Answer
a) $\overline {x}=\dfrac{1}{m}\int_{C} x \rho(x,y,z) ds$; $\overline {y}=\dfrac{1}{m}\int_{C} y \rho(x,y,z) ds$; $\overline {z}=\dfrac{1}{m}\int_{C} z \rho(x,y,z) ds$
b) The center of mass of the helix is: $(\overline {x}, \overline {y}, \overline {z})=(0,0, 3\pi)$
Work Step by Step
(a) $\overline {x}=\dfrac{1}{m}\int_{C} x \rho(x,y,z) ds; \\\overline {y}=\dfrac{1}{m}\int_{C} y \rho(x,y,z) ds ; \\ \overline {z}=\dfrac{1}{m}\int_{C} z \rho(x,y,z) ds$
b) Here, we have the mass of the wire:
$m=\int_C \rho(x,y) ds=k \int_{0}^{2\pi} \sqrt{13} dt=(2 \pi k) (\sqrt {13}) $
Now, we have $\overline {x}=\dfrac{1}{m}\int_{C} x \rho(x,y,z) ds=\dfrac{k\sqrt{13}}{m}\int_{0}^{2\pi} x dt$
and $(\dfrac{1}{2 \pi}) [-2 \cos t]_{0}^{2 \pi} =0$
$\overline {y}=\dfrac{1}{m}\int_{C} y \rho(x,y,z) ds$
and $\dfrac{k\sqrt{13}}{m}\int_{0}^{2\pi} y dt=(\dfrac{1}{2 \pi}) [2 \sin t]_{0}^{2 \pi} =0$
and $\overline {z}=\dfrac{1}{m}\int_{C} z \rho(x,y,z) ds$
and $\dfrac{k\sqrt{13}}{m}\int_{0}^{2\pi} z dt=(\dfrac{1}{2 \pi})[3\dfrac{t^2}{2}]_{0}^{2 \pi} =3 \pi$
Thus, the center of mass of the helix is: $(\overline {x}, \overline {y}, \overline {z})=(0,0, 3\pi)$
