1. The problem is to correctly identify the upper limit of the integral in part 8 (i), which is given as $\sqrt{2ax - x^2}$, not $2a \times \sqrt{2ax - x^2}$.\n\n2. This means the integral's upper limit is the square root of the expression $2ax - x^2$, which is a function of $x$ and $a$.\n\n3. The expression $\sqrt{2ax - x^2}$ can be interpreted as the radius or boundary in a geometric or physical context, depending on the problem.\n\n4. To work with this limit, ensure you substitute it correctly in the integral or equation, without multiplying by $2a$.\n\n5. For example, if the integral is $\int_0^{\sqrt{2ax - x^2}} f(t) dt$, then the upper limit is exactly $\sqrt{2ax - x^2}$.\n\n6. Always double-check the limits in integrals to avoid errors in evaluation or interpretation.\n\nFinal answer: The correct upper limit is $\sqrt{2ax - x^2}$, not $2a \times \sqrt{2ax - x^2}$.