Категория:Задачник Кузнецова Интегралы Задача 20

В этой категории варианты Задачи 20, раздела "Интегралы" в задачнике Кузнецова Л.А.

Варианты[править]

Вычислить объемы тел, ограниченных поверхностями.

Задача	Условие	Задача	Условие
20-1	${\displaystyle {\frac {x^{2}}{9}}+y^{2}=1,\;z=y,\;z=0\;(y\geq 0)}$	20-2	${\displaystyle z=x^{2}+4y^{2},\;z=2}$
20-3	${\displaystyle {\frac {x^{2}}{9}}+{\frac {y^{2}}{4}}-z^{2}=1,\;z=0,\;z=3}$	20-4	${\displaystyle {\frac {x^{2}}{9}}+{\frac {y^{2}}{4}}-{\frac {z^{2}}{36}}=-1,\;z=12}$
20-5	${\displaystyle {\frac {x^{2}}{16}}+{\frac {y^{2}}{9}}+{\frac {z^{2}}{4}}=1,\;z=1,\;z=0}$	20-6	${\displaystyle x^{2}+y^{2}=9,\;z=y,\;z=0\;(y\geq 0)}$
20-7	${\displaystyle z=x^{2}+9y^{2},\;z=3}$	20-8	${\displaystyle {\frac {x^{2}}{4}}+y^{2}-z^{2}=1,\;z=0,\;z=3}$
20-9	${\displaystyle {\frac {x^{2}}{9}}+{\frac {y^{2}}{16}}-{\frac {z^{2}}{64}}=-1,\;z=16}$	20-10	${\displaystyle {\frac {x^{2}}{16}}+{\frac {y^{2}}{9}}+{\frac {z^{2}}{16}}=1,\;z=2,\;z=0}$
20-11	${\displaystyle {\frac {x^{2}}{3}}+{\frac {y^{2}}{4}}=1,\;z=y{\sqrt {3}},\;z=0\;(y\geq 0)}$	20-12	${\displaystyle z=2x^{2}+8y^{2},\;z=4}$
20-13	${\displaystyle {\frac {x^{2}}{81}}+{\frac {y^{2}}{25}}-z^{2}=1,\;z=0,\;z=2}$	20-14	${\displaystyle {\frac {x^{2}}{4}}+{\frac {y^{2}}{9}}-{\frac {z^{2}}{36}}=-1,\;z=12}$
20-15	${\displaystyle {\frac {x^{2}}{16}}+{\frac {y^{2}}{9}}+{\frac {z^{2}}{36}}=1,\;z=3,\;z=0}$	20-16	${\displaystyle {\frac {x^{2}}{3}}+{\frac {y^{2}}{16}}=1,\;z=y{\sqrt {3}},\;z=0\;(y\geq 0)}$
20-17	${\displaystyle z=x^{2}+5y^{2},\;z=5}$	20-18	${\displaystyle {\frac {x^{2}}{9}}+{\frac {y^{2}}{4}}-z^{2}=1,\;z=0,\;z=4}$
20-19	${\displaystyle {\frac {x^{2}}{9}}+{\frac {y^{2}}{25}}-{\frac {z^{2}}{100}}=-1,\;z=20}$	20-20	${\displaystyle {\frac {x^{2}}{16}}+{\frac {y^{2}}{9}}+{\frac {z^{2}}{64}}=1,\;z=4,\;z=0}$
20-21	${\displaystyle {\frac {x^{2}}{27}}+{\frac {y^{2}}{25}}=1,\;z={\frac {y}{\sqrt {3}}},\;z=0\;(y\geq 0)}$	20-22	${\displaystyle z=4x^{2}+9y^{2},\;z=6}$
20-23	${\displaystyle x^{2}+{\frac {y^{2}}{4}}-z^{2}=1,\;z=0,\;z=3}$	20-24	${\displaystyle {\frac {x^{2}}{25}}+{\frac {y^{2}}{9}}-{\frac {z^{2}}{100}}=-1,\;z=20}$
20-25	${\displaystyle {\frac {x^{2}}{16}}+{\frac {y^{2}}{9}}+{\frac {z^{2}}{100}}=1,\;z=5,\;z=0}$	20-26	${\displaystyle {\frac {x^{2}}{27}}+y^{2}=1,\;z={\frac {y}{\sqrt {3}}},\;z=0\;(y\geq 0)}$
20-27	${\displaystyle z=2x^{2}+18y^{2},\;z=6}$	20-28	${\displaystyle {\frac {x^{2}}{25}}+{\frac {y^{2}}{9}}-z^{2}=1,\;z=0,\;z=2}$
20-29	${\displaystyle {\frac {x^{2}}{16}}+{\frac {y^{2}}{9}}-{\frac {z^{2}}{64}}=-1,\;z=16}$	20-30	${\displaystyle {\frac {x^{2}}{16}}+{\frac {y^{2}}{9}}+{\frac {z^{2}}{144}}=1,\;z=6,\;z=0}$
20-31	${\displaystyle {\frac {x^{2}}{16}}+{\frac {y^{2}}{9}}+{\frac {z^{2}}{196}}=1,\;z=7,\;z=0}$