Q. 69 - Calculus

Question

The $x$-axis is rotated around on the interval $[-1,1]$ by the graph of the function $f\left(x\right)=e^{4x}$ Set up and evaluate the definite integral to ascertain the precise area of the surface of rotation.

Step-by-Step Solution

Answer

 The surface area determined by rotating the graph of $f\left(x\right)={e^4}^x$around the $x$-axis on the range $[1,-1]$ is

$\frac{\pi }{16}\left[4e^4\sqrt{16e^8+1}-4e^{-4}\sqrt{16e^{-8}+1}+\ln \left|\frac{4e^4+\sqrt{16e^8+1}}{4e^{-4}+\sqrt{16e^{-8}+1}}\right|\right]$

Step 1 : Given information

The function $f\left(x\right)=e^{4x}$ on the interval  $[-1,1]$

Step 2 : Calculation

Remember that the surface area of a solid of revolution is given by rotating a function's graph around the x-axis from point a to point b using a definite integral: 

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

Keep in mind that the function$f\left(x\right)$has a continuous derivative in the range $[-1,1]$ and is differentiable. Utilize the chain rule of differentiation to differentiate the function with regard to $x$.

$\begin{array}{rcl}f\left(x\right)& =& e^{4x}\\ f^{\text{'}}\left(x\right)& =& e^{4x}·4\\ & =& 4e^{4x}\end{array}$

In order to evaluate the resultant integral, use the derivative in the integral on the right side of the equation $\left(1\right).$

$\begin{array}{rcl}S& =& 2\pi {\int }^1{e}^{4x}\sqrt{1+{\left(4e^{4x}\right)}^2}dx\\ & =& 2\pi {\int }_{-1}^1{e}^{4x}\sqrt{1+16e^{8x}}dx\end{array}$

Considered as values as,

$\begin{array}{rcl}4e^{4x}& =& u\\ 16e^{4x}dx& =& du\\ e^{4x}dx& =& \frac{du}{16}\end{array}$

And substitute the values.

$\begin{array}{rcl}S& =& 2\pi {\int }_{e^{+}}^{4e^4}\sqrt{1+u^2}\left(\frac{1}{16}du\right)\\ & =& \frac{\pi }{8}{\int }_{4e^4}^{4e^4}\sqrt{1+u^2}du\\ & =& \frac{\pi }{8}{\left[\frac{u}{2}\sqrt{1+u^2}+\frac{1}{2}\ln \left|u+\sqrt{1+u^2}\right|\right]}_{4e^4}^{4e^4}\end{array}$

Further, reduce the aforementioned statement to obtain.

$$\begin{gathered} S=\frac{\pi }{16}\left[4e^4\sqrt{16e^8+1}+\ln \left|4e^4+\sqrt{16e^8+1}\right|-4e^{-4}\sqrt{16e^{-8}+1}-\ln \left|4e^{-4}+\sqrt{16e^{-8}+1}\right|\right] \\ =\frac{\pi }{16}\left[4e^4\sqrt{16e^8+1}-4e^{-4}\sqrt{16e^{-8}+1}+\ln \left|\frac{4e^4+\sqrt{16e^8+1}}{4e^{-4}+\sqrt{16e^{-8}+1}}\right|\right] \end{gathered}$$

Consequently, the surface area determined by rotating the graph of $f\left(x\right)={e^4}^x$ $$" width="9" style="max-width: none;" >around the$x$-axis on the range$[1,-1]$ is

$S=\frac{\pi }{16}\left[4e^4\sqrt{16e^8+1}-4e^{-4}\sqrt{16e^{-8}+1}+\ln \left|\frac{4e^4+\sqrt{16e^8+1}}{4e^{-4}+\sqrt{16e^{-8}+1}}\right|\right]$