Дополнительная задача 8 Глава 4 ГДЗ Мордкович Семенов 11 класс (Алгебра)

Задача

Рассмотрим вариант решения задания из учебника Мордкович, Семенов, Александрова 11 класс, Просвещение:

Подробный ответ

а) $$f(x)=\begin{cases}x+1,&-1\le x\le 0,\\1-x,&0\le x\le 1.\end{cases}$$
$$V=\pi\int_{-1}^{0}(x+1)^2\,dx+\pi\int_{0}^{1}(1-x)^2\,dx$$
$$=\frac{\pi}{3}(x+1)^3\Big|_{-1}^{0}-\frac{\pi}{3}(1-x)^3\Big|_{0}^{1}$$
$$=\frac{\pi}{3}\cdot 1^3-\frac{\pi}{3}\cdot 0^3-\frac{\pi}{3}\cdot 0^3+\frac{\pi}{3}\cdot 1^3=\frac{2\pi}{3}.$$
б) $$f(x)=\begin{cases}x+1,&-1\le x\le 0,\\1,&0\le x\le 1.\end{cases}$$
$$V=\pi\int_{-1}^{0}(x+1)^2\,dx+\pi\int_{0}^{1}1\,dx$$
$$=\frac{\pi}{3}(x+1)^3\Big|_{-1}^{0}+\pi x\Big|_{0}^{1}$$
$$=\frac{\pi}{3}\cdot 1^3-\frac{\pi}{3}\cdot 0^3+\pi=\frac{4\pi}{3}.$$
в) $$f(x)=\begin{cases}1,&-1\le x\le 0,\\1-x,&0\le x\le 1.\end{cases}$$
$$V=\pi\int_{-1}^{0}1\,dx+\pi\int_{0}^{1}(1-x)^2\,dx$$
$$=\pi x\Big|_{-1}^{0}-\frac{\pi}{3}(1-x)^3\Big|_{0}^{1}$$
$$=\pi(0-(-1))+\frac{\pi}{3}\cdot 1^3=\frac{4\pi}{3}.$$
г) $$f(x)=\begin{cases}x+1,&-1\le x\le 0{,}5,\\3-3x,&0{,}5\le x\le 1.\end{cases}$$
$$V=\pi\int_{-1}^{0{,}5}(x+1)^2\,dx+\pi\int_{0{,}5}^{1}(3-3x)^2\,dx$$
$$=\frac{\pi}{3}(x+1)^3\Big|_{-1}^{0{,}5}+9\pi\int_{0{,}5}^{1}(1-x)^2\,dx$$
$$=\frac{\pi}{3}\left(\frac{3}{2}\right)^3+3\pi(1-x)^3\Big|_{0{,}5}^{1}$$
$$=\frac{9\pi}{8}+\frac{3\pi}{8}=\frac{12\pi}{8}=\frac{3\pi}{2}.$$
д) $$f(x)=\begin{cases}x+1,&-1\le x\le 0,\\\sqrt{1-x^2},&0\le x\le 1.\end{cases}$$
$$V=\pi\int_{-1}^{0}(x+1)^2\,dx+\pi\int_{0}^{1}(1-x^2)\,dx$$
$$=\frac{\pi}{3}(x+1)^3\Big|_{-1}^{0}+\pi\left(x-\frac{x^3}{3}\right)\Big|_{0}^{1}$$
$$=\frac{\pi}{3}+\pi\left(1-\frac{1}{3}\right)=\pi.$$
е) $$f(x)=\begin{cases}\sqrt{1+x},&-1\le x\le 0,\\\sqrt{1-x^2},&0\le x\le 1.\end{cases}$$
$$V=\pi\int_{-1}^{0}(1+x)\,dx+\pi\int_{0}^{1}(1-x^2)\,dx$$
$$=\pi\left(x+\frac{x^2}{2}\right)\Big|_{-1}^{0}+\pi\left(x-\frac{x^3}{3}\right)\Big|_{0}^{1}$$
$$=\pi\left(0-\left(-1+\frac12\right)\right)+\pi\left(1-\frac13\right)=\frac{7\pi}{6}.$$