Q. 85 - Calculus

Question

One of Dr. Geek’s favorite beakers is exactly like the shape obtained by revolving the graph of

$y=2{\left(\frac{\ln x}{x}\right)}^{1/2}$

from $x=1 \mathrm{to} x=10$ around the x-axis, as shown in the figure and measured in inches. Given that the volume of the shape obtained by revolving f around the x-axis on [a,b] can be calculated with the formula $\pi {\int }_a^b(f{\left(x\right))}^{2}dx$, about how much liquid can the beaker hold?

Step-by-Step Solution

Answer

$33.22 \mathrm{cubic} \mathrm{inches}$ liquid the beaker can hold.

Step 1. Given Information

One of Dr. Geek’s favorite beakers is exactly like the shape obtained by revolving the graph of

$y=2{\left(\frac{\ln x}{x}\right)}^{1/2}$

from $x=1 \mathrm{to} x=10$ around the x-axis, as shown in the figure and  measured in inches. Given that the volume of the shape obtained by revolving f around the x-axis on [a, b] can be calculated with the formula $\pi {\int }_a^b(f{\left(x\right))}^{2}dx$, about how much liquid can the beaker hold?

Step 2. We have to calculate the formula π ∫ a b ( f ( x ) ) 2 d x

As we know 

$$\begin{gathered} y=f\left(x\right) \\ f\left(x\right)=2{\left(\frac{\ln x}{x}\right)}^{1/2} \end{gathered}$$

$$\begin{gathered} \pi {\int }_a^b(f{\left(x\right))}^{2}dx=\pi {\int }_1^{10}{\left[2{\left(\frac{\ln x}{x}\right)}^{1/2}\right]}^{2}dx \\ \pi {\int }_a^b(f{\left(x\right))}^{2}dx=\pi {\int }_1^{10}4\left(\frac{\ln x}{x}\right)dx \\ \pi {\int }_a^b(f{\left(x\right))}^{2}dx=4\pi {\int }_1^{10}\left(\frac{\ln x}{x}\right)dx \end{gathered}$$

Step 3. Using the substitution method.

Let

$$\begin{gathered} u=\ln x \\ \frac{du}{dx}=\frac{1}{x} \\ du=\frac{1}{x}dx \end{gathered}$$

Step 4. Now the integral is

$$\begin{gathered} \pi {\int }_a^b(f{\left(x\right))}^{2}dx=4\pi {\int }_1^{10}udu \\ \pi {\int }_a^b(f{\left(x\right))}^{2}dx=4\pi {\left[\frac{u^{1+1}}{1+1}\right]}_1^{10} \\ \pi {\int }_a^b(f{\left(x\right))}^{2}dx=4\pi {\left[\frac{u^2}{2}\right]}_1^{10} \\ \pi {\int }_a^b(f{\left(x\right))}^{2}dx=4\pi ·\frac{1}{2}{\left[{\left(\ln x\right)}^2\right]}_1^{10} \\ \pi {\int }_a^b(f{\left(x\right))}^{2}dx=2\pi {\left[{\left(\ln x\right)}^2\right]}_1^{10} \end{gathered}$$

Step 5. Now simplifying the integral.

$$\begin{gathered} \pi {\int }_a^b(f{\left(x\right))}^{2}dx=2\pi {\left[{\left(\ln x\right)}^2\right]}_1^{10} \\ \pi {\int }_a^b(f{\left(x\right))}^{2}dx=2\times 3.14\left[{\left(\ln 10\right)}^2-{\left(\ln 1\right)}^2\right] \\ \pi {\int }_a^b(f{\left(x\right))}^{2}dx=6.28\left[{(2.30)}^2-0\right] \\ \pi {\int }_a^b(f{\left(x\right))}^{2}dx=6.28\times 5.29 \\ \pi {\int }_a^b(f{\left(x\right))}^{2}dx=33.22 \end{gathered}$$