SurfaceArea - 3-masala yechimi | MathModel.uz

15.9-bo'lim : Er yuzasi

3. $\displaystyle z = 3 + 2y + \frac{1}{4}{x^4}$ ning $y = {x^5}$, $x = 1$ va $x$ o'qi bilan chegaralangan $xy$ tekisligidagi mintaqadan yuqorida joylashgan qismining sirt maydonini aniqlang.

Yechim

Xo'sh, keling, biz nima bilan shug'ullanayotganimizni his qilishimiz uchun sirtning qisqacha eskizidan boshlaylik.

$A three dimensional sketch of the surface drawn with the traditional axes. The orange surface is the portion of $z = 3 + 2y + \frac{1}{4}{x^4}$ above the xy-plane, which sags down towards the middle and curls up at the corners. The greenish region beneath it in the xy-plane is the region $D$, bounded by the red curve $y = {x^5}$, the line $x = 1$ and the x-axis.$

$A three dimensional sketch of the same surface $z = 3 + 2y + \frac{1}{4}{x^4}$ and the region $D$ beneath it, this time drawn with a set of “box” axes to make the surface easier to visualize.$

Biz "an'anaviy" o'qlar to'plami bilan bir qatorda "quti" o'qlari to'plami bilan eskizlarni berdik, bu sirtni ingl.

Biz izlayotgan sirt $xy$ tekisligi ustida joylashgan to'q sariq qism va $xy$ tekisligidagi yashil rangli mintaqa biz sirtni grafigini chizayotgan hududdir. i.e. bu biz integralda ishlatadigan $D$ mintaqasi.

2-qadam

Sirt maydoni uchun integral,

\[S = \iint\limits_{D}{{\sqrt {{{\left[ {{x^3}} \right]}^2} + {{\left[ 2 \right]}^2} + 1} \,dA}} = \iint\limits_{D}{{\sqrt {{x^6} + 5} \,dA}}\]

3-qadam

Endi, 1-bosqichda aytib o'tilganidek, $D$ hududi sirt eskizlarida ko'rsatilgan. Mana, to'liqlik uchun $D$ ning 2D eskizi.

$A two dimensional sketch of the region $D$. It is bounded above by the curve $y = {x^5}$, on the right by the vertical line $x = 1$ and below by the x-axis. The curve hugs the x-axis and then climbs steeply near $x = 1$ to reach $\left( {1,1} \right)$, and the thin region between it and the x-axis is shaded green.$

Ushbu mintaqa uchun limitlar:

\[\begin{array}{c}0 \le x \le 1\\ 0 \le y \le {x^5}\end{array}\]

Shuni ham yodda tutingki, integrand bizdan integratsiyani shu tartibda bajarishni talab qiladi.

4-qadam

3-bosqichdagi chegaralar bilan integral shunday bo'ladi:

\[S = \iint\limits_{D}{{\sqrt {{x^6} + 5} \,dA}} = \int_{0}^{1}{{\int_{0}^{{{x^5}}}{{\sqrt {{x^6} + 5} \,dy}}\,dx}}\]

5-qadam

OK, biz qilishimiz kerak bo'lgan narsa - bu integralni baholash.

\[\begin{align*}S & = \int_{0}^{1}{{\int_{0}^{{{x^5}}}{{\sqrt {{x^6} + 5} \,dy}}\,dx}}\\ & = \int_{0}^{1}{{\left. {\left( {y\sqrt {{x^6} + 5} } \right)} \right|_0^{{x^5}}\,dx}}\\ & = \int_{0}^{1}{{{x^5}\sqrt {{x^6} + 5} \,dx}}\\ & = \left. {\frac{1}{9}{{\left( {{x^6} + 5} \right)}^{\frac{3}{2}}}} \right|_0^1 = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{1}{9}\left( {{6^{\frac{3}{2}}} - {5^{\frac{3}{2}}}} \right) = 0.3907}}\end{align*}\]