15.3-bo'lim : Umumiy hududlar bo'yicha ikki karrali integrallar
3. \( \displaystyle \iint\limits_{D}{{10{x^2}{y^3} - 6\,dA}}\) ni baholang, bu erda \(D\) \(x = - 2{y^2}\) va \(x = {y^3}\) bilan chegaralangan mintaqadir.
Yechim
Quyida \(D\) hududining qisqacha eskizi keltirilgan.
![A sketch of the region \(D\). It is the thin sliver in the third quadrant between the curve \(x = - 2{y^2}\), which can also be written \(y = - \sqrt { - \frac{1}{2}x} \), and the curve \(x = {y^3}\), which can also be written \(y = \sqrt[3]{x}\). The two meet at the origin and again at \(\left( { - 8, - 2} \right)\) and the region between them is shaded green.](/Solutions/CalcIII/DIGeneralRegion//Solutions/CalcIII/DIGeneralRegion/DIGeneralRegion_Files/image003.webp)
E'tibor bering, biz keyingi bosqichda yordam berish uchun har bir egri chiziq uchun tenglamaning ikkala shaklini ham berdik.
Umuman olganda, bu eskiz ko'pincha integralni to'g'ri o'rnatish uchun muhimdir. Biz integratsiya tartibini aniqlashimiz kerak va ko'pincha mintaqa ma'lum bir tartibni "majburlaydi". Ko'pgina hududlarni integratsiyaning ma'lum bir tartibini bajarish orqali osonlik bilan hal qilish mumkin va ba'zida buni haqiqatan ham ko'rishning yagona yo'li \(D\) eskiziga ega bo'lishdir.
Agar siz integralni ikkala tartibda ham bajarishingiz mumkin bo'lsa ham, \(D\) eskizi ko'pincha integrallar uchun chegaralarni o'rnatishda yordam beradi.
2-qadam
Endi, bu muammo bilan, mintaqa integratsiyadagi qiyinchilikni real o'zgartirmasdan, integratsiya tartibini amalga oshirishga imkon beradi.
Shunday qilib, biz foydalanishimiz mumkin bo'lgan har bir integratsiya tartibi uchun limitlar mavjud.
\[\begin{array}{*{20}{c}}\begin{array}{c} - 8 \le x \le 0\\ \sqrt[3]{x} \le y \le - \displaystyle \sqrt { - \frac{1}{2}x} \end{array}&{\hspace{0.25in}{\mbox{OR}}\hspace{0.25in}}&\begin{array}{c} - 2 \le y \le 0\\ - 2{y^2} \le x \le {y^3}\end{array}\end{array}\]3-qadam
Yuqorida ta'kidlab o'tilganidek, har ikkala buyurtma ham qiyinchilikda haqiqiy o'zgarishsiz bajarilishi mumkin. Biroq, birinchi chegaralar to'plami uchun \(y\) chegaralarining ikkalasi ham ildizdir va bu ikkinchi integratsiyani biroz chigallashtirishi mumkin. Yuqori \(y\) chegarasi ham barcha minus belgilari bilan biroz chalkash.
Shunday qilib, ushbu muammoni hal qilish uchun avval \(x\) ga integratsiya qilaylik. Bu erda birinchi navbatda \(x\) integratsiyasi uchun o'rnatilgan integral.
\[\iint\limits_{D}{{10{x^2}{y^3} - 6\,dA}} = \int_{{ - 2}}^{0}{{\int_{{ - 2{y^2}}}^{{{y^3}}}{{10{x^2}{y^3} - 6\,dx}}\,dy}}\]4-qadam
Mana \(x\) integratsiyasi.
\[\begin{align*}\iint\limits_{D}{{10{x^2}{y^3} - 6\,dA}} & = \int_{{ - 2}}^{0}{{\int_{{ - 2{y^2}}}^{{{y^3}}}{{10{x^2}{y^3} - 6\,dx}}\,dy}}\\ & = \int_{{ - 2}}^{0}{{\left. {\left( {\frac{{10}}{3}{x^3}{y^3} - 6x} \right)} \right|_{ - 2{y^2}}^{{y^3}}\,dy}}\\ & = \int_{{ - 2}}^{0}{{\frac{{10}}{3}{y^9}\left( {{y^3}} \right) - 6{y^3} - \left[ {\frac{{10}}{3}\left( { - 8{y^6}} \right)\left( {{y^3}} \right) - 6\left( { - 2{y^2}} \right)} \right]\,dy}}\\ & = \int_{{ - 2}}^{0}{{\frac{{10}}{3}{y^{12}} + \frac{{80}}{3}{y^9} - 6{y^3} - 12{y^2}\,dy}}\end{align*}\]5-qadam
Bunday holda, biz integralning murakkabligini kamaytirish uchun qila oladigan juda oddiy soddalashtirishning kichik miqdori mavjud va shuning uchun biz buni qildik.
Faqat \(y\) integratsiyasini bajarish qoladi. Mana o'sha ish.
\[\begin{align*}\iint\limits_{D}{{10{x^2}{y^3} - 6\,dA}} & = \int_{{ - 2}}^{0}{{\frac{{10}}{3}{y^{12}} + \frac{{80}}{3}{y^9} - 6{y^3} - 12{y^2}\,dy}}\\ & = \left. {\left( {\frac{{10}}{{39}}{y^{13}} + \frac{8}{3}{y^{10}} - \frac{3}{2}{y^4} - 4{y^3}} \right)} \right|_{ - 2}^0 = \require{bbox} \bbox[2pt,border:1px solid black]{{ - \frac{{8296}}{{13}} = - 638.1538}}\end{align*}\]