(\displaystyle y = - \frac{5}{2}x + 11:\hspace{0.25in}\frac{1}{4}\left( {3u + v + 12} \right) = - \frac{5}{2}\left( {\frac{1}{2}} \right)\left( {u - v} \right) + 11\,\,\,\,\,\,\,\,\,\, \to \,\,\,\,\,\,\,v = 2u - 8\)

\(\displaystyle y = - \frac{1}{2}x + 3:\hspace{0.25in}\frac{1}{4}\left( {3u + v + 12} \right) = - \frac{1}{2}\left( {\frac{1}{2}} \right)\left( {u - v} \right) + 3\,\hspace{0.25in}\to \,\,\,\,\,\,\,u = 0\)

E'tibor bering, bu holda biz yuqorida ko'rib chiqqan birinchi va oxirgi chegara tenglamasi mos ravishda \(u\) o'qiga va \(v\) o'qiga aylantirildi. Bu ba'zida sodir bo'ladi va integralni baholashda hayotimizni biroz osonlashtirishi mumkin.

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:

\[\begin{array}{c}0 \le u \le 4\\ 2u - 8 \le v \le 0\end{array}\]

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.

\[\frac{{\partial \left( {x,y} \right)}}{{\partial \left( {u,v} \right)}} = \left| {\begin{array}{*{20}{c}}{\displaystyle \frac{{\partial x}}{{\partial u}}}&\displaystyle {\frac{{\partial x}}{{\partial v}}}\\\displaystyle {\frac{{\partial y}}{{\partial u}}}&\displaystyle {\frac{{\partial y}}{{\partial v}}}\end{array}} \right| = \left| {\begin{array}{*{20}{c}}{\displaystyle \frac{1}{2}}&{\displaystyle - \frac{1}{2}}\\{\displaystyle \frac{3}{4}}&{\displaystyle \frac{1}{4}}\end{array}} \right| = \frac{1}{8} - \left( { - \frac{3}{8}} \right) = \frac{1}{2}\]

3-qadam

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

\[\begin{align*}\iint\limits_{R}{{x + 2y\,dA}} & = \int_{0}^{4}{{\int_{{2u - 8}}^{0}{{\left[ {\frac{1}{2}\left( {u - v} \right) + 2\left( {\frac{1}{4}} \right)\left( {3u + v + 12} \right)} \right]\left| {\frac{1}{2}} \right|\,dv}}\,du}}\\ & = \int_{0}^{4}{{\int_{{2u - 8}}^{0}{{u + 3\,dv}}\,du}}\end{align*}\]

Yakobiyni qo'shishni unutmang va undagi mutlaq qiymat satrini unutmang. Bunday holda biz mutlaq qiymat satrlarini tashlab qo'yishimiz mumkin, chunki bizda mutlaq qiymatda faqat ijobiy 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.

\[\begin{align*}\iint\limits_{R}{{x + 2y\,dA}} & = \int_{0}^{4}{{\int_{{2u - 8}}^{0}{{u + 3\,dv}}\,du}}\\ & = \int_{0}^{4}{{\left. {\left( {u + 3} \right)v} \right|_{2u - 8}^0\,du}}\\ & = \int_{0}^{4}{{\left( {u + 3} \right)\left( { - 2u + 8} \right)\,du}}\\ & = \int_{0}^{4}{{24 + 2u - 2{u^2}\,du}}\\ & = \left. {\left( {24u + {u^2} - \frac{2}{3}{u^3}} \right)} \right|_0^4\\ & = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{208}}{3}}}\end{align*}\]