Question

In Problems 17 to 30, for the curve $\mathit{y}\mathbf{=}\sqrt{\mathbf{x}}$, between $\mathit{x}\mathbf{=}\mathbf{0}$and $\mathit{x}\mathbf{=}\mathbf{2}$ , find:

The volume of the solid generated when the area is revolved about the axis.

Short answer

The volume of the solid is $2\mathrm{\pi }$.

Step-by-step solution

Definition of the volume of solid

The space taken by a particular solid is measured by a special quantity known as the volume of the solid. The cubic units required to fill the space is the measure of the volume.

Representation of the area bounded by the curve

Draw the bounded region for the curve $y=\sqrt{x}$ , between $x=0$ and $x=2$.

Determination of the volume of the solid generated when the area is revolved about the  x -axis

Determine the value of elementary volume from the above figure.

$$\begin{gathered} dV=\mathrm{\pi }{\left(\mathrm{f}\left(\mathrm{x}\right)\right)}^2 \\ =\mathrm{\pi x} \end{gathered}$$

Determine the value of total volume.

$$\begin{gathered} V={\int }_0^2\mathrm{\pi xdx} \\ =\frac{\mathrm{\pi }}{2}{\left[{\mathrm{x}}^2\right]}_0^2 \\ =2\mathrm{\pi } \end{gathered}$$

Thus, the volume of the solid is $2\mathrm{\pi }$ .