Computing mass with a denstiy function

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SUMMARY

The discussion focuses on computing the total mass of a wire bent in a quarter circle defined by the parametric equations x=cos(t) and y=sin(t) for 0 ≤ t ≤ π/2, with a density function ρ(x,y) = x² + y². The correct approach involves integrating the density function along the curve, but the density simplifies to a constant value of 1. The mass is calculated as 1 times the length of the curve, which is 1/4 of the circumference of a circle with radius 1, yielding a total mass of π/2.

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Homework Statement


Compute the total mass of a wire bent in a quarter circle with parametric equations: x=cos(t), y=sin(t), 0\leq t \leq \pi/2
and density function \rho(x,y) = x^2+y^2


Homework Equations



not exactly too sure which equations if any i need to use. maybe the jacobian

The Attempt at a Solution



i simply substituted the x and y into the density function to get

(6cost)^2 + (6sint)^2 and took the integral of that with the bounds of integration from 0 to pi/2. the answer i am getting is 56.549 and is wrong and I'm not sure if there's an extra step i need to do
 
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You want, of course, to integrate the density function along the length of the given curve. The differential of length, when the curve is given by parametric equations, x= x(t), y= y(t), is
\sqrt{\left(\frac{dx}{dt}\right)^2+ \left(\frac{dy}{dx}\right)^2}dt.

However, here, you should be able to see that the density at point (x,y) is cos^2(t)+ sin^2(t)= 1. (I have no idea where you got the "6" in your formula). Since that density is constant around the arc, the mass is just 1 times the length of the curve. And that is simply 1/4 the circumference of a circle of radius 1.
 

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