Q. 65 - Calculus

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]. 

$f\left(x\right)=\sqrt{x}, [a,b]=[0,4]$

Step-by-Step Solution

Answer

The exact area of the surface of the revolution obtained by revolving the curve $f\left(x\right)=\sqrt{x}$ around the x-axis on the interval $\left[0,4\right]$ is $S=\frac{\mathrm{\pi }}{6}\left(17\sqrt{17}-1\right).$

Step 1. Given Information.

The given curve is $f\left(x\right)=\sqrt{x}$ and the interval is $\left[0,4\right].$

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,

$$\begin{gathered} S=2\pi {\int }_0^4\sqrt{x}\sqrt{1+{\left(\frac{1}{2\sqrt{x}}\right)}^2}dx \\ S=2\pi {\int }_0^4\sqrt{x}\sqrt{1+\frac{1}{4x}}dx \\ S=\pi {\int }_0^4\sqrt{1+4x}dx \\ S=\pi {\left[\frac{2}{3}{\left(1+4x\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}·\frac{1}{4}\right]}_0^4 \\ S=\frac{\mathrm{\pi }}{6}\left[{\left(17\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-1\right] \\ S=\frac{\mathrm{\pi }}{6}\left(17\sqrt{17}-1\right) \end{gathered}$$

Thus, the exact area is $S=\frac{\mathrm{\pi }}{6}\left(17\sqrt{17}-1\right).$