Предмет: Геометрия, автор: 4emzan9t

Параллелограмм ABCD вращается вокруг прямой, проходящей через вершину А параллельно меньшей диагонали BD. Найдите объем тела вращения, если в данном параллелограмме угол А = 60 градусов, большая сторона = 6 дм, а меньшая диагональ перпендикулярна стороне.

Ответы

Автор ответа: tyrin
0
Задача: найти объем тела вращения.
1)Найти объем внутреннего конуса.
2)Найти объем внешнего усеченного конуса.
Решение:
sin30=D1A1 / 6
D1A1=6*sin30=3
D1A1-радиус внутреннего конуса
AA1=√(6²-3²)=3√3=H
AA1-высота внешнего усеченного конуса
Vвнутр=π3²*3√3 / 3 = 9√3π
Vвнеш=π/3*3√3(6²+6*3+3²)=63√3π
Vтела=Vвнеш-Vвнутр=63√3π-9√3π=54√3π
Ответ:54√3π
Автор ответа: 4emzan9t
0
хм. Почему то думал, что через интегралы надо. Спасибо, дам лучший.
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