17.6-bo'lim : Divergensiya teoremasi
2. \( \displaystyle \iint\limits_{S}{{\vec F\centerdot d\vec S}}\) ni baholash uchun Divergensiya teoremasidan foydalaning, bunda \(\vec F = \sin \left( {\pi x} \right)\,\vec i + z{y^3}\,\vec j + \left( {{z^2} + 4x} \right)\,\vec k\) va \(S\) \( - 1 \le x \le 2\), \(0 \le y \le 1\) va \(1 \le z \le 4\) bilan qutining yuzasi. E'tibor bering, qutining barcha olti tomoni \(S\)-ga kiritilgan.
Yechim
Keling, ushbu muammoni hal qilishda biz ishlayotgan sirtning qisqacha eskizi bilan boshlaylik.
Biz an'anaviy o'qlari bo'lgan eskizni va sirtni ingl.
Shuni ham yodda tutingki, biz yuqorida ko'rsatilgan qutining barcha olti tomonini o'z ichiga olganimiz sababli, sirt qutini o'rab oladi (yoki agar siz ushbu terminologiyadan foydalanmoqchi bo'lsangiz, chegara egri chizig'i).
2-qadam
Divergensiya teoremasidan quyidagi yo'nalishda foydalanamiz.
\[\iint\limits_{S}{{\vec F\centerdot d\vec S}} = \iiint\limits_{E}{{{\mathop{\rm div}\nolimits} \vec F\,dV}}\]bu erda \(E\) 1-qadamdagi eskizlarda ko'rsatilgan qattiq narsadir.
\(E\) bu shunchaki quti va muammo bayonida berilgan joyda uni belgilaydigan chegaralar. Bizning integralimiz uchun chegaralar shunday bo'ladi:
\[\begin{array}{c} - 1 \le x \le 2\\ 0 \le y \le 1\\1 \le z \le 4\end{array}\]Bizga vektor maydonining divergentsiyasi ham kerak bo'ladi, shuning uchun bu erda.
\[{\mathop{\rm div}\nolimits} \vec F = \frac{\partial }{{\partial x}}\left( {\sin \left( {\pi x} \right)} \right) + \frac{\partial }{{\partial y}}\left( {z{y^3}\,} \right) + \frac{\partial }{{\partial z}}\left( {{z^2} + 4x} \right) = \pi \cos \left( {\pi x} \right) + 3z{y^2} + 2z\]3-qadam
Endi integralga Divergentsiya teoremasini qo'llaymiz va uni uch karrali integralga aylantiramiz.
\[\begin{align*}\iint\limits_{S}{{\vec F\centerdot d\vec S}} & = \iiint\limits_{E}{{{\mathop{\rm div}\nolimits} \vec F\,dV}}\\ & = \iiint\limits_{E}{{\pi \cos \left( {\pi x} \right) + 3z{y^2} + 2z\,dV}}\\ & = \int_{{ - 1}}^{2}{{\int_{0}^{1}{{\int_{1}^{4}{{\pi \cos \left( {\pi x} \right) + 3z{y^2} + 2z\,dz}}\,dy}}\,dx}}\end{align*}\]4-qadam
Keyin integralni baholashda qilishimiz kerak.
\[\begin{align*}\iint\limits_{S}{{\vec F\centerdot d\vec S}} & = \int_{{ - 1}}^{2}{{\int_{0}^{1}{{\int_{1}^{4}{{\pi \cos \left( {\pi x} \right) + 3z{y^2} + 2z\,dz}}\,dy}}\,dx}}\\ & = \int_{{ - 1}}^{2}{{\int_{0}^{1}{{\left. {\left( {\pi z\cos \left( {\pi x} \right) + \frac{3}{2}{z^2}{y^2} + {z^2}} \right)} \right|_1^4\,dy}}\,dx}}\\ & = \int_{{ - 1}}^{2}{{\int_{0}^{1}{{3\pi \cos \left( {\pi x} \right) + \frac{{45}}{2}{y^2} + 15\,dy}}\,dx}}\\ & = \int_{{ - 1}}^{2}{{\left. {\left( {3y\pi \cos \left( {\pi x} \right) + \frac{{15}}{2}{y^3} + 15y} \right)} \right|_0^1\,dx}}\\ & = \int_{{ - 1}}^{2}{{3\pi \cos \left( {\pi x} \right) + \frac{{45}}{2}\,dx}}\\ & = \left. {\left( {3\sin \left( {\pi x} \right) + \frac{{45}}{2}x} \right)} \right|_{ - 1 }^2 = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{135}}{2}}}\end{align*}\]