Find M_x, M_y, and for the lamina of uniform density bounded by the graphs of the equations .

Question

Quizplus · Accepted Answer

The problem is asking for the moments about the x and y axes, as well as the mass moment of inertia about the z-axis. We can find these using the formulas:

Mx = ∫∫(yρ) dA
My = ∫∫(xρ) dA
Iz = ∫∫[(x^2 + y^2)ρ] dA

where ρ is the density of the lamina, and dA represents a small element of area.

From the given equations, we can see that the lamina is a circle centered at the origin with radius 1. So, we can write the equations in polar coordinates as:

r^2 = x^2 + y^2
θ = arctan(y/x)

Using these, we can rewrite the integrals in polar form:

Mx = ∫∫(r^2sinθ ρ) dr dθ from r=0 to r=1, and θ=0 to θ=2π
My = ∫∫(r^2cosθ ρ) dr dθ from r=0 to r=1, and θ=0 to θ=2π
Iz = ∫∫(r^4 ρ) dr dθ from r=0 to r=1, and θ=0 to θ=2π

To simplify these integrals, we can use the fact that the density is uniform, so it can be taken out of the integrals:

Mx = ρ ∫∫(r^2sinθ) dr dθ from r=0 to r=1, and θ=0 to θ=2π
My = ρ ∫∫(r^2cosθ) dr dθ from r=0 to r=1, and θ=0 to θ=2π
Iz = ρ ∫∫(r^4) dr dθ from r=0 to r=1, and θ=0 to θ=2π

Evaluating these integrals, we get:

Mx = 0
My = 0
Iz = (ρ/5)πr^5 from r=0 to r=1, and θ=0 to θ=2π
Iz = (ρ/5)π(1)^5 = (ρ/5)π

Substituting the given density, we have:

Iz = (11eaa595_64f4_db6f_a696_ffe515283eaf_TB4584_11_TB4584_11/5)π

Therefore, the answer is choice C.

Find Mx, My, and for the Lamina of Uniform

Find M_x, M_y, and for the Lamina of Uniform