SurfaceIntegrals - 1-masala yechimi | MathModel.uz

17.3-bo'lim : Yuzaki integrallar

1. $ \displaystyle \iint\limits_{S}{{z + 3y - {x^2}\,dS}}$ ni baholang, bu erda $S$ $z = 2 - 3y + {x^2}$ ning $xy$ tekisligida uchburchak ustida joylashgan qismi $\left( {0,0} \right)$, $\left( {2,0} \right)$ va $\left( {2, - 4} \right)$ cho'qqilari bilan.

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 surface is $z = 2 - 3y + {x^2}$, which slopes up steeply towards negative $y$, and the greenish triangle in the xy-plane below it is the region $D$, the triangle with vertices $\left( {0,0} \right)$, $\left( {2,0} \right)$ and $\left( {2, - 4} \right)$.$

$A three dimensional sketch of the same surface $z = 2 - 3y + {x^2}$ and the triangle $D$ beneath it, this time drawn with a set of “box” axes to make the surface easier to visualize.$

Biz an'anaviy o'qlari bo'lgan eskizni va sirtni ingl.

To'q sariq sirt biz ushbu muammoda ishlayotgan $z = 2 - 3y + {x^2}$ eskizidir. Sirt ostidagi yashil rangli uchburchak - bu sirt ostida joylashgan muammo bayonotida ko'rsatilgan uchburchak. Ushbu uchburchak ushbu muammo uchun $D$ mintaqasi bo'ladi.

Mana, yuqoridagi eskizdagi asosan xiralashgan ko'rinishdan yaxshiroq ko'rish uchun $D$ ning qisqacha eskizi.

$A two dimensional sketch of the region $D$. It is the triangle with vertices $\left( {0,0} \right)$, $\left( {2,0} \right)$ and $\left( {2, - 4} \right)$, bounded above by the x-axis, on the right by the line $x = 2$ and below by the line $y = - 2x$. The inside is shaded green.$

$D$ ni tavsiflash uchun quyidagi limitlar to'plamidan foydalanishimiz mumkin.

Qaysi to‘plamni integralda ishlatishni o‘rnatganimizdan so‘ng hal qilamiz.

2-qadam

Keling, endi integralni o'rnatamiz. Bu holda sirt shaklda bo'ladi,

\[z = g\left( {x,y} \right) = 2 - 3y + {x^2}\]

shuning uchun sirt integrali uchun quyidagi formuladan foydalanamiz.

\[\iint\limits_{S}{{f\left( {x,y,z} \right)\,dS}} = \iint\limits_{D}{{f\left( {x,y,g\left( {x,y} \right)} \right)\sqrt {{{\left( {\frac{{\partial g}}{{\partial x}}} \right)}^2} + {{\left( {\frac{{\partial g}}{{\partial y}}} \right)}^2} + 1} \,dA}}\]

U holda integral,

\[\begin{align*}\iint\limits_{S}{{z + 3y - {x^2}\,dS}} & = \iint\limits_{D}{{\left[ {\left( {2 - 3y + {x^2}} \right) + 3y - {x^2}} \right]\sqrt {{{\left( {2x} \right)}^2} + {{\left( { - 3} \right)}^2} + 1} \,dA}}\\ & = \iint\limits_{D}{{2\sqrt {4{x^2} + 10} \,dA}}\end{align*}\]

Sirt tenglamasini integranddagi $z$ ga ulashni unutmang va $D$ 1-bosqichda chizilgan uchburchak ekanligini unutmang.

3-qadam

Endi biz qilishimiz kerak bo'lgan yagona narsa - ikki karrali integralni baholash va bu hozir juda qiyin bo'lmasligi kerak.

Birinchidan, shuni esda tutingki, integraldan biz avval $y$ ga nisbatan integratsiya qilishni xohlashimiz aniq bo'lishi kerak (agar siz trigni almashtirishni xohlamasangiz ....). Shunday qilib, integral bo'ladi,

\[\begin{align*}\iint\limits_{S}{{z + 3y - {x^2}\,dS}} & = \iint\limits_{D}{{2\sqrt {4{x^2} + 10} \,dA}}\\ & = \int_{0}^{2}{{\int_{{ - 2x}}^{0}{{2\sqrt {4{x^2} + 10} \,dy}}\,dx}}\\ & = \int_{0}^{2}{{\left. {\left( {2y\sqrt {4{x^2} + 10} } \right)} \right|_{ - 2x}^0\,dx}}\\ & = \int_{0}^{2}{{4x\sqrt {4{x^2} + 10} \,dx}}\\ & = \left. {\frac{1}{3}{{\left( {4{x^2} + 10} \right)}^{\frac{3}{2}}}} \right|_0^2 = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{1}{3}\left( {{{26}^{\frac{3}{2}}} - {{10}^{\frac{3}{2}}}} \right) = 33.6506}}\end{align*}\]