Chọn phương án A.

Phương trình hoành độ giao điểm $$\begin{aligned}
\sqrt{x}=x^2&\Leftrightarrow x=x^4\Leftrightarrow x-x^4=0\\
&\Leftrightarrow x\left(1-x^3\right)=0\\
&\Leftrightarrow\left[\begin{array}{l}
x=0\\
1-x^3=0
\end{array}\right.\\
&\Leftrightarrow\left[\begin{array}{l}
x=0\\
x=1.
\end{array}\right.
\end{aligned}$$
Xét dấu \(\left(x^2\right)^2-\left(\sqrt{x}\right)^2=x^4-x=x\left(x^3-1\right)\):

Suy ra \(\left(x^2\right)^2-\left(\sqrt{x}\right)^2\leq0\) khi \(x\in[0;1]\).

Vậy: $$\begin{aligned}V&=\pi\displaystyle\int\limits_{0}^{1}\left[\left(\sqrt{x}\right)^2-\left(x^2\right)^2\right]\mathrm{\,d}x\\
&=\pi\displaystyle\int\limits_{0}^{1}\left(x-x^4\right)\mathrm{\,d}x\\
&=\pi\left(\dfrac{x^2}{2}-\dfrac{x^5}{5}\right)\bigg|_0^1\\
&=\dfrac{3\pi}{10}.\end{aligned}$$