Find Mass and C.O.M using a line integral

[Smazmbazm]

Homework Statement

Using line integrals, find the mass and the position of the center of mass of a thin wire in the shape of a half-circle x^{2} + y^{2} = r^{2}, x ≥ 0 and -r ≤ y≤ r if the linear density is ρ(x,y) = x^{2} + y^{2}

The mass is given by the integral of the density along the curve while the center of mass is defined as

\frac{∫_{C} x ρ(x,y)ds}{∫_{C}ρ(x,y)ds}

for the x co-ordinate and similarly for the y co-ordinate.

The Attempt at a Solution

My attempt was to substitute x^{2} = y^{2} - r^{2} and y^{2} = x^{2} - r^{2} into the density function to get ρ(x,y) = x^{2} + y^{2} - 2r^{2} then do the double integral

∫^{x}_{0}∫^{r}_{-r} x^{2} + y^{2} - 2r^{2} dy dx

from that I got an answer of \frac{2rx^{2}}{3}

EDIT: I get \frac{2rx^{2}}{3} - \frac{10r^3x}{3} rather then \frac{2rx^{2}}{3}

Is that correct? Or am I supposed to follow Example 1 on this page, http://tutorial.math.lamar.edu/Classes/CalcIII/LineIntegralsPtI.aspx ?

 

[LCKurtz]

Homework Statement

Using line integrals, find the mass and the position of the center of mass of a thin wire in the shape of a half-circle x^{2} + y^{2} = r^{2}, x ≥ 0 and -r ≤ y≤ r if the linear density is ρ(x,y) = x^{2} + y^{2}

The mass is given by the integral of the density along the curve while the center of mass is defined as

\frac{∫_{C} x ρ(x,y)ds}{∫_{C}ρ(x,y)ds}

for the x co-ordinate and similarly for the y co-ordinate.

The Attempt at a Solution

My attempt was to substitute x^{2} = y^{2} - r^{2} and y^{2} = x^{2} - r^{2} into the density function to get ρ(x,y) = x^{2} + y^{2} - 2r^{2} then do the double integral

Stop right there! Those are line integrals and you don't evaluate them with double integrals. Do the line integrals. Polar coordinates might be a good idea.

 

[ParoxysmX]

Hah, physics 211 I assume?

 

[Smazmbazm]

Physics 211 indeed, ParoxysmX =]. LCKrutz, ok so,

x = r cos(t), y = r sin(t)

ds = \sqrt{((dx/dt)^{2} + (dy/dt)^{2})} <br /> = \sqrt{(r^{2}sin^{2}(t) +r^{2}cos^{2}(t))}dt<br /> = \sqrt{r^2}dt = r dt

Then substitute x and y values into the density function to get

Mass = ∫_{C}(r^{2}cos^{2}(t) + r^{2}sin^{2}(t))(r)dt
Mass =∫_{C}r^{3}(cos^{2}(t)+sin^{2}(t))dt = ∫_{C}r^{3}dt
Mass =r^{3}t |^{\pi}_{0} = r^{3}\pi

Correct?

Then for the x and y, do the above process but multiply by r cost and r sint and divide by r^{3}\pi to get x and y which are x = 0 and y = 2r^{4}/r^{3}\pi.