Question

In Exercises 63–72, set up and solve a definite integral to find the exact area of each surface of revolution obtained by revolving the curve y = f(x) around the x-axis on the interval [a, b].

$f\left(x\right)=\sqrt{3x+1},[a,b]=[0,3]$

Short answer

The exact area of the surface of the revolution obtained by revolving the curve$f\left(x\right)=\sqrt{3x+1}$ around the x-axis on the interval $\left[0,3\right]$is$\frac{\mathrm{\pi }}{18}\left(343-13\sqrt{13}\right).$

Step-by-step solution

Step 1. Given Information.

The given curve is $f\left(x\right)=\sqrt{3x+1}$and the interval is$\left[0,3\right].$

Step 2. Find the exact area.

To find the area, we will use the formula of surface area as a definite integral which is$S=2\pi {\int }_a^{b}f\left(x\right)\sqrt{1+{(f\text{'}(x\left)\right)}^2}dx.$

So,

$$\begin{gathered} S=2\pi {\int }_0^3\sqrt{3x+1}\sqrt{1+{\left(\frac{3}{2\sqrt{3x+1}}\right)}^2}dx \\ S=2\pi {\int }_0^3\sqrt{3x+1}\sqrt{1+\frac{9}{4\left(3x+1\right)}}dx \\ S=\pi {\int }_0^3\sqrt{12x+13}dx \\ S=\pi {\left[\frac{2}{3}{\left(12x+13\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}·\frac{1}{12}\right]}_0^3 \\ S=\frac{\mathrm{\pi }}{18}\left[{\left(49\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-{\left(13\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right] \\ S=\frac{\mathrm{\pi }}{18}\left(343-13\sqrt{13}\right) \end{gathered}$$

Thus, the exact area is$\frac{\mathrm{\pi }}{18}\left(343-13\sqrt{13}\right).$