Q. 13 - Calculus

Question

Set up and solve definite integrals to find each volume, surface area, or arc length that follows. Solve each volume problem both with disks/washers and with shells, if possible .

The area of the surface obtained by revolving the curve $f\left(x\right)=\sin \left(\mathrm{\pi x}\right)$ around the $x$-axis on $[-1, 1]$.

Step-by-Step Solution

Answer

0

Step 1. Given Information.

The area of the surface obtained by revolving the curve $f\left(x\right)=\sin \left(\mathrm{\pi x}\right)$ around the $x$-axis on $\left[-1,1\right]$.

Step 2. Formulation.

Let $f\left(x\right)$ be a nonnegative smooth function over the interval $\left[a,b\right]$ . Then, the surface area of the surface of revolution formed by revolving the graph of $f\left(x\right)$ around the $x$-axis .

The surface area of the curve is given by

 $S=2\pi {\int }_a^{b}f\left(x\right)\sqrt{(f\text{'}{\left(x\right))}^2+1} dx$.

Step 3. Calculation.

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

Then, the function will be

$$\begin{gathered} S=2\pi {\int }_{-1}^1\sin \left(\mathrm{\pi x}\right)\sqrt{{\left(\mathrm{\pi cos}\left(\mathrm{\pi x}\right)\right)}^2+1} dx \\ S=2\pi {\int }_{-1}^1\sin \left(\mathrm{\pi x}\right)\sqrt{\left({\pi }^2{\cos }^2\left(\mathrm{\pi x}\right)\right)+1} dx \\ Let \cos \left(\mathrm{\pi x}\right)=t \Rightarrow -\mathrm{\pi sin}\left(\mathrm{\pi x}\right)\mathrm{dx} =\mathrm{dt} \\ \mathrm{If} \mathrm{x}=-1 , \mathrm{t}=-1 \\ \mathrm{x}=1 , \mathrm{t}=-1 \\ \mathrm{S}=-{\int }_{-1}^{-1}\sqrt{\left({\mathrm{\pi }}^2{\mathrm{t}}^2\right)+1} \mathrm{dt} \\ \mathrm{S}=0 \end{gathered}$$