Задача 54958 Найти массу кривой...
Условие
Решение
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]