SurfaceArea - 2-masala yechimi | MathModel.uz

15.9-bo'lim : Er yuzasi

2. $z = 13 - 4{x^2} - 4{y^2}$ ning $x \le 0$ va $y \le 0$ bilan $z = 1$ ustidagi 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 the elliptic paraboloid $z = 13 - 4{x^2} - 4{y^2}$ that lies above $z = 1$ and over the third quadrant of the xy-plane, so it is a quarter of a dome. The bluish “walls” underneath it are only a frame of reference and are not part of the surface.$

A three dimensional sketch of the same quarter dome, this time drawn with a set of “box” axes to make the surface easier to visualize. The orange surface is the paraboloid and the bluish walls are only a frame of reference.

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

Biz kutayotgan sirt elliptik paraboloidning (eskizdagi apelsin yuzasi) $z = 1$ dan yuqori va 3-banddagi qismidir.^rd $xy$ tekisligi kvadranti. Ko'k rangli "devorlar" shunchaki sirtni ko'rishga yordam beradigan mos yozuvlar ramkasini ta'minlash uchun mavjud va aslida bizni qiziqtiradigan sirtning bir qismi emas.

2-qadam

Sirt maydoni uchun integral,

\[S = \iint\limits_{D}{{\sqrt {{{\left[ { - 8x} \right]}^2} + {{\left[ { - 8y} \right]}^2} + 1} \,dA}} = \iint\limits_{D}{{\sqrt {64{x^2} + 64{y^2} + 1} \,dA}}\]

3-qadam

Endi, oxirgi bosqichda aytilganidek, $D$ $xy$ tekisligida bo'lishi kerak va u aylana mintaqa bo'lishi (umid qilamanki) juda aniq bo'lishi kerak. Agar siz to'g'ridan-to'g'ri yuqoridan sirtga qarasangiz, quyidagi mintaqani ko'rasiz.

$A sketch of the region $D$ as seen looking straight down at the surface from above. It is the quarter of the disk ${x^2} + {y^2} = 3$ that lies in the third quadrant, with $x \le 0$ and $y \le 0$, and it is shaded orange with the circular arc drawn in red.$

Doira tenglamasini aniqlash uchun biz faqat $z = 1$ ga teng elliptik paraboloid tenglamasini o'rnatishimiz kerak:

\[1 = 13 - 4{x^2} - 4{y^2}\hspace{0.25in} \to \hspace{0.25in}4{x^2} + 4{y^2} = 12\hspace{0.25in} \to \hspace{0.25in}{x^2} + {y^2} = 3 \]

Demak, bu $\sqrt 3 $ radiusi doirasining 3 dagi qismi^rd kvadrant.

Shu nuqtada, biz integralni qutb koordinatalari nuqtai nazaridan baholashimiz kerakligi ham aniq bo'lishi kerak. Qutbli koordinatalar nuqtai nazaridan $D$ uchun chegaralar:

\[\begin{array}{c} \displaystyle \pi \le \theta \le \frac{{3\pi }}{2}\\ 0 \le r \le \sqrt 3 \end{array}\]

4-qadam

Endi integralni qutb koordinatalariga aylantiramiz.

\[S = \iint\limits_{D}{{\sqrt {64{x^2} + 64{y^2} + 1} \,dA}} = \int_{\pi }^{{\frac{3}{2}\pi }}{{\int_{0}^{{\sqrt 3 }}{{r\sqrt {64{r^2} + 1} \,dr}}\,d\theta }}\]

$dA$ ni qutb koordinatalariga aylantirish uchun qo'shimcha $r$ ni olishni unutmang. Agar sizga integrallarni qutbli koordinatalarga aylantirish bo'yicha yangilanish kerak bo'lsa, siz borishingiz kerak orqaga va ushbu bo'limdagi ba'zi muammolarni hal qiling.

5-qadam

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

\[\begin{align*}S & = \int_{\pi }^{{\frac{3}{2}\pi }}{{\int_{0}^{{\sqrt 3 }}{{r\sqrt {64{r^2} + 1} \,dr}}\,d\theta }}\\ & = \int_{\pi }^{{\frac{3}{2}\pi }}{{\left. {\frac{1}{{192}}{{\left( {64{r^2} + 1} \right)}^{\frac{3}{2}}}} \right|_0^{\sqrt 3 }\,d\theta }}\\ & = \int_{\pi }^{{\frac{3}{2}\pi }}{{\frac{1}{{192}}\left( {{{193}^{\frac{3}{2}}} - 1} \right)\,d\theta }}\\ & = \left. {\frac{1}{{192}}\left( {{{193}^{\frac{3}{2}}} - 1} \right)\theta } \right|_\pi ^{\frac{3}{2}\pi } = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{\pi }{{384}}\left( {{{193}^{\frac{3}{2}}} - 1} \right) = 21.9277}}\end{align*}\]