SurfaceIntegrals - 5-masala yechimi | MathModel.uz

17.3-bo'lim : Yuzaki integrallar

5. $ \displaystyle \iint\limits_{S}{{yz + 4xy\,dS}}$ ni baholang, bu erda $S$ - $4x + 2y + z = 8$, $z = 0$, $y = 0$ va $x = 0$ bilan chegaralangan qattiq jismning yuzasi. E'tibor bering, ushbu qattiq jismning barcha to'rtta yuzasi $S$ tarkibiga kiritilgan.

Yechim

Keling, ushbu muammoni hal qilishda biz ishlayotgan sirtning qisqacha eskizi bilan boshlaylik.

$A three dimensional sketch of the surface drawn with the traditional axes. The orange face is the plane $4x + 2y + z = 8$ forming the top of the solid and the dark green face on the left is the plane $x = 0$. The other two bounding planes, $z = 0$ and $y = 0$, are also part of the surface but are not visible in this view.$

$A three dimensional sketch of the same solid, this time drawn with a set of “box” axes to make the surfaces easier to visualize. The orange face is the plane $4x + 2y + z = 8$ and the dark green face is the plane $x = 0$.$

Xo'sh, muammo bayonida ta'kidlanganidek, eskizdagi barcha to'rtta sirt (ikkitasi ko'rsatilmagan) $S$ ning bir qismidir, shuning uchun ularning har birini quyidagicha aniqlaymiz.

${S_1}:{\mbox{ Plane given by }}4x + 2y + z = 8{\mbox{ }}\left( {i.e{\mbox{ the top of the solid}}} \right)$

${S_2}:{\mbox{ Plane given by }}y = 0{\mbox{ }}\left( {i.e{\mbox{ the triangle on right side of the solid}}} \right)$

${S_3}:{\mbox{ Plane given by }}x = 0{\mbox{ }}\left( {i.e{\mbox{ the triangle at back of the solid - not shown in sketch}}} \right)$

${S_4}:{\mbox{ Plane given by }}z = 0{\mbox{ }}\left( {i.e{\mbox{ the triangle on bottom of the solid - not shown in sketch}}} \right)$

Yuqoridagi ta'riflarda ta'kidlanganidek, dastlabki ikkita sirt eskizda ko'rsatilgan, ammo oxirgi ikkitasi qattiq jismning yo'nalishi tufayli ko'rsatilmagan. Quyida koordinatalar tekisliklariga mos keladigan uchta sirtning har birining eskizlari keltirilgan.

$Three two dimensional sketches side by side, one for each of the surfaces lying in a coordinate plane, with limits listed underneath each. On the left is ${S_2}$, the triangle in the xz-plane bounded by the line $4x + z = 8$, with $0 \le x \le 2$ and $0 \le z \le 8 - 4x$. In the middle is ${S_3}$, the triangle in the yz-plane bounded by $2y + z = 8$, with $0 \le y \le 4$ and $0 \le z \le 8 - 2y$. On the right is ${S_4}$, the triangle in the xy-plane bounded by $4x + 2y = 8$, with $0 \le x \le 2$ and $0 \le y \le 4 - 2x$. All three are shaded green.$

Har bir eskizda biz ularning har biri uchun o'zgaruvchilarga limitlar berdik, chunki har bir sirt bo'ylab sirt integralini bajarishni boshlaganimizda, oxir-oqibat bizga kerak bo'ladi.

Endi biz ushbu sirtlarning har biri uchun integralni ko'rib chiqishimiz va bajarishimiz kerak va biz ularni ushbu bo'limdagi birinchi bir nechta muammolarga qaraganda biroz tezroq o'tkazamiz.

2-qadam

${S_1}$ bilan boshlaylik. Bunday holda, sirt $z$ uchun osongina echilishi mumkin,

\[z = 8 - 4x - 2y\]

Shu tarzda yozilgan sirt tenglamasi bilan $D$ mintaqasi ichida bo'ladi xy‑samolyot va agar siz bu haqda o'ylab ko'rsangiz, aslida $D$ ${S_4}$ dan boshqa narsa emasligini ko'rasiz!

Bu holda integral,

\[\begin{align*}\iint\limits_{{{S_1}}}{{yz + 4xy\,dS}} & = \iint\limits_{D}{{\left[ {y\left( {8 - 4x - 2y} \right) + 4xy} \right]\sqrt {{{\left( { - 4} \right)}^2} + {{\left( { - 2} \right)}^2} + 1} \,dA}}\\ & = \sqrt {21} \iint\limits_{D}{{8y - 2{y^2}\,dA}}\end{align*}\]

Sirt tenglamasini integranda $z$ ga ulashni unutmang va ildizni hisoblashda sirt tenglamasidan foydalanishni unutmang!

Endi, yuqorida aytib o'tilganidek, bu sirt uchun $D$ ${S_4}$ dan boshqa narsa emas va shuning uchun biz 1-bosqichda ${S_4}$ eskizidagi chegaralardan foydalanishimiz mumkin.

Endi bu sirt uchun integralni hisoblaymiz.

\[\begin{align*}\iint\limits_{{{S_1}}}{{yz + 4xy\,dS}} & = \sqrt {21} \int_{0}^{2}{{\int_{0}^{{4 - 2x}}{{8y - 2{y^2}\,dy}}\,dx}}\\ & = \sqrt {21} \int_{0}^{2}{{\left. {\left[ {4{y^2} - \frac{2}{3}{y^3}} \right]} \right|_0^{4 - 2x}\,dx}}\\ & = \sqrt {21} \int_{0}^{2}{{4{{\left( {4 - 2x} \right)}^2} - \frac{2}{3}{{\left( {4 - 2x} \right)}^3}\,dx}}\\ & = \sqrt {21} \left. {\left[ { - \frac{2}{3}{{\left( {4 - 2x} \right)}^3} + \frac{1}{{12}}{{\left( {4 - 2x} \right)}^4}} \right]} \right|_0^2 = \underline {\frac{{64\sqrt {21} }}{3} = 97.7616} \end{align*}\]

3-qadam

Keyinchalik biz ${S_2}$ bilan shug'ullanamiz. Bu holda sirt uchun tenglama oddiygina $y = 0$ va $D$ 1-bosqichdagi ${S_2}$ eskizida berilgan.

Bu holda integral,

\[\iint\limits_{{{S_2}}}{{yz + 4xy\,dS}} = \iint\limits_{D}{{0\sqrt {{{\left( 0 \right)}^2} + 1 + {{\left( 0 \right)}^2}} \,dA}} = \iint\limits_{D}{{0\,dA}} = \underline 0 \]

Shunday qilib, bu holda biz integralni hisoblashimiz shart emas edi. Ba'zida bizga omad kulib boqadi, garchi bu tez-tez sodir bo'lmasa ham.

4-qadam

Endi biz ${S_3}$-ga g'amxo'rlik qilishimiz mumkin. Bu holda sirt uchun tenglama oddiygina $x = 0$ va $D$ 1-bosqichdagi ${S_3}$ eskizida berilgan.

Bu holda integral,

\[\iint\limits_{{{S_3}}}{{yz + 4xy\,dS}} = \iint\limits_{D}{{\left[ {yz + 4\left( 0 \right)y} \right]\sqrt {1 + {{\left( 0 \right)}^2} + {{\left( 0 \right)}^2}} \,dA}} = \iint\limits_{D}{{yz\,dA}}\]

Sirt tenglamasini integranddagi $x$ ga ulashni unutmang va ildizni hisoblashda sirt tenglamasidan foydalanishni unutmang (garchi bu holda ildiz faqat bittaga baholanadi)!

1-bosqichdagi eskizdagi $D$ chegaralaridan foydalanib, biz ushbu sirt uchun integralni tezda baholashimiz mumkin.

\[\begin{align*}\iint\limits_{{{S_3}}}{{yz + 4xy\,dS}} & = \int_{0}^{4}{{\int_{0}^{{8 - 2y}}{{yz\,dz}}\,dy}}\\ & = \int_{0}^{4}{{\left. {\left[ {\frac{1}{2}y{z^2}} \right]} \right|_0^{8 - 2y}\,dy}}\\ & = \int_{0}^{4}{{32y - 16{y^2} + 2{y^3}\,dy}}\\ & = \left. {\left[ {16{y^2} - \frac{{16}}{3}{y^3} + \frac{1}{2}{y^4}} \right]} \right|_0^4 = \underline {\frac{{128}}{3} = 42.6667} \end{align*}\]

5-qadam

Va nihoyat, ${S_4}$-ga g'amxo'rlik qilaylik. Bu holda sirt uchun tenglama oddiygina $z = 0$ va $D$ 1-bosqichdagi ${S_4}$ eskizida berilgan.

Bu holda integral,

\[\iint\limits_{{{S_4}}}{{yz + 4xy\,dS}} = \iint\limits_{D}{{\left[ {y\left( 0 \right) + 4xy} \right]\sqrt {{{\left( 0 \right)}^2} + {{\left( 0 \right)}^2} + 1} \,dA}} = \iint\limits_{D}{{4xy\,dA}}\]

Sirt tenglamasini integranddagi $z$ ga ulashni unutmang va ildizni hisoblashda sirt tenglamasidan foydalanishni unutmang (garchi bu holda ildiz faqat bittaga baholanadi)!

1-bosqichdagi eskizdagi $D$ chegaralaridan foydalanib, biz ushbu sirt uchun integralni tezda baholashimiz mumkin.

\[\begin{align*}\iint\limits_{{{S_4}}}{{yz + 4xy\,dS}} & = \int_{0}^{2}{{\int_{0}^{{4 - 2x}}{{4xy\,dy}}\,dx}}\\ & = \int_{0}^{2}{{\left. {\left[ {2x{y^2}} \right]} \right|_0^{4 - 2x}\,dx}}\\ & = \int_{0}^{2}{{32x - 32{x^2} + 8{x^3}\,dx}}\\ & = \left. {\left[ {16{x^2} - \frac{{32}}{3}{x^3} + 2{x^4}} \right]} \right|_0^2 = \underline {\frac{{32}}{3} = 10.6667} \end{align*}\]

6-qadam

Endi to'liq sirt bo'yicha integralning qiymatini olish uchun biz qilishimiz kerak bo'lgan narsa - yuqoridagi to'rtta sirt ustidagi har bir integralning qiymatlarini jamlash. Shunday qilib,

\[\iint\limits_{S}{{yz + 4xy\,dS}} = \left( {\frac{{64\sqrt {21} }}{3}} \right) + \left( 0 \right) + \left( {\frac{{128}}{3}} \right) + \left( {\frac{{32}}{3}} \right) = \frac{{64\sqrt {21} }}{3} + \frac{{160}}{3} = \require{bbox} \bbox[2pt,border:1px solid black]{{151.0949}}\]

Biz har bir alohida integral qiymatlar qayerdan kelganini ko'rsatish uchun qavslar qo'yamiz. Umuman olganda, bular, albatta, kerak emas.