Answer
$\int_Cyzcosxds=6\sqrt {10}$
Work Step by Step
$ds=\sqrt {(\frac{dx}{dt})^{2}+(\frac{dy}{dt})^{2}+(\frac{dz}{dt})^{2}}dt$
$ds=\sqrt {{1}^{2}+({-3sint})^{2}+({3cost})^{2}}dt$
$ds=\sqrt {1+9}dt$
$ds=\sqrt {10}dt$
Now, $\int_Cyzcosxds=\int _{0}^{\pi}(3cost)(3sint)cos(t)\sqrt {10}dt$
$=9\sqrt {10}\int _{0}^{\pi}cos^{2}tsintdt$
Substitute $cost=u$ and $-sintdt=du$
Limits of integration will change from $\int _{0}^{\pi}$ to$\int _{1}^{-1}$.
$=-9\sqrt {10}\int _{1}^{-1}u^{2}du$
$=-9\sqrt {10}[\frac{u^{3}}{3}] _{1}^{-1}$
$=-9\sqrt {10}[\frac{-1}{3}-\frac{1}{3}]$
Hence, $\int_Cyzcosxds=6\sqrt {10}$
