1. **State the problem:** We want to evaluate the definite integral $$\int_0^a \sqrt{b^2 - \frac{x^2 b^2}{a^2}} \, dx.$$ 2. **Simplify the integrand:** Factor out $b^2$ inside the square root: $$\sqrt{b^2 - \frac{x^2 b^2}{a^2}} = \sqrt{b^2 \left(1 - \frac{x^2}{a^2}\right)} = b \sqrt{1 - \frac{x^2}{a^2}}.$$ 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. 7. **Evaluate the integral:** The formula for $$\int \sqrt{1 - t^2} \, dt$$ is $$\frac{t}{2} \sqrt{1 - t^2} + \frac{\arcsin(t)}{2} + C.$$ 8. **Apply limits:** $$\int_0^1 \sqrt{1 - t^2} \, dt = \left[ \frac{t}{2} \sqrt{1 - t^2} + \frac{\arcsin(t)}{2} \right]_0^1.$$ 9. **Calculate at $t=1$:** $$\frac{1}{2} \cdot 0 + \frac{\arcsin(1)}{2} = 0 + \frac{\pi/2}{2} = \frac{\pi}{4}.$$ 10. **Calculate at $t=0$:** $$0 + \frac{\arcsin(0)}{2} = 0.$$ 11. **Subtract:** $$\frac{\pi}{4} - 0 = \frac{\pi}{4}.$$ 12. **Final answer:** $$a b \cdot \frac{\pi}{4} = \frac{a b \pi}{4}.$$