Chapter 7.4, Problem 29E

To determine

To calculate: The region's area is equal $y=\sqrt[4]{x}$ from $x=0$ to $x=16$ .

Answer to Problem 29E

Wolfram Alpha will be used as there is no straightforward way to calculate previously integral.

  $L\approx 16.6469$

Explanation of Solution

Given Information: To determine the length of the specified curve

  $y=\sqrt[4]{x}$

For each $x\in \left[0,16\right]$ .

Calculation:

Use the well-known formula to get the smooth curve's length (page 421). Nevertheless, take note that the derivative

  $\begin{array}{c}\frac{dy}{dx}=\frac{d}{dx}\left(\sqrt[4]{x}\right)\\ =\frac{1}{4}{x}^{1/4-1}\\ =\frac{1}{4\sqrt[4]{x^3}}\end{array}$

Consider the given curve as a function of the $x$ in terms of $y$ since is not defined at $x=0$ . Therefore, $y\in \left[0,2\right]$ for any obtain $x=y^4$ .

Studying the $x=y^4$ in the interval $\left[0,2\right]$ because the given curve (1) is in the first quadrant.

Using the definition as a guide

  $\begin{array}{c}L={{\displaystyle \underset{0}{\overset{2}{\int }}\sqrt{1+\left(\frac{dx}{dy}\right)}}}^{2}dy\\ ={{\displaystyle \underset{0}{\overset{2}{\int }}\sqrt{1+\left(\frac{d}{dy}\left(y^4\right)\right)}}}^{2}dy\end{array}$

To evaluate the following derivative for the previous step:

  $\begin{array}{c}\frac{d}{dy}\left(y^4\right)=4y^{4-1}\\ =4y^{}\end{array}$

When returning to the formula

  $\begin{array}{c}L={\displaystyle \underset{0}{\overset{2}{\int }}\sqrt{1+{\left(4y^3\right)}^2}dy}\\ ={\displaystyle \underset{0}{\overset{2}{\int }}\sqrt{1+16y^6}}dy\\ \end{array}$

Since there is no straightforward way to calculate the prior integral, utilise the Wolfram Alpha function to evaluate this. Consequently, it is clear that $\left(1\right)$ is the length of the curve.

  $L\approx 16.6469$