You're set up is spot on.
Indeed, we're finding the area bounded below the curve $y = \sqrt{2x - x^2}\,$ and above the curve $y = x^2,\;$ between $x = 0$ and $x = 1$:
So indeed, we need to integrate:
$$\begin{align} I & = \int_0^1 \left(\sqrt{2x - x^2} - x^2\right)\,dx \\ \\
& = \int_0^1 \sqrt{1 - 1 + 2x - x^2} \,dx - \int_0^1 x^2\,dx\\ \\
& = \int_0^1 \sqrt{1 - (x - 1)^2} \,dx - \int_0^1 x^2\,dx
\end{align}$$
Now, for the first integral, we use the trigonometric substitution $(x - 1) = \sin\theta$, so that $dx = \cos\theta\,d\theta$.
Finding the new bounds of integration for the first integral (so we can save ourselves the task of "back substitution" at the very end):
At $x = 0, \sin\theta = -1 \implies \theta = -\pi/2.\;$ At $x = 1, \sin\theta = 0 \implies \theta = 0$.
This gives you, after substituting for the first integral:
$$I = \int_{-\pi/2}^{0} \sqrt{1 - \sin^2 \theta}\cos\theta\,d\theta - \int_0^1 x^2\,dx $$
We will use the identities $$\begin{align}\;1 - \sin^2\theta & = \cos^2 \theta\tag{1} \\ \cos^2 \theta & = \dfrac {1 + \cos (2\theta)}{2}\tag{2}\end{align}$$
$$I =\int_{-\pi/2}^0 \left(\sqrt{\cos^2\theta}\right)\cos\theta\,d\theta = \int_{-\pi/2}^0 \cos^2\theta\,d\theta - \int_0^1 x^2\,dx \tag{1}$$
$$I = \dfrac 12\int_{-\pi/2}^0 \left(1 + \cos(2\theta)\right) \,d\theta - \int_0^1 x^2 \,dx\tag{2}$$
Integration should now be relatively straightforward:
$$I = \left[\dfrac \theta2 + \dfrac 14\sin(2\theta)\right]\Big|_{-\pi/2}^0 \;- \;\dfrac{x^3}3\Big|_0^1\quad = \quad \frac\pi4-\dfrac 13$$
The curve $y=x^2$ is indeed the outer curve, so in terms of $y$ the outer curve has equation $x=\sqrt{y}$.
Similarly, in terms of $y$, the inner curve has equation $x=y^2$.
So the volume is
$$\int_{y=0}^1 \pi\left[(\sqrt{y})^2 -(y^2)^2\right]\,dy.$$
Best Answer
Hint: We get from $$\frac{x^2}{2}=\frac{16}{x^3}$$ the equation $$x^5=32$$ And we have to integrate $$\int_{0}^{2}\frac{x^2}{2}dx+\int_{2}^{4}\frac{16}{x^3}dx$$