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15.3-bo'lim : Umumiy hududlar bo'yicha ikki karrali integrallar

7. \( \displaystyle \iint\limits_{D}{{3 - 6xy\,dA}}\) ni baholang, bu erda \(D\) quyida ko'rsatilgan mintaqadir.

The region \(D\) is shaded in. It is bounded above by the horizontal line \(y = 1\) and below by the horizontal line \(y = - 4\). The upper part of the region is the area inside the upward opening parabola \(y = {x^2}\), running from \(\left( { - 1,1} \right)\) across to \(\left( {1,1} \right)\), and the lower part is the area inside the downward opening parabola \(y = - {x^2}\), running from \(\left( { - 2, - 4} \right)\) across to \(\left( {2, - 4} \right)\). The two parts meet at the origin where the region pinches down to a single point.
Yechim

Birinchidan, quyida ko'rsatilganidek, \(D\)-dagi ikkita kichik mintaqani belgilaymiz.

The region \(D\) is shaded in. It is bounded above by the horizontal line \(y = 1\) and below by the horizontal line \(y = - 4\). The upper part of the region is the area inside the upward opening parabola \(y = {x^2}\), running from \(\left( { - 1,1} \right)\) across to \(\left( {1,1} \right)\), and the lower part is the area inside the downward opening parabola \(y = - {x^2}\), running from \(\left( { - 2, - 4} \right)\) across to \(\left( {2, - 4} \right)\). The two parts meet at the origin where the region pinches down to a single point. The upper region is labled \(D_{1}\) and the lower region is labeled \(D_{2}\).
2-qadam

Umid qilamizki, biz \(D\) ni to'liq tavsiflaydigan bitta chegara to'plamini ololmasligimiz aniq, shuning uchun biz integralni quyidagicha ajratishimiz kerak.

\[\iint\limits_{D}{{3 - 6xy\,dA}} = \iint\limits_{{{D_{\,1}}}}{{3 - 6xy\,dA}} + \iint\limits_{{{D_{\,2}}}}{{3 - 6xy\,dA}}\]
3-qadam

Mintaqa, shuningdek, (umid qilamanki) biz \(y\) ni ikkala integral bilan birlashtirishimiz kerakligini aniq ko'rsatishi kerak. Shunday qilib, bu erda har bir integral uchun chegaralar mavjud.

\[\begin{array}{*{20}{c}}\begin{array}{c}{D_{\,1}}\,\\ - 1 \le x \le 1\\ {x^2} \le y \le 1\end{array}&{\hspace{0.5in}}&\begin{array}{c}{D_{\,2}}\\ - 2 \le x \le 2\\ - 4 \le y \le - {x^2}\end{array}\end{array}\]

U holda integrallar,

\[\iint\limits_{D}{{3 - 6xy\,dA}} = \int_{{ - 1}}^{1}{{\int_{{{x^2}}}^{1}{{3 - 6xy\,dy}}\,dx}} + \int_{{ - 2}}^{2}{{\int_{{ - 4}}^{{ - {x^2}}}{{3 - 6xy\,dy}}\,dx}}\]
4-qadam

Endi integrallarni bajarishdan boshqa ko'p narsa qilish kerak emas. Mana ikkalasi uchun \(y\) integratsiyasi.

\[\begin{align*}\iint\limits_{D}{{3 - 6xy\,dA}} & = \int_{{ - 1}}^{1}{{\left. {\left( {3y - 3x{y^2}} \right)} \right|_{{x^2}}^1\,dx}} + \int_{{ - 2}}^{2}{{\left. {\left( {3y - 3x{y^2}} \right)} \right|_{ - 4}^{ - {x^2}}\,dx}}\\ & = \int_{{ - 1}}^{1}{{3{x^5} - 3{x^2} - 3x + 3\,dx}} + \int_{{ - 2}}^{2}{{12 + 48x - 3{x^2} - 3{x^5}\,dx}}\end{align*}\]
5-qadam

Va nihoyat, ikkala integral uchun \(x\) integratsiyasi.

\[\begin{align*}\iint\limits_{D}{{3 - 6xy\,dA}} & = \int_{{ - 1}}^{1}{{3{x^5} - 3{x^2} - 3x + 3\,dx}} + \int_{{ - 2}}^{2}{{12 + 48x - 3{x^2} - 3{x^5}\,dx}}\\ & = \left. {\left( {\frac{1}{2}{x^6} - {x^3} - \frac{3}{2}{x^2} + 3x} \right)} \right|_{ - 1}^1 + \left. {\left( {12x + 24{x^2} - {x^3} - \frac{1}{2}{x^6}} \right)} \right|_{ - 2}^2\\ & = \hspace{1.15in}4\hspace{1.4in} + \hspace{1.15in}32\\ & = \require{bbox} \bbox[2pt,border:1px solid black]{{36}}\end{align*}\]

Har doim mintaqadagi har bir integral bitta integral bilan bajarilishini kutmang. Vaqti-vaqti bilan siz integralni bo'lishingiz va alohida kichik mintaqalar bo'yicha haqiqiy integratsiyani amalga oshirishingiz kerak bo'ladi. Bunday holda, bu aniq edi, lekin ba'zida muammoga duch kelmaguningizcha va uni kichik hududlarda qilish osonroq bo'lishini tushunmaguningizcha, bu unchalik aniq bo'lmasligi mumkin.