Question

Помогите решить любой один пример из этого задания.

Answer 1

$2.5.\; a)\; \; x^2+y^2=ax\; \; \to \; \; (x-\frac{a}{2})^2+y^2=(\frac{a}{2})^2\\\\okryznost\; ,\; \; R=\frac{a}{2}\; ,\; \; centr\; (\frac{a}{2},0)\\\\x=\rho \, cos\phi \; ,\; \; y=\rho \, sin\phi \; ,\; \; dx\, dy=\rho \, d\rho d\phi \; ,\; \; x^2+y^2=\rho ^2\\\\\rho ^2=a\, \rho \, cos\phi \; ,\; \; \rho =a\, cos\phi \\\\b)\; \; x^2+y^2=2ax\; \; \to \; \; (x-a)^2+y^2=a^2\\\\okryznost\; ,\; \; R=a\; ,\; \; centr\; (a,0)\\\\\rho ^2=2a\rho \, cos\phi \; ,\; \; \rho =2a\, cos\phi \\\\c)\; \; y=0\; ,\; y\ \textless \ 0\;\; \to \\\\\rho \, sin\phi \leq 0\; ,\; \; sin\phi \leq 0\; ,\; \pi \leq \phi \leq 2\pi$

$d)\; \; Peresechenie:\; \; a\, cos\phi =2acos\phi \; ,\; \; a\, cos\phi =0\; ,\; cos\phi =0\\\\\phi =\frac{\pi}{2}+2\pi n,\, n\in Z\; ;\; \; \; \frac{3\pi }{2}\in [\pi ,2\pi ]$

$e)\; \; \iint\limits_{D}\, (x^2+y^2)\, dx\, dy=\int\limits^{2\pi }_{3\pi /2}\, d\phi \int\limits^{acos\phi }_{2acos\phi }\, \underbrace {\rho ^2\cdot \rho }_{\rho ^3}\, d\rho =\\\\=\int\limits^{2\pi }_{3\pi /2}\, d\phi \Big ( \frac{\rho ^4}{4}\Big |_{acos\phi }^{2acos\phi }\Big )=\frac{1}{4}\int\limits^{2\pi }_{3\pi /2}\, (16a^4cos^4\phi -a^4cos^4\phi )d\phi =\\\\= \frac{15a^4}{4}\int\limits^{2\pi }_{3\pi /2}\, (cos^2\phi )^2d\phi =[\, (cos^2\phi )^2=\Big (\frac{1+cos2\phi }{2}\Big )^2=\frac{1}{4}(1+2cos2\phi +cos^22\phi )=$

$= \frac{1}{4}(1+2cos2\phi +\frac{1+cos4\phi }{2})=\frac{1}{4}(\frac{3}{2}+2cos2\phi +\frac{1}{2}cos4\phi )\, ]=\\\\=\frac{15a^4}{16}\int\limits^{2\pi }_ {3\pi /2}(\frac{3}{2}+2cos2\phi +\frac{1}{2}cos4\phi )d\phi =\\\\=\frac{15a^4}{16}(\frac{3}{2}\phi +sin2\phi +\frac{1}{8}sin4\phi )\Big |_{3\pi /2}^{2\pi }=\frac{15a^4}{16}(3\pi -\frac{9\pi}{4})=\frac{45\, \pi \, a^4}{64}$