To find the total charge on the disk, we need to integrate the charge density over the entire disk.
The charge density at each point (x, y) on the disk is given by σ(x, y) = 16 + x^2 + y^2.
To integrate this charge density over the disk, we can use a double integral, with the bounds of integration being the disk itself.
The disk can be described by the inequality x^2 + y^2 ≤ 5, which represents a disk of radius √5 centered at the origin.
Therefore, the integral becomes:
∬[x^2+y^2≤5] (16 + x^2 + y^2) dA,
where dA represents the differential area element.
Converting to polar coordinates, we have:
∬[x^2+y^2≤5] (16 + x^2 + y^2) dA = ∫[0 to 2π] ∫[0 to √5] (16 + r^2) r dr dθ,
where r represents the radial distance and θ represents the angle.
Evaluating this double integral, we get:
∫[0 to 2π] ∫[0 to √5] (16 + r^2) r dr dθ
= 2π ∫[0 to √5] (16 + r^2) r dr
= 2π [8r^2 + (1/3)r^4] evaluated from 0 to √5
= 2π [8(√5)^2 + (1/3)(√5)^4] - 2π [8(0)^2 + (1/3)(0)^4]
= 2π [8(5) + (1/3)(25)]
= 2π [40 + 25/3]
= 2π [120/3 + 25/3]
= 2π (145/3)
= 290π/3.
Therefore, the total charge on the disk is 290π/3 coulombs.
Electric charge is distributed over the disk
x^2+y^2≤5 so that the charge density at (x,y) is σ(x,y)=16+x^2+y^2 coulombs per square meter.
Find the total charge on the disk.
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