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 curved area of this solid.

Short answer

The curved area of the solid is$\frac{13\mathrm{\pi }}{3}$

Step-by-step solution

Definition of double integral

Double integral of $\mathbf{f}\left(x,y\right)$over the area, A in the $\left(x,y\right)$plane as the limit of this sum, and it is written as $\mathbf{\int }{\mathbf{\int }}_{\mathbf{A}}\mathit{f}\left(x,y\right)\mathit{d}\mathit{x}\mathit{d}\mathit{y}$

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 surface element of the solid generated when the area is revolved about the x-axis.

Determine the surface element from the figure.

$$\begin{gathered} dS=2\mathrm{\pi f}\left(\mathrm{x}\right)\mathrm{dL} \\ =2\mathrm{\pi f}\left(\mathrm{x}\right)\sqrt{1+{\left(\mathrm{f}\text{'}\left(\mathrm{x}\right)\right)}^2}\mathrm{dx} \\ =2\mathrm{\pi }\sqrt{\mathrm{x}}\sqrt{1+\frac{1}{4\mathrm{x}}}\mathrm{dx} \\ =2\mathrm{\pi }\sqrt{\mathrm{x}+\frac{1}{4}}\mathrm{dx} \end{gathered}$$

 Step 4: Determination of the total surface of the solid generated when the area is revolved about the axis.

Determine the total surface of the solid generated.

$$\begin{gathered} S=2\mathrm{\pi }{\int }_0^2\sqrt{\mathrm{x}+\frac{1}{4}}\mathrm{d}\left(\mathrm{x}+\frac{1}{4}\right) \\ {\overline{)=\frac{4\mathrm{\pi }}{3}{\left(\mathrm{x}+\frac{1}{4}\right)}^{3/2}}}_0^2 \\ =\frac{13\mathrm{\pi }}{3} \end{gathered}$$

Thus, the value of curved area of the solid is$\frac{13\mathrm{\pi }}{3}$ .