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Piotrek1919

Rejestracja: 20 Feb 2017
Offline Ostatnio: Feb 20 2017 11:59
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#126588 Oblicz pole powierzchni płata

Napisane przez bb314 w 08.02.2016 - 20:56

Odpowiedz sobie sama :)

 

x^2=y^2+z^2 \gr\ \Rightarrow\  z=\sq{x^2-y^2}\gr\ \Rightarrow\  \{ \fr{\partial z}{\partial x}=\fr{x}{\sq{x^2-y^2}}\\ \fr{\partial z}{\partial y}=-\fr{y}{\sq{x^2-y^2}}
 
\bl\fr18P=\iint_D\sq{1+\(\fr{\partial z}{\partial x}\)^2+\(\fr{\partial z}{\partial y}\)^2}dydx=\iint_D\sq{1+\fr{x^2}{x^2-y^2}+\fr{y^2}{x^2-y^2}}dydx=\iint_D\fr{\sq2x}{\sq{x^2-y^2}}dydx=
\ \ \ \ \ =\[x=r\cos\varphi\\y=r\sin\varphi\\dxdy=rdrd\varphi\]=\sq2\int_0^{\fr\p4}\int_0^R\fr{r\cos\varphi}{\sq{r^2\cos^2\varphi-r^2\sin^2\varphi}}rdrd\varphi=\sq2\int_0^{\fr\p4}\int_0^R\fr{r}{\sq{1-tg^2\varphi}}drd\varphi=
 
\ \ \ \ \ =\sq2\cd\fr{R^2}{2}\int_0^{\fr\p4}\fr{1}{\sq{1-tg^2\varphi}}d\varphi=\[t^2=1-tg^2\varphi\\2tdt=-2tg\varphi(1+tg^2\varphi)d\varphi\\tg\varphi=\sq{1-t^2}\\\varphi=0\to t=1\\\varphi=\fr\p4\to t=0\]=\sq2\cd\fr{R^2}{2}\int_0^1\fr{dt}{\sq{1-t^2}(2-t^2)}=
 
\ \ \ \ \ =\fr{\sq2}{2}R^2\|\ \\\fr{\sq2}{2}arctg\fr{t}{\sq{2-2t^2}}\\\ \|_0^1=\fr{\sq2}{2}R^2\cd\fr{\sq2}{2}\cd\fr\p2\bl=\fr\p4R^2
 
\fr18P=\fr\p4R^2\gr\ \Rightarrow\ \re P=2\p R^2
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ :shifty: \ :shifty:
 

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