The surface density (mass/area) of a circular disc of radius a depends on the distance from the centre as `rho(r)=A+Br.` Find its moment of inertia about the line perpendicular to the plane of the disc through its centre.

Explanation

To find the moment of inertia \( I \) of a circular disc with a surface density that varies with the distance from the center, we can follow these steps:

### Step 1: Define the surface density
The surface density \( \rho(r) \) is given by:
\[ \rho(r) = A + Br \]
where \( A \) and \( B \) are constants, and \( r \) is the distance from the center of the disc.

### Step 2: Consider an elemental ring
We consider a small elemental ring of radius \( r \) and thickness \( dr \). The area of this ring is given by:
\[ dA = 2\pi r \, dr \]

### Step 3: Find the mass of the elemental ring
The mass \( dm \) of the elemental ring can be expressed as:
\[ dm = \rho(r) \cdot dA = (A + Br) \cdot (2\pi r \, dr) \]

### Step 4: Express the moment of inertia of the ring
The moment of inertia \( dI \) of the elemental ring about the axis perpendicular to the plane of the disc through its center is:
\[ dI = r^2 \, dm = r^2 \cdot (A + Br) \cdot (2\pi r \, dr) \]
This simplifies to:
\[ dI = 2\pi r^2 (A + Br) r \, dr = 2\pi (A r^3 + B r^4) \, dr \]

### Step 5: Integrate to find the total moment of inertia
To find the total moment of inertia \( I \), we integrate \( dI \) from \( r = 0 \) to \( r = a \):
\[ I = \int_0^a dI = \int_0^a 2\pi (A r^3 + B r^4) \, dr \]

### Step 6: Calculate the integrals
We can separate the integral:
\[ I = 2\pi \left( \int_0^a A r^3 \, dr + \int_0^a B r^4 \, dr \right) \]
Calculating each integral:
1. For \( \int_0^a r^3 \, dr \):
\[ \int_0^a r^3 \, dr = \left[ \frac{r^4}{4} \right]_0^a = \frac{a^4}{4} \]
2. For \( \int_0^a r^4 \, dr \):
\[ \int_0^a r^4 \, dr = \left[ \frac{r^5}{5} \right]_0^a = \frac{a^5}{5} \]

### Step 7: Substitute back into the equation for \( I \)
Substituting the results of the integrals back into the equation for \( I \):
\[ I = 2\pi \left( A \cdot \frac{a^4}{4} + B \cdot \frac{a^5}{5} \right) \]

### Step 8: Final expression for the moment of inertia
Thus, the moment of inertia \( I \) of the circular disc about the axis perpendicular to the plane through its center is:
\[ I = \frac{2\pi a^4 A}{4} + \frac{2\pi a^5 B}{5} \]
This can be simplified to:
\[ I = \frac{\pi a^4 A}{2} + \frac{2\pi a^5 B}{5} \]