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Step-by-Step Solution
Step 1: Express Mass Element in Terms of r
The surface mass density of the disc is given by $ \sigma(r) = A + Br $. For a thin ring at a distance $ r $ from the center with thickness $ dr $, the area of this ring is $ dA = 2\pi r\,dr $. Hence, the mass of this thin ring element is:
$$
dm = \sigma(r)\,dA = (A + Br)\,\bigl(2\pi r\,dr\bigr).
$$
Step 2: Write the Moment of Inertia Expression
The moment of inertia $ dI $ of the thin ring about the axis perpendicular to the plane (through the center) is:
$$
dI = r^2 \, dm = r^2 \,(A + Br)\,\bigl(2\pi r\,dr\bigr).
$$
Simplify inside the integrand:
$$
dI = 2\pi (A + Br)\,r^3 \,dr.
$$
Step 3: Integrate Over the Entire Radius
To get the total moment of inertia, integrate from $ r = 0 $ to $ r = a $:
$$
I = \int_{0}^{a} dI = \int_{0}^{a} \bigl[ 2\pi (A + Br)\,r^3 \bigr] \,dr.
$$
Factor out the constant $ 2\pi $:
$$
I = 2\pi \int_{0}^{a} \bigl( A r^3 + B r^4 \bigr)\,dr.
$$
Step 4: Perform the Integration
Now integrate term by term:
$$
I = 2\pi \left[ A \int_{0}^{a} r^3 \,dr + B \int_{0}^{a} r^4 \,dr \right].
$$
The integrals are straightforward power integrals:
$$
\int_{0}^{a} r^3 \,dr = \frac{a^4}{4},
\quad
\int_{0}^{a} r^4 \,dr = \frac{a^5}{5}.
$$
Thus,
$$
I = 2\pi \left[ A \frac{a^4}{4} + B \frac{a^5}{5} \right].
$$
Factor out $ a^4 $:
$$
I = 2\pi a^4 \left[ \frac{A}{4} + \frac{aB}{5} \right].
$$
Step 5: Final Result
Hence, the moment of inertia of the disc is:
$$
I = 2\pi a^4 \left( \frac{A}{4} + \frac{aB}{5} \right).
$$
Solution Reference Image
