Evaluate ∫C ysin(z)ds, where C is the circular helix given by the equations x = cos(t), y = sin(t), z = t, 0 ≤ t ≤ 2π. SOLUTION The formula for a line integral in space gives the following. ∫y sin(z)ds = sin2(t) dt = (sin(t))2√ (cos(t))2 + (sin(t))2 + 1dt = 1 2 (1 - cos(2t))dt = √2 2 =

The line integral is[tex]\displaystyle\int_Cy\sin z\,\mathrm ds=\int_0^{2\pi}y(t)\sin z(t)\,\sqrt{\left(\frac{\mathrm dx}{\mathrm dt}\right)^2+\left(\frac{\mathrm dy}{\mathrm dt}\right)^2+\left(\frac{\mathrm dz}{\mathrm dt}\right)^2}\,\mathrm dt[/tex]We have[tex]x=\cos t\implies\dfrac{\mathrm dx}{\mathrm dt}=-\sin t[/tex][tex]y=\sin t\implies\dfrac{\mathrm dy}{\mathrm dt}=\cos t[/tex][tex]z=t\implies\dfrac{\mathrm dz}{\mathrm dt}=1[/tex]so the integral reduces to[tex]\displaystyle\int_0^{2\pi}\sin^2t\sqrt{(-\sin t)^2+\cos^2t+1^2}\,\mathrm dt=\frac{\sqrt2}2\int_0^{2\pi}(1-\cos2t)\,\mathrm dt=\boxed{\frac\pi{\sqrt2}}[/tex]