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Multivariable calculus mass density question

  1. Oct 2, 2014 #1
    1. The problem statement, all variables and given/known data
    Let the path C traverse part of the circle or radius 3 at the origin, in a clockwise direction, from (0,-3) to (3,0). Calculate the total mass of a wire in shape C, if the mass density of the wire is u=x^2+4y

    2. Relevant equations
    mass of plate equation= double integral u(x,y) dx dy
    3. The attempt at a solution
    I converted the wire into polar coordinates as its a circle, with x=3cos(theta) and y=3sin(theta) and as it travels from -pi/2 to 2pi, 0<r<3 and -pi/2<theta<2pi, after doing that i subbed x=3cos(theta) and y=3sin(theta) into the mass density equation (u) to obtain u=9cos^2(theta) + 12sin(theta) and as the mass of plate equation is double integral u(x,y) dx dy I subbed the vaules into this equation but with respect to polar coordinates to get:

    double integral 9cos^2(theta) + 12sin(theta) dtheta dr with 0<r<3 and -pi/2<theta<2pi

    solving this ended up getting 135*pi/4 -36 to be the answer, but i'm a little confused as i think i worked out the mass for 3/4 of the circle, instead of the wire and am now thinking i might need to work out a ratio for area of circumference/total area of circle and multiply by this ratio to get the right answer.

    Any Help would be much appreciated!
     
  2. jcsd
  3. Oct 2, 2014 #2

    Ray Vickson

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    Why are you doing a dxdy integral? Isn't a wire a one-dimensional object?
     
  4. Oct 2, 2014 #3
    i'm not sure to be honest i thought it would be a two-dimensional object as it would have both a x and y direction but i am probably wrong, if it is a one dimensional object, how would i go about solving this problem?, would it just be x=3cost and y=3sint where -pi/2<t<2pi and hence you would get the integral: 9cos^2(theta) + 12sin(theta) dt where -pi/2<t<2pi and solving this would get you 45pi/4 -12, thanks for the fast reply :)
     
  5. Oct 2, 2014 #4

    Mark44

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    I haven't done any work on the problem, but the density is given as a function of x and y, and the wire follows a curved path (a quarter circle), so an iterated integral seems reasonable to me.
     
  6. Oct 2, 2014 #5

    Ray Vickson

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    My interpretation is that the wire is "infinitely thin", so is just a curved line in 2 dimensions. For that reason you need two coordinates ##(x,y)## to specify a point on the wire. The "density" ##\rho(x,y)## would then be the mass per unit length at the point ##(x,y)##; that is, the mass ##\Delta m## of a little bit of wire of length ##\Delta s## located at ##(x,y)## would be ##\Delta m = \rho(x,y) \Delta s##. But that is just my interpretation.
     
  7. Oct 2, 2014 #6

    pasmith

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    That would also be my interpretation. Otherwise I would have expected the problem to ask for the mass of a "plate" and for the boundary of the plate to be a closed curve.
     
  8. Oct 2, 2014 #7
    yeah that makes sense, thanks for that, because of that i would only need to integrate once then and hence got x=3cost and y=3sint where -pi/2<t<2pi and then the integral: 9cos^2(theta) + 12sin(theta) dt where -pi/2<t<2pi and solving this would get you 45pi/4 -12, would this be correct or would i need to do something else?
     
  9. Oct 2, 2014 #8

    LCKurtz

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    And, just in case anyone is still confused about 1 dimensional or 2 dimensional object, remember that even though the wire exists in 3 dimensions, it only takes one parameter to locate a point on it. That's why you use ##\vec r(t)=\langle x(t),y(t),z(t)\rangle## to describe it. Same for a surface. It exists in 3 dimensional space but you only need two parameters to describe it, so you use ##\vec r(u,v)##.
     
  10. Oct 2, 2014 #9

    Ray Vickson

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    It looks wrong. You should show your work in detail, because I cannot figure out how you got your answer.
     
  11. Oct 3, 2014 #10

    RUber

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    It looks like that would be the right integral, but the bounds for theta might be incorrect. From your initial problem, you wrote "Clockwise," which would indicate a negative change in theta, from 3*pi/2 to 0.
     
  12. Oct 3, 2014 #11

    Ray Vickson

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    The integral is not quite correct.
     
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