A parabola x = y² with 0 ≤ x ≤ 1 is shown in the figure. The volume of the solid of rotation obtained by rotating the shaded area by 360° around the x–axis is

The correct option is π/2.

To find the volume of the solid of rotation obtained by rotating the shaded area 360° around the x-axis, we use the method of disks/washers.

The given equation of the parabola is:

$$x=y^2$$

Since the rotation is about the x-axis, we express the radius of the cross-sectional disks in terms of x. Rearranging the equation for y, we get:

$$y^2=x$$

The shaded region is bounded by the curve, the x-axis, and the vertical line from x = 0 to x = 1. The formula for the volume $V$ of a solid of revolution rotated around the x-axis is given by:

$$V={\int }_a^b\pi \left[f\right(x){]}^{2}dx$$

Here, the radius squared is $\left[f\right(x){]}^2=y^2=x$, and the limits of integration are from $x=0$ to $x=1$.

Substituting these values into the volume formula:

$$V={\int }_0^1\pi xdx$$

We can pull the constant π outside the integral:

$$V=\pi {\int }_0^{1}xdx$$

Integrating $x$ with respect to $x$ gives:

$$V=\pi {\left[\frac{x^2}{2}\right]}_0^1$$

Evaluating this at the limits 1 and 0:

$$V=\pi (\frac{1^2}{2}-\frac{0^2}{2})$$

$$V=\pi \left(\frac{1}{2}\right)=\frac{\pi }{2}$$

Thus, the volume of the solid of rotation is indeed π/2.