DIGeneralRegion - 13-masala yechimi | MathModel.uz

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

13. $xy$-tekisligi va $f\left( {x,y} \right) = 2 + \cos \left( {{x^2}} \right)$ orasidagi va $\left( {0,0} \right)$, $\left( {6,0} \right)$ va $\left( {6,2} \right)$ uchburchak ustida joylashgan hudud hajmini aniqlash uchun ikki karrali integraldan foydalaning.

Yechim

Keling, avval funksiyaning va uning ostida joylashgan uchburchakning eskizini olamiz.

$A three dimensional sketch of the surface $f\left( {x,y} \right) = 2 + \cos \left( {{x^2}} \right)$ drawn with the traditional axes. The surface ripples up and down in the x direction with the ripples getting closer and closer together as $x$ increases, and it is constant in the y direction. The greenish triangle drawn underneath the surface on the xy-plane is the triangle with vertices $\left( {0,0} \right)$, $\left( {6,0} \right)$ and $\left( {6,2} \right)$.$

$A three dimensional sketch of the same surface $f\left( {x,y} \right) = 2 + \cos \left( {{x^2}} \right)$ and the triangle beneath it, this time drawn with a “box frame” set of axes to make the surface easier to see.$

Sirt an'anaviy o'qlar to'plami va shuningdek, "quti ramkasi" o'qlari to'plami bilan chizilgan. Ba'zan ikkala eskiz ham mavjud bo'lganda, sirt bilan nima sodir bo'layotganini ko'rish osonroq bo'ladi.

Sirt ostidagi yashil rangli uchburchak muammo bayonotida ko'rsatilgan uchburchakdir.

2-qadam

Endi biz kutayotgan hajm quyidagi integral bilan berilgan:

\[V = \iint\limits_{D}{{2 + \cos \left( {{x^2}} \right)\,dA}}\]

bu erda $D$ - muammo bayonotida ko'rsatilgan uchburchak.

Shunday qilib, integralni baholash uchun bizga $D$ eskizi kerak bo'ladi, shunda biz integrallanish tartibini va integrallar uchun chegaralarni aniqlashimiz mumkin.

$A sketch of the region $D$. It is the triangle with vertices $\left( {0,0} \right)$, $\left( {6,0} \right)$ and $\left( {6,2} \right)$, with the bottom side along the x-axis, the right side vertical at $x = 6$ and the upper side the line $y = \frac{1}{3}x$. The inside is shaded green.$

3-qadam

$D$ mintaqasini integratsiyaning har ikkala tartibi uchun osongina tasvirlash mumkin. Biroq, integralni birinchi navbatda $x$ bilan birlashtirib bo'lmasligi aniq bo'lishi kerak va shuning uchun biz avval $y$ ga nisbatan integrallashimiz kerak.

Mana bu tartib bilan integral uchun chegaralar.

\[\begin{array}{c} 0 \le x \le 6\\ 0 \le y \le \displaystyle \frac{1}{3}x\end{array}\]

Keyin hajm uchun integrali,

\[V = \iint\limits_{D}{{2 + \cos \left( {{x^2}} \right)\,dA}} = \int_{0}^{6}{{\int_{0}^{{\frac{1}{3}x}}{{2 + \cos \left( {{x^2}} \right)\,dy}}\,dx}}\]

4-qadam

Endi biz qilishimiz kerak bo'lgan narsa - bu integralni baholash. Mana $y$ integratsiyasi.

\[V = \int_{0}^{6}{{\left. {\left( {2y + y\cos \left( {{x^2}} \right)} \right)} \right|_0^{\frac{1}{3}x}\,dx}} = \int_{0}^{6}{{\frac{2}{3}x + \frac{1}{3}x\cos \left( {{x^2}} \right)\,dx}}\]

5-qadam

Va nihoyat, $x$ integratsiyasi va shuning uchun hajm,

\[V = \left. {\left( {\frac{1}{3}{x^2} + \frac{1}{6}\sin \left( {{x^2}} \right)\,} \right)} \right|_0^6 = \require{bbox} \bbox[2pt,border:1px solid black]{{12 + \frac{1}{6}\sin \left( {36} \right) = 11.8347}}\]

Agar siz o'nli kasrlarga o'tayotgan bo'lsangiz, kalkulyatoringizni radianga o'rnatishni unutmang!