Задача 54958 14. Знайти масу кривої [m] x =...

5f958e4af4c6ff700820662e

25.10.2020 18:03:43

14. Знайти масу кривої [m] x = \frac{R}{\sqrt{2}} \cos t, [/m]

[m] y = \frac{R}{\sqrt{2}} \cos t, z = R \sin t, 0 \leq t \leq \frac{\pi}{2} [/m]

з густиною [m] \mu = x + y. [/m]

5f3ea7e3faf909182968ddd9

25.10.2020 21:08:48

★

[m] m=\int _{L} μ (x;y)ds[/m]

Кривая задана параметрически:

значит [m] ds=\sqrt{(x`_{t})^2+(y`_{t})^2+(z`_{t})^2}dt[/m]

[m] x`_{t}=\frac{R}{\sqrt{2}}\cdot (-sint)[/m] ⇒ [m] (x`_{t})^2=\frac{R^2}{2}\cdot (sin^2t)[/m]

[m] y`_{t}=\frac{R}{\sqrt{2}}\cdot (-sint)[/m]⇒ [m] (y`_{t})^2=\frac{R^2}{2}\cdot (sin^2t)[/m]

[m] z`_{t}=R \cdot (cost)[/m]⇒ [m] (z`_{t})^2=R^2\cdot (cos^2t)[/m]

[m]x`_{t})^2+(y`_{t})^2+(z`_{t})^2=R^2\cdot (sin^2t+cos^2t)=R^2[/m]

[m] ds=Rdt[/m]




[m] m=\int^{\frac{\pi}{2}} _{0} (\frac{R}{\sqrt{2}}\cdot cost +\frac{R}{\sqrt{2}}\cdot cost)\cdot R dt=[/m]

[m]=R^2\sqrt{2}\int^{\frac{\pi}{2}} _{0} cos^2tdt=R^2\sqrt{2}\int^{\frac{\pi}{2}} _{0}\frac {1+cos2t}{2}dt=[/m]

[m]=R^2\sqrt{2}\cdot (\frac{1}{2}t+\frac{sin2t}{4})|^{\frac{\pi}{2}} _{0}=R^2\sqrt{2} \pi[/m]