1. **State the problem:** We want to evaluate the definite integral $$\int_0^a \sqrt{b^2 - \frac{b^2 x^2}{a^2}} \, dx.$$ 2. **Simplify the integrand:** Factor out $b^2$ inside the square root: $$\sqrt{b^2 - \frac{b^2 x^2}{a^2}} = \sqrt{b^2 \left(1 - \frac{x^2}{a^2}\right)} = b \sqrt{1 - \frac{x^2}{a^2}}.$$ 3. **Rewrite the integral:** $$\int_0^a b \sqrt{1 - \frac{x^2}{a^2}} \, dx = b \int_0^a \sqrt{1 - \frac{x^2}{a^2}} \, dx.$$ 4. **Use substitution:** Let $$t = \frac{x}{a} \implies x = a t, \quad dx = a dt.$$ When $x=0$, $t=0$; when $x=a$, $t=1$. 5. **Rewrite integral in terms of $t$:** $$b \int_0^1 \sqrt{1 - t^2} \cdot a \, dt = a b \int_0^1 \sqrt{1 - t^2} \, dt.$$ 6. **Recognize the integral:** The integral $$\int_0^1 \sqrt{1 - t^2} \, dt$$ represents the area of a quarter circle of radius 1. Its value is known: $$\int_0^1 \sqrt{1 - t^2} \, dt = \frac{\pi}{4}.$$ 7. **Combine results:** $$a b \cdot \frac{\pi}{4} = \frac{a b \pi}{4}.$$ **Final answer:** $$\boxed{\frac{a b \pi}{4}}.$$