## Calculus 8th Edition

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: $(\overline {x}, \overline {y}, \overline {z})=(0,0, 3\pi)$
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, $m=\int_C \rho(x,y) ds=k \int_{0}^{2\pi} \sqrt{13} dt=2 \pi k\sqrt {13}$ Now, $\overline {x}=\dfrac{1}{m}\int_{C} x \rho(x,y,z) ds=(k\sqrt{13}/m) \times \int_{0}^{2\pi} x dt=\dfrac{1}{2 \pi} [-2 \cos t]_{0}^{2 \pi} =0$ $\overline {y}=\dfrac{1}{m}\int_{C} y \rho(x,y,z) ds=(k\sqrt{13}/m) \times \int_{0}^{2\pi} y dt=\dfrac{1}{2 \pi} [2 \sin t]_{0}^{2 \pi} =0$ $\overline {z}=\dfrac{1}{m}\int_{C} z \rho(x,y,z) ds=(k\sqrt{13}/m) \times \int_{0}^{2\pi} z dt=(1/2 \pi) [\dfrac{3t^2}{2}]_{0}^{2 \pi} =3 \pi$ Therefore, the center of mass of the helix is: $(\overline {x}, \overline {y}, \overline {z})=(0,0, 3\pi)$