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Пр.7)   Израчунати запремину тела које настаје обртањем површи

            између кривих ${x^2} + {y^2} = 16$ и ${y^2} = 4x + 4$ око $x$-осе.

$C\left( {0;0} \right),r = 4$

$y = 0;4x + 4 = 0$

\[\begin{array}{*{20}{c}}
{{x^2} + {y^2} = 16}&{}&{{y^2} = 4x + 4} \\
{{y^2} = 16 - {x^2}}&{}&{y = \sqrt {4x + 4} } \\
{y = \sqrt {16 - {x^2}} }&{}&{}
\end{array}\]

${x^2} + {y^2} = 16$

$\underline {{y^2} = 4x + 4} $

${x^2} + 4x + 4 - 16 = 0$

${x^2} + 4x - 12 = 0$

${x_{1,2}} = \frac{{ - 4 \pm \sqrt {16 + 48} }}{2} = \frac{{ - 4 \pm 8}}{2}$

\[\begin{array}{*{20}{c}}
{{x_1} = - 6}&{}&{{x_2} = 2} \\
{{y^2} = 4\left( { - 6} \right) + 4}&{}&{{y^2} = 4 \cdot 2 + 4} \\
{{y^2} = - 20}&{}&{{y^2} = 12} \\
\bot &{}&{\begin{array}{*{20}{c}}
{y = 2\sqrt 3 }&{y = - 2\sqrt 3 }
\end{array}}
\end{array}\]

$A\left( {2;2\sqrt 3 } \right),B\left( {2; - 2\sqrt 3 } \right)$

$V = \pi \left( {\int\limits_{ - 1}^2 {{{\left( {\sqrt {4x + 4} } \right)}^2}dx}  + \int\limits_2^4 {{{\left( {\sqrt {16 - {x^2}} } \right)}^2}dx} } \right) =$

$= \pi \left( {\left. {\left( {4\frac{{{x^2}}}{2} + 4x} \right)} \right|_{ - 1}^2 + \left. {\left( {16x - \frac{{{x^3}}}{3}} \right)} \right|_2^4} \right) = $

$ = \pi \left( {2 \cdot {2^2} + 4 \cdot 2 - \left( {2 \cdot {{\left( { - 1} \right)}^2} + 4 \cdot \left( { - 1} \right)} \right) + 16 \cdot 4 - \frac{{{4^3}}}{3} - \left( {16 \cdot 2 - \frac{{{2^3}}}{3}} \right)} \right) = $

$ = \pi \left( {8 + 8 - 2 + 4 + 64 - \frac{{64}}{3} - 32 + \frac{8}{3}} \right) = \pi \left( {50 - \frac{{56}}{3}} \right) = \frac{{94}}{3}\pi $