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.
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.
To find the total charge on the disk, we need to integrate the charge density over the disk area.
The given charge density function is σ(x, y) = 16 + x^2 + y^2.
The disk is defined by the inequality x^2 + y^2 ≤ 5.
To set up the integral, we will integrate the charge density function over the disk area:
Q = ∬ σ(x, y) dA
where Q is the total charge on the disk and dA is the differential area element.
To evaluate this integral, we will convert to polar coordinates.
In polar coordinates, x = r cos(θ) and y = r sin(θ), where r is the radial distance from the origin and θ is the angle.
The disk can be represented as 0 ≤ r ≤ √5 and 0 ≤ θ ≤ 2π.
Now, we need to express the charge density function in terms of r and θ.
σ(x, y) = 16 + x^2 + y^2
= 16 + (r cos(θ))^2 + (r sin(θ))^2
= 16 + r^2 cos^2(θ) + r^2 sin^2(θ)
= 16 + r^2 (cos^2(θ) + sin^2(θ))
= 16 + r^2
Now, we can rewrite the integral as:
Q = ∬ σ(x, y) dA
= ∬ (16 + r^2) r dr dθ
To evaluate this integral, we first integrate with respect to r, then with respect to θ.
∬ (16 + r^2) r dr dθ = ∫[0 to 2π] ∫[0 to √5] (16r + r^3) dr dθ
Integrating with respect to r:
= ∫[0 to 2π] [(8r^2 + (1/4)r^4)]|[0 to √5] dθ
= ∫[0 to 2π] (8(√5)^2 + (1/4)(√5)^4 - 8(0)^2 - (1/4)(0)^4) dθ
= ∫[0 to 2π] (8(5) + (1/4)(25)) dθ
= ∫[0 to 2π] (40 + 25/4) dθ
= ∫[0 to 2π] (160/4 + 25/4) dθ
= ∫[0 to 2π] (185/4) dθ
= (185/4) θ |[0 to 2π]
= (185/4)(2π - 0)
= (185/4)(2π)
= 370π/4
= 92.5π
Therefore, the total charge on the disk is 92.5π coulombs.