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Примена одређеног интеграла 5

Одређени интеграл. Примена одређеног интеграла на израчунавање површине. Сложенији примери.

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Пр.5)   Израчунати површину између кривих ${x^2} + {y^2} = 9$ и ${x^2} - 8y = 0$.

${\left( {x - a} \right)^2} + {\left( {y - b} \right)^2} = {r^2}$

$C\left( {a,b} \right)$

${x^2} + {y^2} = 9$

$C\left( {0;0} \right)$

${y^2} = 9 - {x^2}$

$y = \sqrt {9 - {x^2}} $

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${x^2} = 8$

$y = \frac{1}{8}{x^2}$

483 png

\[\left\{ \begin{gathered}
{x^2} + {y^2} = 9 \hfill \\
{x^2} = 8y \hfill \\
\end{gathered} \right.\left\{ \begin{gathered}
{y_1} = - 9;\underline {{y_2} = 1} \hfill \\
{x_1} = 2\sqrt 2 ;{x_2} = - 2\sqrt 2 \hfill \\
\end{gathered} \right.\]

\[\begin{array}{*{20}{c}}
\begin{gathered}
{y_1} = - 9 \hfill \\
x_1^2 = 8 \cdot \left( { - 9} \right) \hfill \\
{x_1} = - 72 \hfill \\
\end{gathered} &{}&\begin{gathered}
{y_2} = 1 \hfill \\
x_2^2 = 8 \hfill \\
{x_1} = 2\sqrt 2 ;{x_2} = - 2\sqrt 2 ; \hfill \\
\end{gathered}
\end{array}\]

$A\left( {2\sqrt 2 ;1} \right),B\left( { - 2\sqrt 2 ;1} \right)$

484 png

$P = 2{P_1} = 2\left( {\int\limits_0^{2\sqrt 2 } {\sqrt {9 - {x^2}} dx}  - \int\limits_0^{2\sqrt 2 } {\frac{1}{8}{x^2}dx} } \right) = $

$\underbrace {\int {\sqrt {9 - {x^2}} dx} }_I = $

\[\begin{array}{*{20}{c}}
\begin{gathered}
{\left. {u = \sqrt {9 - {x^2}} } \right|^\prime } \hfill \\
du = {\left( {\sqrt {9 - {x^2}} } \right)^\prime }dx \hfill \\
du = {\left( {{{\left( {9 - {x^2}} \right)}^{\frac{1}{2}}}} \right)^\prime }dx \hfill \\
du = \frac{1}{2}{\left( {9 - {x^2}} \right)^{ - \frac{1}{2}}}{\left( {9 - {x^2}} \right)^\prime }dx \hfill \\
du = \frac{1}{{2\sqrt {9 - {x^2}} }}\left( { - 2x} \right)dx \hfill \\
du = \frac{{ - x}}{{\sqrt {9 - {x^2}} }}dx \hfill \\
\end{gathered} &{}&\begin{gathered}
dv = dx \hfill \\
v = x \hfill \\
\hfill \\
\hfill \\
\hfill \\
\hfill \\
\hfill \\
\hfill \\
\hfill \\
\hfill \\
\hfill \\
\hfill \\
\end{gathered}
\end{array}\]

$ = x\sqrt {9 - {x^2}}  - \int {x\frac{{ - xdx}}{{\sqrt {9 - {x^2}} }}}  =$

$= x\sqrt {9 - {x^2}}  - \int {\frac{{9 - {x^2} - 9}}{{\sqrt {9 - {x^2}} }}} dx =$

$= x\sqrt {9 - {x^2}}  - \left( {\int {\frac{{{{\sqrt {9 - {x^2}} }^2}}}{{\sqrt {9 - {x^2}} }}} dx - \int {\frac{9}{{\sqrt {9 - {x^2}} }}} dx} \right) = $

$ = x\sqrt {9 - {x^2}}  - \int {\sqrt {9 - {x^2}} } dx + 9\int {\frac{{dx}}{{\sqrt {9 - {x^2}} }}}  =$

$= x\sqrt {9 - {x^2}}  - \int {\sqrt {9 - {x^2}} } dx + 9\int {\frac{{dx}}{{3\sqrt {1 - {{\left( {\frac{x}{3}} \right)}^2}} }}} $

 

$\frac{x}{3} = t|'$

$\frac{1}{3}dx = dt$

$dx = 3dt$

$ = x\sqrt {9 - {x^2}}  - \int {\sqrt {9 - {x^2}} } dx + 9\int {\frac{{3dt}}{{3\sqrt {1 - {t^2}} }}}  =$

$= x\sqrt {9 - {x^2}}  - \underbrace {\int {\sqrt {9 - {x^2}} } dx}_I + 9\arcsin \frac{x}{3}$

 

$2I = x\sqrt {9 - {x^2}}  + 9\arcsin \frac{x}{3}$

$I = \frac{{x\sqrt {9 - {x^2}} }}{2} + \frac{9}{2}\arcsin \frac{x}{3}$

 

$P = 2{P_1} = 2\left( {\int\limits_0^{2\sqrt 2 } {\sqrt {9 - {x^2}} dx}  - \int\limits_0^{2\sqrt 2 } {\frac{1}{8}{x^2}dx} } \right) =$

$= 2\left( {\left. {\left( {\frac{{x\sqrt {9 - {x^2}} }}{2} + \frac{9}{2}\arcsin \frac{x}{3}} \right)} \right|_0^{2\sqrt 2 } - \left. {\frac{1}{8} \cdot \frac{{{x^3}}}{3}} \right|_0^{2\sqrt 2 }} \right) = $

$ = 2\left( {\frac{{2\sqrt 2 \sqrt {9 - {{\left( {2\sqrt 2 } \right)}^2}} }}{2} + \frac{9}{2}\arcsin \frac{{2\sqrt 2 }}{3} - 0 - \frac{9}{2}\arcsin 0 - \left( {\frac{1}{8} \cdot \frac{{{{\left( {2\sqrt 2 } \right)}^3}}}{3} - 0} \right)} \right) = $

$ = 2\left( {\sqrt 2  + \frac{9}{2}\arcsin \frac{{2\sqrt 2 }}{3} - \frac{{8 \cdot 2\sqrt 2 }}{{24}}} \right) = 2\left( {\frac{{\sqrt 2 }}{3} + \frac{9}{2}\arcsin \frac{{2\sqrt 2 }}{3}} \right) =$

$= \frac{{2\sqrt 2 }}{3} + 9\arcsin \frac{{2\sqrt 2 }}{3}$