DIGeneralRegion - Amaliy mashqlar | MathModel.uz

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

$ \displaystyle \iint\limits_{D}{{42{y^2} - 12x\,dA}}$ ni baholang, bu erda $D = \left\{ {\left( {x,y} \right)|0 \le x \le 4,{{\left( {x - 2} \right)}^2} \le y \le 6} \right\}$ Yechim
$ \displaystyle \iint\limits_{D}{{2y{x^2} + 9{y^3}\,dA}}$ ni baholang, bu erda $D$ $ \displaystyle y = \frac{2}{3}x$ va $y = 2\sqrt x $ bilan chegaralangan mintaqadir. Yechim
$ \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
$ \displaystyle \iint\limits_{D}{{x\left( {y - 1} \right)\,dA}}$ ni baholang, bu erda $D$ $y = 1 - {x^2}$ va $y = {x^2} - 3$ bilan chegaralangan hududdir. Yechim
$ \displaystyle \iint\limits_{D}{{5{x^3}\cos \left( {{y^3}} \right)\,dA}}$ ni baholang, bu erda $D$ $y = 2$, $ \displaystyle y = \frac{1}{4}{x^2}$ va $y$ o'qi bilan chegaralangan hududdir. Yechim
$ \displaystyle \iint\limits_{D}{{\frac{1}{{{y^{\frac{1}{3}}}\left( {{x^3} + 1} \right)}}\,dA}}$ ni baholang, bu erda $D$ $x = - {y^{\frac{1}{3}}}$, $x = 3$ va $x$ o'qi bilan chegaralangan mintaqadir. Yechim
$ \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
$\iint\limits_{D}{{{{\bf{e}}^{\,{y^{\,4}}}}\,dA}}$ ni baholang, bu erda $D$ quyida ko'rsatilgan mintaqadir.
$The region $D$ is shaded in. It is the region between the curve $x = {y^3}$ and the y-axis, running from $y = - 1$ at the bottom up to $y = 1$ at the top. Above the x-axis the region lies to the right of the y-axis and to the left of the curve, widening out to the segment from $\left( {0,1} \right)$ to $\left( {1,1} \right)$ at the top. Below the x-axis it is the mirror image through the origin, lying to the left of the y-axis and widening out to the segment from $\left( { - 1, - 1} \right)$ to $\left( {0, - 1} \right)$ at the bottom. The two halves meet at the origin.$
Yechim
$ \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.
1. Avval $x$, keyin esa $y$ ga nisbatan integratsiya qiling.
2. Avval $y$, keyin esa $x$ ga nisbatan integratsiya qiling.
Yechim

10 va 11-masalalar uchun avvalo integrallash tartibini teskari qilib, berilgan integralni baholang.

$ \displaystyle \int_{0}^{3}{{\int_{{2x}}^{6}{{\sqrt {{y^2} + 2} \,dy}}\,dx}}$ Yechim
$ \displaystyle \int_{0}^{1}{{\int_{{ - \sqrt y }}^{{{y^{\,2}}}}{{6x - y\,dx}}\,dy}}$ Yechim
$y = 1 - {x^2}$ va $y = {x^2} - 3$ bilan chegaralangan mintaqa maydonini aniqlash uchun ikki karrali integraldan foydalaning. Yechim
$xy$-tekisligi va $f\left( {x,y} \right) = 2 + \cos \left( {{x^2}} \right)$ orasidagi va $\left( {0,0} \right)$, $\left( {6,0} \right)$ va $\left( {6,2} \right)$ uchlari bo‘lgan uchburchak ustida joylashgan hudud hajmini aniqlash uchun ikki karrali integraldan foydalaning. Yechim
$z = 6 - 5{x^2}$ va $y = 2x$, $y = 2$, $x = 0$ va $xy$ tekisliklari bilan chegaralangan mintaqa hajmini aniqlash uchun ikki karrali integraldan foydalaning. Yechim
Ikki tsilindr ${x^2} + {y^2} = 4$ va ${x^2} + {z^2} = 4$ kesishmasidan hosil bo'lgan hudud hajmini aniqlash uchun ikki karrali integraldan foydalaning. Yechim