ChangeOfVariables - 7-masala yechimi | MathModel.uz

15.8-bo'lim : O'zgaruvchilarni almashtirish

7. $\displaystyle \iint\limits_{R}{{x{y^3}\,dA}}$ ni baholang, bunda $R$ $xy = 1$, $xy = 3$, $y = 2$ va $y = 6$ bilan chegaralangan mintaqa bo‘lib, $\displaystyle x = \frac{v}{{6u}}$, $y = 2u$ transformatsiyasi yordamida.

Yechim

Biz qilishimiz kerak bo'lgan birinchi narsa $R$ konvertatsiyasini aniqlashdir. Biz avvalgi misolda $R$ konvertatsiyasini aniqladik. Biroq, keling, ushbu muammoni hal qilish uchun jarayonni qaytadan ko'rib chiqamiz (bir nechta tafsilotlar o'tkazib yuborilgan).

Birinchidan, $R$ eskizi.

$A sketch of the region $R$. It is bounded on the left by the curve $xy = 1$, on the right by the curve $xy = 3$, above by the line $y = 6$ and below by the line $y = 2$, and the inside is shaded green. Dotted extensions of the two curves are drawn to show more of each curve.$

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

$y = 6:\hspace{0.25in}2u = 6\hspace{0.5in} \to \hspace{0.25in}\,\,\,u = 3$

$\displaystyle xy = 3:\,\,\,\,\,\,\,\,\,\left( {\frac{v}{{6u}}} \right)\left( {2u} \right) = 3\,\hspace{0.25in}\to \hspace{0.25in}\,\,\,v = 9$

$y = 2:\hspace{0.25in}2u = 2\hspace{0.5in} \to \hspace{0.25in}\,\,\,u = 1$

$\displaystyle xy = 1:\,\,\,\,\,\,\,\,\,\left( {\frac{v}{{6u}}} \right)\left( {2u} \right) = 1\,\,\hspace{0.25in}\to \hspace{0.25in}\,\,\,v = 3$

Mana o'zgartirilgan hududning eskizi.

$A sketch of the transformed region, drawn in the uv-plane. It is a rectangle whose sides are the lines $u = 1$ on the left, $u = 3$ on the right, $v = 9$ across the top and $v = 3$ across the bottom, and the inside is shaded.$

Shunday qilib, o'zgartirilgan mintaqa uchun chegaralar:

\[\begin{array}{c}1 \le u \le 3\\ 3 \le v \le 9\end{array}\]

2-qadam

Bizga bu o'zgarishning yakobiyi kerak bo'ladi.

\[\frac{{\partial \left( {x,y} \right)}}{{\partial \left( {u,v} \right)}} = \left| {\begin{array}{*{20}{c}}{\displaystyle \frac{{\partial x}}{{\partial u}}}&\displaystyle {\frac{{\partial x}}{{\partial v}}}\\\displaystyle {\frac{{\partial y}}{{\partial u}}}&\displaystyle {\frac{{\partial y}}{{\partial v}}}\end{array}} \right| = \left| {\begin{array}{*{20}{c}}{\displaystyle - \frac{v}{{6{u^2}}}}&{\displaystyle \frac{1}{{6u}}}\\ & \\2&0\end{array}} \right| = 0 - \frac{2}{{6u}} = - \frac{1}{{3u}}\]

3-qadam

Endi biz integralni “yangi” $uv$ koordinatalar tizimi nuqtai nazaridan yozishimiz mumkin.

\[\begin{align*}\iint\limits_{R}{{x{y^3}\,dA}} & = \int_{1}^{3}{{\int_{3}^{9}{{\left( {\frac{v}{{6u}}} \right){{\left( {2u} \right)}^3}\left| { - \frac{1}{{3u}}} \right|\,dv}}\,du}}\\ & = \int_{1}^{3}{{\int_{3}^{9}{{\frac{4}{9}v\,u\,dv}}\,du}}\end{align*}\]

Jacobian-ni qo'shishni unutmang va unga mutlaq qiymat chiziqlari kerakligini unutmang. Bu holda biz ishlayotgan $u$ diapazoni (1-qadamda berilgan) ijobiy ekanligini bilamiz, biz mutlaq qiymatlar satridagi miqdor manfiy ekanligini bilamiz va shuning uchun biz minus belgisini tashlab, mutlaq qiymat satrlarini ham tushirishimiz mumkin.

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.

\[\begin{align*}\iint\limits_{R}{{x{y^3}\,dA}} & = \int_{1}^{3}{{\int_{3}^{9}{{\frac{4}{9}v\,u\,dv}}\,du}}\\ & = \int_{1}^{3}{{\left. {\left( {\frac{2}{9}{v^2}u} \right)} \right|_3^9\,du}}\\ & = \int_{1}^{3}{{16u\,du}}\\ & = \left. {\left( {8{u^2}} \right)} \right|_1^3\\ & = \require{bbox} \bbox[2pt,border:1px solid black]{{64}}\end{align*}\]