Question

криволинейные интегралы и скалярные поля

Answer 1

Ответ:

1. Криволинейный интеграл в точке M равен $\boldsymbol{\dfrac{\sqrt{2} }{6}}$

$\boldsymbol{\boxed{ \int\limits_\Gamma {f(M)} \, dl = \frac{\sqrt{2} }{6} }}$

2. Производная по направлению скалярного поля в точке M равна 3

$\boldsymbol{\boxed{{\dfrac{\partial U}{\partial l} \bigg (M \bigg) = 3}}}$

Пошаговое объяснение:

1.

$f(M) = xy$

$\Gamma:$

$x + y = 1 \Longrightarrow y = 1-x$

$0 \leq x\leq 1$

$\displaystyle \int\limits_\Gamma {f(M)} \, dl = \int\limits^1_0 {x(1-x)\sqrt{1+ ((1-x)')^{2}} } \, dx = \int\limits^1_0 {(x-x^{2} )\sqrt{1+ (-1)^{2}} } \, dx$

$\displaystyle = \int\limits^1_0 {(x-x^{2} )\sqrt{1+1} } \, dx = \sqrt{2} \int\limits^1_0 {(x-x^{2} ) } \, dx = \sqrt{2} \bigg(\frac{x^{2} }{2} - \frac{x^{3}}{3} \bigg) \bigg|_{0}^{1} =$

$\displaystyle =\sqrt{2} \bigg( \bigg(\frac{1^{2} }{2} - \frac{1^{3}}{3} \bigg) - \bigg(\frac{0^{2} }{2} - \frac{0^{3}}{3} \bigg) \bigg) = \sqrt{2} \bigg( \frac{1}{2} - \frac{1}{3} \bigg) = \sqrt{2} \bigg( \frac{3-2}{6} \bigg) = \frac{\sqrt{2} }{6}$

2.

$U(x;y;z) =\sqrt{(x^{2} + y^{2} + z^{2})^{3}}$

$\overrightarrow{l}(1;-1;1),M(1;1;1)$

Частные производные:

$\dfrac{\partial U}{\partial x} =\dfrac{\partial}{\partial x} \bigg(\sqrt{(x^{2} + y^{2} + z^{2})^{3}} \bigg) = \dfrac{\dfrac{\partial}{\partial x} \bigg((x^{2} + y^{2} + z^{2})^{3} \bigg)}{2\sqrt{(x^{2} + y^{2} + z^{2})^{3}}} =$

$= \dfrac{3(x^{2} + y^{2} + z^{2})^{2}\dfrac{\partial}{\partial x} \bigg((x^{2} + y^{2} + z^{2}) \bigg)}{2\sqrt{(x^{2} + y^{2} + z^{2})^{3}}} = \dfrac{6x(x^{2} + y^{2} + z^{2})^{2}}{2\sqrt{(x^{2} + y^{2} + z^{2})^{3}}} =$

$= \dfrac{3x(x^{2} + y^{2} + z^{2})^{2}}{(x^{2} + y^{2} + z^{2})\sqrt{(x^{2} + y^{2} + z^{2})}} = \dfrac{3x(x^{2} + y^{2} + z^{2})}{\sqrt{(x^{2} + y^{2} + z^{2})}}$

$\dfrac{\partial U}{\partial y} =\dfrac{\partial}{\partial y} \bigg(\sqrt{(x^{2} + y^{2} + z^{2})^{3}} \bigg) = \dfrac{\dfrac{\partial}{\partial y} \bigg((x^{2} + y^{2} + z^{2})^{3} \bigg)}{2\sqrt{(x^{2} + y^{2} + z^{2})^{3}}} =$

$= \dfrac{3(x^{2} + y^{2} + z^{2})^{2}\dfrac{\partial}{\partial y} \bigg((x^{2} + y^{2} + z^{2}) \bigg)}{2\sqrt{(x^{2} + y^{2} + z^{2})^{3}}} = \dfrac{6y(x^{2} + y^{2} + z^{2})^{2}}{2\sqrt{(x^{2} + y^{2} + z^{2})^{3}}} =$

$= \dfrac{3y(x^{2} + y^{2} + z^{2})^{2}}{(x^{2} + y^{2} + z^{2})\sqrt{(x^{2} + y^{2} + z^{2})}} = \dfrac{3y(x^{2} + y^{2} + z^{2})}{\sqrt{(x^{2} + y^{2} + z^{2})}}$

$\dfrac{\partial U}{\partial z} =\dfrac{\partial}{\partial z} \bigg(\sqrt{(x^{2} + y^{2} + z^{2})^{3}} \bigg) = \dfrac{\dfrac{\partial}{\partial z} \bigg((x^{2} + y^{2} + z^{2})^{3} \bigg)}{2\sqrt{(x^{2} + y^{2} + z^{2})^{3}}} =$

$= \dfrac{3(x^{2} + y^{2} + z^{2})^{2}\dfrac{\partial}{\partial z} \bigg((x^{2} + y^{2} + z^{2}) \bigg)}{2\sqrt{(x^{2} + y^{2} + z^{2})^{3}}} = \dfrac{6z(x^{2} + y^{2} + z^{2})^{2}}{2\sqrt{(x^{2} + y^{2} + z^{2})^{3}}} =$

$= \dfrac{3z(x^{2} + y^{2} + z^{2})^{2}}{(x^{2} + y^{2} + z^{2})\sqrt{(x^{2} + y^{2} + z^{2})}} = \dfrac{3z(x^{2} + y^{2} + z^{2})}{\sqrt{(x^{2} + y^{2} + z^{2})}}$

Модуль вектора $\overrightarrow{l}:$

$|\overrightarrow{l}| = \sqrt{x_{l}^{2} + y_{l}^{2}+z_{l}^{2}} = \sqrt{1^{2} + (-1)^{2}+1^{2}} = \sqrt{1+1+1} = \sqrt{3}$

Направляющие косинусы вектора $\overrightarrow{l}:$

$\cos \alpha = \dfrac{x_{l}}{|\overrightarrow{l}|} = \dfrac{1}{\sqrt{3} }$

$\cos \beta = \dfrac{y_{l}}{|\overrightarrow{l}|} = \dfrac{-1}{\sqrt{3} } = -\dfrac{1}{\sqrt{3} }$

$\cos \gamma = \dfrac{z_{l}}{|\overrightarrow{l}|} = \dfrac{1}{\sqrt{3} }$

Производная по направления вектора $\overrightarrow{l}$ в точке $M:$

$\dfrac{\partial U}{\partial l} \bigg (M \bigg) = \dfrac{\partial U}{\partial x} \cos \alpha +\dfrac{\partial U}{\partial y} \cos \beta + \dfrac{\partial U}{\partial z} \cos \gamma=$

$= \dfrac{3x(x^{2} + y^{2} + z^{2})}{\sqrt{3} \cdot\sqrt{(x^{2} + y^{2} + z^{2})}} - \dfrac{3y(x^{2} + y^{2} + z^{2})}{\sqrt{3} \cdot\sqrt{(x^{2} + y^{2} + z^{2})}} + \dfrac{3z(x^{2} + y^{2} + z^{2})}{\sqrt{3} \cdot\sqrt{(x^{2} + y^{2} + z^{2})}}=$

$= \dfrac{3 \cdot 1(1^{2} + 1^{2} + 1^{2})}{\sqrt{3} \cdot\sqrt{(1^{2} + 1^{2} + 1^{2})}} - \dfrac{3 \cdot 1(1^{2} + 1^{2} + 1^{2})}{\sqrt{3} \cdot\sqrt{(1^{2} + 1^{2} + 1^{2})}} + \dfrac{3 \cdot 1(1^{2} + 1^{2} + 1^{2})}{\sqrt{3} \cdot\sqrt{(1^{2} + 1^{2} + 1^{2})}}=$

$= \dfrac{3 \cdot 1(1^{2} + 1^{2} + 1^{2})}{\sqrt{3} \cdot\sqrt{(1^{2} + 1^{2} + 1^{2})}}=\dfrac{3 \cdot 3}{\sqrt{3} \sqrt{3} } = 3$.

#SPJ1