(i) The area of the shaded region is given by the integral:

\(\frac{1}{2} \int_0^1 (x^4 - 1) \, dx\)

Integrating, we have:

\(\frac{1}{2} \left[ \frac{x^5}{5} - x \right]_0^1\)

\(= \frac{1}{2} \left( \frac{1}{5} - 1 \right) = -\frac{2}{5}\)

The area is \(\frac{2}{5}\).

(ii) The volume when rotated about the x-axis is:

\(\pi \int_0^1 y^2 \, dx = \frac{\pi}{4} \int_0^1 (x^8 - 2x^4 + 1) \, dx\)

Integrating, we have:

\(\frac{\pi}{4} \left[ \frac{x^9}{9} - \frac{2x^5}{5} + x \right]_0^1\)

\(= \frac{\pi}{4} \left( \frac{1}{9} - \frac{2}{5} + 1 \right) = \frac{8\pi}{45}\)

The volume is \(\frac{8\pi}{45}\) or 0.559.

(iii) The volume when rotated about the y-axis is:

\(\pi \int_{-\frac{1}{2}}^0 x^2 \, dy = (\pi) \int_{-\frac{1}{2}}^0 (2y+1)^{1/2} \, dy\)

Integrating, we have:

\((\pi) \left[ \frac{(2y+1)^{3/2}}{3/2} \right]_{-\frac{1}{2}}^0\)

\(= (\pi) \left[ \frac{1}{3} - 0 \right] = \frac{\pi}{3}\)

The volume is \(\frac{\pi}{3}\) or 1.05.