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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.

Respuesta :

jsebguerrero

Answer:

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Explanation:

With the given information we have:

[tex]x(t) = 1 + cos(t)\\ y(t)=sin(t)\\ \rho(x,y) = 3x[/tex]

the interval is [tex][0,\pi ][/tex]

now the mass [tex]m[/tex] has the given expression:

[tex]m = \int \rho(x,y) dS[/tex]

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

[tex]dS=\sqrt{x'(t)^2 + y'(t)^2}=\sqrt{cos(t)^2 + sin(t)^2}dt=dt[/tex]

therefore we have:

[tex]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[/tex]

we solve the integral:

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

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