To find the total charge on the square metal sheet with given surface charge density, we need to integrate the surface charge density over the entire area of the sheet.

The metal sheet is positioned on the x-y plane with vertices at \((-2, -2)\), \((2, -2)\), \((-2, 2)\), and \((2, 2)\). The surface charge density is given by:

\rho_s(x, y) = 4|y| \, \mu C/m^2

Let's calculate the total charge Q on the sheet using the formula:

Q = \int \int \rho_s(x, y) \, dx \, dy

For this problem, we integrate over the area of the square from \(-2\) to \(2\) for both x and y:

Q = \int_{-2}^{2} \int_{-2}^{2} 4|y| \, dx \, dy

First, integrate with respect to \(x\):

\int_{-2}^{2} 4|y| \, dx = 4|y| \times (2 - (-2)) = 16|y|

Next, integrate with respect to \(y\):

Q = \int_{-2}^{2} 16|y| \, dy

Since \(|y|\) depends on the value of \(y\), we need to split it:

• For \(y\) from \(-2\) to \(0\), \(|y| = -y\)
• For \(y\) from \(0\) to \(2\), \(|y| = y\)

Thus, the integral becomes:

Q = \int_{-2}^{0} 16(-y) \, dy + \int_{0}^{2} 16y \, dy

Calculate each integral separately:

• \int_{-2}^{0} 16(-y) \, dy = 16 \left[ \frac{-y^2}{2} \right]_{-2}^{0} = 16 \left( 0 - \frac{-4}{2} \right) = 32
• \int_{0}^{2} 16y \, dy = 16 \left[ \frac{y^2}{2} \right]_{0}^{2} = 16 \left( \frac{4}{2} - 0 \right) = 32

Adding both integrals gives the total charge:

Q = 32 + 32 = 64 \, \mu C

Thus, the total charge on the sheet is \(64\) \(\mu C\), rounded off to the nearest integer.

The correct answer is 64.