Multivariable calculus mass density question

[jamesdocherty]

Homework Statement

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

Homework Equations

mass of plate equation= double integral u(x,y) dx dy

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!

 

[Ray Vickson]

Homework Statement

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

Homework Equations

mass of plate equation= double integral u(x,y) dx dy

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!

Why are you doing a dxdy integral? Isn't a wire a one-dimensional object?

 

[jamesdocherty]

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 :)

 

[Mark44]

Why are you doing a dxdy integral? Isn't a wire a one-dimensional object?

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.

 

[Ray Vickson]

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 :)

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.

 

[pasmith]

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.

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.

 

[jamesdocherty]

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?

 

[LCKurtz]

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 $\vec r(u,v)$.

 

[Ray Vickson]

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?

It looks wrong. You should show your work in detail, because I cannot figure out how you got your answer.

 

[RUber]

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 :)

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.

 

[Ray Vickson]

The integral is not quite correct.