Q. 13 - Calculus

Question

The area of the surface obtained by revolving the curve

$f\left(x\right)=\sin \left(\pi x\right)$ around the x-axis on $\left[-1, 1\right]$.

Step-by-Step Solution

Answer

The area of surface obtained by revolving the given curve is $4\sqrt{1+{\mathrm{\pi }}^2}+\frac{4}{\mathrm{\pi }}\ln \left(\sqrt{1+{\mathrm{\pi }}^2}+\mathrm{\pi }\right)$

Step 1. Given information

The given curve is $f\left(x\right)=\sin \left(\mathrm{\pi x}\right)$ around the x-axis $\left[-1, 1\right]$ 

Step 2. differentiate given function

$$\begin{gathered} f\left(x\right)=\sin \left(\mathrm{\pi x}\right) \\ f\text{'}\left(x\right)=\mathrm{\pi cos}\left(\mathrm{\pi x}\right) \end{gathered}$$

The required surface area is given by

$$\begin{gathered} S=4\mathrm{\pi }{\int }_0^1\mathrm{f}\left(\mathrm{x}\right)\sqrt{1+{\left(\mathrm{f}\text{'}\left(\mathrm{x}\right)\right)}^2}\mathrm{dx} \\ =4\mathrm{\pi }{\int }_0^1\sin \left(\mathrm{\pi x}\right)\sqrt{1+{\mathrm{\pi }}^2{\cos }^2\left(\mathrm{\pi x}\right)}\mathrm{dx} \\ \mathrm{u}=\mathrm{\pi cos}\left(\mathrm{\pi x}\right) \\ \sin \left(\mathrm{\pi x}\right)\mathrm{dx}=\frac{1}{{\mathrm{\pi }}^2}\mathrm{du} \\ \mathrm{S}=4\mathrm{\pi }{\int }_{\mathrm{\pi }}^{-\mathrm{\pi }}\sqrt{1+{\mathrm{u}}^2}·\left(-\frac{1}{{\mathrm{\pi }}^2}\right)\mathrm{du} \\ =4\mathrm{\pi }{\int }_{\mathrm{\pi }}^{-\mathrm{\pi }}\sqrt{1+{\mathrm{u}}^2 \mathrm{du}} \\ =-\frac{4}{\mathrm{\pi }}{\left[\frac{\mathrm{u}}{2}\sqrt{1+{\mathrm{u}}^2}+\frac{1}{2}\ln \left(\left|\mathrm{u}+\sqrt{1+{\mathrm{u}}^2}\right|\right)\right]}_{\mathrm{\pi }}^{-\mathrm{\pi }} \\ =-\frac{2}{\mathrm{\pi }}\left[-\mathrm{\pi }\sqrt{1+{\mathrm{\pi }}^2}+\ln \left(\left|-\mathrm{\pi }+\sqrt{1+{\mathrm{\pi }}^2}-\mathrm{\pi }\sqrt{1+{\mathrm{\pi }}^2}-\ln \left(\left|\mathrm{\pi }+\sqrt{1+{\mathrm{\pi }}^2}\right|\right)\right|\right)\right] \\ =\ln \left(\mathrm{x}\right)-\ln \left(\mathrm{y}\right)=\ln \left(\frac{\mathrm{x}}{\mathrm{y}}\right) \\ \mathrm{S}=\frac{2}{\mathrm{\pi }}\left[2\mathrm{\pi }\sqrt{1+{\mathrm{\pi }}^2}+\ln \left(\left|\frac{\sqrt{1+{\mathrm{\pi }}^2}+\mathrm{\pi }}{\sqrt{1+{\mathrm{\pi }}^2}-\mathrm{\pi }}\right|\right)\right] \\ =4\sqrt{1+{\mathrm{\pi }}^2}+\frac{4}{\mathrm{\pi }}\ln \left(\mathrm{\pi }+\sqrt{1+{\mathrm{\pi }}^2}\right) \end{gathered}$$

Step 3. The solution

The area of surface obtained by revolving the graph of $f\left(x\right)=\sin \left(\mathrm{\pi x}\right)$ around x-axis on the interval $\left[-1, 1\right]$ is $4\sqrt{1+{\mathrm{\pi }}^2}+\frac{4}{\mathrm{\pi }}\ln \left(\sqrt{1+{\mathrm{\pi }}^2}+\mathrm{\pi }\right)$