Question

If rho(x,y) is the density of a wire (mass per unit length), then

Answer 1

Answer:

See description

Explanation:

With the given information we have:

$x(t) = 1 + cos(t)\\ y(t)=sin(t)\\ \rho(x,y) = 3x$

the interval is $[0,\pi ]$

now the mass $m$ has the given expression:

$m = \int \rho(x,y) dS$

we will use the formula for a line integral and let:

$dS=\sqrt{x'(t)^2 + y'(t)^2}=\sqrt{cos(t)^2 + sin(t)^2}dt=dt$

therefore we have:

$m=\int \rho(x,y)dS=\int\limits^\pi_0 {3*x}dS=\int\limits^\pi _0{3*(1+cos(t))dS\\=\int\limits^\pi _0{3*(1+cos(t))dt$

we solve the integral:

$m=3*\int\limits^\pi _0{(1+cos(t))dt= 3*(t+sin(t))\limits^\pi _0=3*\pi=9.42$