Q2 For $\mathbf{K} \sim \mathbf{ZZ}$ in the following statements, choose the appropriate answer from among (0) $\sim$ (9) at the bottom of this page.
Let $a$ and $t$ be positive real numbers. Let $D$ denote the region of a plane bounded by the graph of the quadratic function in $x$
$$y = \frac{1}{t^2}\left(x - at^2\right)^2$$
the $x$-axis, and the $y$-axis. Let $V_1$ denote the volume of the solid obtained by rotating $D$ once about the $x$-axis, and $V_2$ denote the volume of the solid obtained by rotating $D$ once about the $y$-axis. Now, let us show that for a certain value of $a$, $V_1 = V_2$, independent of the value of $t$.
First, the value of $V_1$ is
$$\begin{aligned}
V_1 &= \pi \int_{\mathbf{K}}^{\mathbf{L}} \frac{1}{t^{\mathbf{M}}}\left(x - at^2\right)^{\mathbf{N}} dx \\
&= \frac{\pi}{\mathbf{O}} a^{\mathbf{P}} t^{\mathbf{Q}}
\end{aligned}$$
Next, the value of $V_2$ is
$$\begin{aligned}
V_2 &= \pi \int_{\mathbf{R}}^{\mathbf{S}} (\cdots) \\
&= \frac{\pi}{\mathbf{T}} a^{\mathbf{W}} t^{\mathbf{W}}
\end{aligned}$$
Hence, when $a = \frac{\mathbf{Y}}{\mathbf{Y}}$, then $V_1 = V_2$, independent of the value of $t$.
Options: (0) 0
(1) 1
(2) 2
(3) 3
(4) 4
(5) 5 (6) 6 (7) $t$ (8) $at^2$ (9) $a^2t^2$