Question

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

m=∫Crho(x,y)ds

is the mass of the wire. Find the mass of a wire having the shape of a semicircle x=1+cos(t),y=sin(t), where t is on the closed interval from 0 to π, if the density at a point P is directly proportional to the distance from the y−axis and the constant of proportionality is 3. Round in the tenths place.

Answer 1

See description

Step-by-step 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=√(x'(t)^2 + y'(t)^2)=√(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`