Question

Dr. Geek also owns some champagne flutes, which are shaped exactly like the shape obtained by revolving the graph of $y=\frac{\ln x}{x}$from $x=0.6$to $x=5$around the $x$-axis, as shown in the figure and measured in inches. Given that the volume of the shape obtained from revolving $f$around the $x$-axis on$[a,b]$can be calculated with the formula $\mathrm{\pi }{\int }_{\mathrm{a}}^{\mathrm{b}}{\left(\mathrm{f}\left(\mathrm{x}\right)\right)}^2\mathrm{dx}$about how much champagne can each glass hold?

Short answer

Each glass can hold $1.582i{n}^3$champagne.

Step-by-step solution

Step 1. Given information

$y=\frac{\ln x}{x}$ from$x=0.6$to$x=5$.

Step 2. The given formula is, src="data:image/svg+xml;base64,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Substitute the known values in the formula.

$\mathrm{\pi }{\int }_{\mathrm{a}}^{\mathrm{b}}{\left(\mathrm{f}\left(\mathrm{x}\right)\right)}^2\mathrm{dx}=\mathrm{\pi }{\int }_{\mathrm{a}}^{\mathrm{b}}{\left(\frac{\ln \mathrm{x}}{x}\right)}^2\mathrm{dx}$.

To solve the integral, let us use the integrate by parts method.

Integrating by parts, $\int f{g}^{\text{'}}=fg-\int f^{\text{'}}g$.

Here, $f={\ln }^{2}x,g^{\text{'}}=\frac{1}{x^2}$.

role="math" localid="1650637610294" $$\begin{gathered} {\mathrm{f}}^{\text{'}}=2\frac{\ln \mathrm{x}}{\mathrm{x}},\mathrm{g}=-\frac{1}{\mathrm{x}} \\ \mathrm{\pi }{\int }_{\mathrm{a}}^{\mathrm{b}}{\left(\frac{\ln \mathrm{x}}{\mathrm{x}}\right)}^2\mathrm{dx}=\mathrm{\pi }\left[-\frac{{\ln }^2\left(\mathrm{x}\right)}{\mathrm{x}}-2\int -\frac{2\ln \left(\mathrm{x}\right)}{{\mathrm{x}}^2}\mathrm{dx}\right] \\ =\mathrm{\pi }\left[-\frac{{\ln }^2\left(\mathrm{x}\right)}{\mathrm{x}}+2\left[-\frac{\ln \left(\mathrm{x}\right)}{\mathrm{x}}-\int -\frac{1}{{\mathrm{x}}^2}\mathrm{dx}\right]\right] \\ =\mathrm{\pi }\left[-\frac{{\ln }^2\left(\mathrm{x}\right)}{\mathrm{x}}-\frac{2\ln \left(\mathrm{x}\right)}{\mathrm{x}}+2\int \frac{1}{{\mathrm{x}}^2}\mathrm{dx}\right] \\ =\mathrm{\pi }\left[-\frac{{\ln }^2\left(\mathrm{x}\right)}{\mathrm{x}}-\frac{2\ln \left(\mathrm{x}\right)}{\mathrm{x}}-\frac{2}{\mathrm{x}}\right]+\mathrm{C} \\ =-\frac{\mathrm{\pi }\left({\ln }^2\left(\mathrm{x}\right)+2\ln \left(\mathrm{x}\right)+2\right)}{\mathrm{x}}+\mathrm{C} \end{gathered}$$

Step 3. Let us use the limit values to find the definite integral.

$$\begin{gathered} \mathrm{\pi }{\int }_{0.6}^5{\left(f\left(x\right)\right)}^{2}dx={\int }_{0.6}^5{\left(\frac{\ln x}{x}\right)}^{2}dx \\ ={\left[-\frac{\mathrm{\pi }\left({\ln }^2\left(\mathrm{x}\right)+2\ln \left(\mathrm{x}\right)+2\right)}{\mathrm{x}}\right]}_{0.6}^5 \\ =\left[\left(-\frac{\mathrm{\pi }\left({\ln }^2\left(0.6\right)+2\ln \left(0.6\right)+2\right)}{0.6}\right)-\left(-\frac{\mathrm{\pi }\left({\ln }^2\left(5\right)+2\ln \left(5\right)+2\right)}{5}\right)\right] \\ =1.582i{n}^3 \end{gathered}$$

Therefore, the required value is,$1.582i{n}^3$.