Find the exact mass of a thin wire in the shape of the helix $x = 2 \sin ( t ) , y = 2 \cos ( t ) , z = 3 t , 0 \leq t \leq 2 \pi$ if the density is 5 .
A) $4 \sqrt { 13 }$
B) $8 \sqrt { 10 } \pi ^ { 2 }$
C) $10 \sqrt { 13 }$
D) $10 \sqrt { 13 } \pi$
E) $\sqrt { 13 } \pi$

Question

Quizplus · Accepted Answer

The mass of the wire can be found using the formula $m = \int_{a}^{b} \rho \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 + \left(\frac{dz}{dt}\right)^2} dt$, where $\rho$ is the density. For the given helix, $x = 2\sin(t)$, $y = 2\cos(t)$, and $z = 3t$, with $\rho = 5$. Differentiating $x$, $y$, and $z$ with respect to $t$ gives $\frac{dx}{dt} = 2\cos(t)$, $\frac{dy}{dt} = -2\sin(t)$, and $\frac{dz}{dt} = 3$. Substituting these into the formula gives $\sqrt{(2\cos(t))^2 + (-2\sin(t))^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13}$. Thus, the integral becomes $m = 5 \int_{0}^{2\pi} \sqrt{13} dt = 5\sqrt{13} [t]_{0}^{2\pi} = 10\pi\sqrt{13}$.

Find the Exact Mass of a Thin Wire in the Shape