Answer:
the mass of the wire is 125/4.
Explanation:
To find the mass of the wire, we need to integrate the density function over the wire. Since the wire has the shape of the first-quadrant part of the circle with center at the origin and radius 5, we can write its equation as:
x^2 + y^2 = 25
Solving for y, we get:
y = sqrt(25 - x^2)
Since the wire is thin, we can assume that its thickness is negligible, so we can treat it as a 2D object. The mass of an infinitesimal element of the wire can be written as:
dm = density * dA
where dA is the infinitesimal area of the element. In polar coordinates, we have:
x = r cos(theta)
y = r sin(theta)
dA = r dr dtheta
Substituting and simplifying, we get:
dm = 2r^3 sin(theta) cos(theta) dr dtheta
To find the total mass of the wire, we need to integrate dm over the first-quadrant part of the circle:
m = ∫∫ 2xy dA
where the limits of integration are:
0 ≤ r ≤ 5
0 ≤ theta ≤ π/2
Substituting the expressions for x and y, we get:
m = ∫[0,π/2] ∫[0,5] 2r^3 sin(theta) cos(theta) dr dtheta
Integrating with respect to r first, we get:
m = ∫[0,π/2] sin(theta) cos(theta) ∫[0,5] 2r^3 dr dtheta
m = ∫[0,π/2] sin(theta) cos(theta) [r^4]_0^5 dtheta
m = ∫[0,π/2] 125 sin(theta) cos(theta) dtheta
m = 125/2 [sin^2(theta)]_0^π/2
m = 125/4
Therefore, the mass of the wire is 125/4.