Show that the length of the curve y = sqrt((1 − x2/3))3 for 0 x 1 is 3/2.

The given curve is $y = (1 - x^{2/3})^{3/2}$.
To find the arc length, we need to calculate $\frac{dy}{dx}$.

The arc length $L$ is given by the formula:

Substitute $\left(\frac{dy}{dx}\right)^2$:

The length of the curve C is indeed $3/2$.

We have $y = (1 - x^{2/3})^{3/2}$ and $\sqrt{1 + \left(\frac{dy}{dx}\right)^2} = x^{-1/3}$.

Let $u = 1 - x^{2/3}$.
Then $du = -\frac{2}{3} x^{-1/3}\,dx$.
So, $x^{-1/3}\,dx = -\frac{3}{2}\,du$.

Change the limits of integration:
When $x = 0$, $u = 1 - 0^{2/3} = 1$.
When $x = 1$, $u = 1 - 1^{2/3} = 0$.

Substitute into the integral:

To reverse the limits, change the sign:

The area of the surface of revolution obtained is $\boxed{\frac{6\pi}{5}}$.

Pappus's Second Theorem states that the surface area $S$ of a surface of revolution generated by rotating a plane curve C about an external axis is equal to the product of the arc length $L$ of C and the distance traveled by the centroid of C.

where $\bar{y}$ is the y-coordinate of the centroid of the arc C, and $2\pi \bar{y}$ is the distance traveled by the centroid.

From part (i), the arc length $L = \frac{3}{2}$.
From part (iii), the surface area $S = \frac{6\pi}{5}$.

Divide both sides by $3\pi$:

The y-coordinate of the centroid of the arc C is $\boxed{\frac{2}{5}}$.

Side length of the square is $s = 2a$.
Area of the square $A = s^2 = (2a)^2 = 4a^2$.
The centroid of the square is its geometric center.

The distance from the center O of the square to the axis of rotation is given as $d = 3a$.
According to Pappus's First Theorem, the volume $V$ of a solid of revolution generated by rotating a plane area $A$ about an external axis is given by:

where $d$ is the distance of the centroid of the area from the axis of rotation.

The volume of the solid swept out by the square is $\boxed{24\pi a^3}$.

The perimeter of the square $P = 4s = 4(2a) = 8a$.
The centroid of the perimeter of the square is also its geometric center.

The distance from the centroid of the perimeter to the axis of rotation is $d = 3a$.
According to Pappus's Second Theorem, the surface area $S$ of a surface of revolution generated by rotating a plane curve (in this case, the perimeter of the square) about an external axis is equal to the product of the length of the curve $P$ and the distance traveled by the centroid of the curve.

The area of the surface of the solid is $\boxed{48\pi a^2}$.