Question

Решите пожалуйста!!!!!

Answer 1

$\displaystyle\\1) S=\int\limits^3_{-3} {9-x^2} \, dx =(9x-\frac{x^3}{3})\mid^3_{-3}=27-\frac{27}{3}-(-27+\frac{27}{3})=27-9+27-9=\\\\\\ =36$

$\displaystyle\\2) V=V_1+V_2=\pi\int\limits^0_{-2} {(x^3-4x)^2} \, dx+\pi\int\limits^2_{0} {(x^3-4x)^2} \, dx \\\\\\\pi\int\limits^0_{-2} {(x^3-4x)^2} \, dx=\pi\int\limits^0_{-2} {x^6-8x^4+16x^2} \, dx=\pi(\frac{x^7}{7}-\frac{8x^5}{5}+\frac{16x^3}{3})\mid^0_{-2}=\\\\\\ =\pi(\frac{0}{7}-\frac{8*0}{5}+\frac{16*0}{3}-(\frac{(-2)^7}{7}-\frac{8*(-2)^5}{5}+\frac{16*(-2)^3}{3}))= \frac{1024}{105}\pi\\\\\\ V_1=V_2\Rightarrow V_2=\frac{1024}{105}\pi$

$\displaystyle\\V=V_1+V_2=\frac{1024}{105}\pi+\frac{1024}{105}\pi=\frac{2048\pi}{105}$