15.3-bo'lim : Umumiy hududlar bo'yicha ikki karrali integrallar
9. \( \displaystyle \iint\limits_{D}{{7{x^2} + 14y\,dA}}\) ni baholang, bu erda \(D\) \(x = 2{y^2}\) va \(x = 8\) bilan chegaralangan mintaqa bo'lib, quyida keltirilgan tartibda.
- Avval \(x\), keyin esa \(y\) ga nisbatan integratsiya qiling.
- Avval \(y\), keyin esa \(x\) ga nisbatan integratsiya qiling.
Yechim
Mana, avval \(x\) bilan integratsiya qilish uchun belgilangan egri chiziqlar bilan mintaqaning qisqacha eskizi.
Avval \(x\) ga nisbatan integrallash uchun chegaralar, keyin:
\[\begin{array}{c} - 2 \le y \le 2\\ \,2{y^2} \le x \le 8\end{array}\]Bu chegaralarni integralga kiritish, keyin,
\[\iint\limits_{D}{{7{x^2} + 14y\,dA}} = \int_{{ - 2}}^{2}{{\int_{{2{y^2}}}^{8}{{7{x^2} + 14y\,dx}}\,dy}}\]2-qadam
Ushbu integral uchun \(x\) integratsiyasi:
\[\begin{align*}\iint\limits_{D}{{7{x^2} + 14y\,dA}} & = \int_{{ - 2}}^{2}{{\int_{{2{y^2}}}^{8}{{7{x^2} + 14y\,dx}}\,dy}}\\ & = \int_{{ - 2}}^{2}{{\left. {\left( {\frac{7}{3}{x^3} + 14xy} \right)} \right|_{2{y^2}}^8\,dy}}\\ & = \int_{{ - 2}}^{2}{{\frac{{3584}}{3} + 112y - 28{y^3} - \frac{{56}}{3}{y^6}\,dy}}\end{align*}\]3-qadam
Va nihoyat, \(y\) integratsiyasi,
\[\begin{align*}\iint\limits_{D}{{7{x^2} + 14y\,dA}} & = \int_{{ - 2}}^{2}{{\frac{{3584}}{3} + 112y - 28{y^3} - \frac{{56}}{3}{y^6}\,dy}}\\ & = \left. {\left( {\frac{{3584}}{3}y + 56{y^2} - 7{y^4} - \frac{8}{3}{y^7}} \right)} \right|_{ - 2}^2 = \require{bbox} \bbox[2pt,border:1px solid black]{{4096}}\end{align*}\]b Avval \(y\), keyin esa \(x\) ga nisbatan integratsiya qiling.
Yechim
Mana, avval \(y\) bilan integratsiya qilish uchun belgilangan egri chiziqlar bilan mintaqaning qisqacha eskizi.
E'tibor bering, \(y\) integratsiyasini amalga oshirish uchun avval \(y\) uchun parabola tenglamasini echishimiz kerak edi, shuning uchun yuqori va pastki egri \(x\) nuqtai nazaridan alohida tenglamalarga ega bo'ladi, biz avval \(y\) ga nisbatan integrallashimiz kerak.
Avval \(y\) ga nisbatan integrallash uchun chegaralar, keyin:
\[\begin{array}{c} \displaystyle 0 \le x \le 8\\ - \sqrt {\frac{1}{2}x} \le y \le \sqrt {\frac{1}{2}x} \end{array}\]Bu chegaralarni integralga kiritish, keyin,
\[\iint\limits_{D}{{7{x^2} + 14y\,dA}} = \int_{0}^{8}{{\int_{{ - \sqrt {\frac{1}{2}x} }}^{{\sqrt {\frac{1}{2}x} }}{{7{x^2} + 14y\,dy}}\,dx}}\]2-qadam
Ushbu integral uchun \(y\) integratsiyasi:
\[\begin{align*}\iint\limits_{D}{{7{x^2} + 14y\,dA}} & = \int_{0}^{8}{{\int_{{ - \sqrt {\frac{1}{2}x} }}^{{\sqrt {\frac{1}{2}x} }}{{7{x^2} + 14y\,dy}}\,dx}}\\ & = \int_{0}^{8}{{\left. {\left( {7{x^2}y + 7{y^2}} \right)} \right|_{ - \sqrt {\frac{1}{2}x} }^{\sqrt {\frac{1}{2}x} }\,dx}}\\ & = \int_{0}^{8}{{\frac{{14}}{{\sqrt 2 }}{x^{\frac{5}{2}}}\,dx}}\end{align*}\]3-qadam
Va nihoyat, \(x\) integratsiyasi,
\[\iint\limits_{D}{{7{x^2} + 14y\,dA}} = \int_{0}^{8}{{\frac{{14}}{{\sqrt 2 }}{x^{\frac{5}{2}}}\,dx}} = \left. {\left( {\frac{4}{{\sqrt 2 }}{x^{\frac{7}{2}}}} \right)} \right|_0^8 = \require{bbox} \bbox[2pt,border:1px solid black]{{4096}}\]Biz bilganimizdek, integratsiyaning birinchi tartibi bilan bir xil natijaga erishdik.