How do I calculate the volume of the revolution about the x-axis for this question?

[commaified]

Find the volume of the solid of revolution obtained by revolving the region bounded by the graphs of the functions about the x-axis:

y = 1/x, y =x, y = 2x

The answer to the question is [64x^(1/2)]/3 pi units^3

 

[Deleted member 4993]

Find the volume of the solid of revolution obtained by revolving the region bounded by the graphs of the functions about the x-axis:

y = 1/x, y =x, y = 2x

The answer to the question is [64x^(1/2)]/3 pi units^3

I would sketch the bounding curves and determine the points of intersections.

Next I would decide whether I want to use disk method of washer method .

Continue ....

 

[blamocur]

Find the volume of the solid of revolution obtained by revolving the region bounded by the graphs of the functions about the x-axis:

y = 1/x, y =x, y = 2x

The answer to the question is [64x^(1/2)]/3 pi units^3

The answer contains an unknown [imath]x[/imath], but I would expect the answer to be a number or an expression with constants only.

 

[TorioSeduto]

First I calculate the intersection points of the three graphs and, doing the calculations I find that they are: [math]x = 0 ;\,\,\, x =\frac{ \sqrt[]{2}}{2} ;\,\,\, x = 1[/math]I would use the Pappus – Guldinus theorem:

[math]V \,=\, 2\pi \cdot \int_{\Omega}{}{y\,dy\,dx}[/math]
I divide the figure into two parts in such a way as to calculate the following integral:

[math]V \,=\, 2 \pi \cdot \Biggl[ \int_{0}^{\frac{\sqrt{2}}{2}}{}\,\int_{x}^{2x}{y \,dy\,dx} \,+\, \int_{\frac{\sqrt{2}}{2}}^{1}{} \int_{x}^{\frac{1}{x}}{ y\,dy\,dx}\Biggr][/math]
Carrying out the calculations, the result obtained is:

[math]\frac{4 \pi}{3} \cdot (\sqrt{2} - 1)[/math]
Check the calculations for safety.