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Find the volume of the solid obtained by revolving the region bounded by the graphs of the functions about the x-axis. Hint: You will need to evaluate two integrals. (Assume x > 0.) y = 1/x , y = x, and y = 2x

Answer :

LammettHash

Using the washer method, the volume is given by

[tex]\displaystyle\pi\left(\int_0^{1/\sqrt2}((2x)^2-x^2)\,\mathrm dx+\int_{1/\sqrt2}^1\left(\left(\frac1x\right)^2-x^2\right)\,\mathrm dx\right)[/tex]

and using the shell method, it is

[tex]2\pi\displaystyle\left(\int_0^1y\left(y-\frac y2\right)\,\mathrm dy+\int_1^{\sqrt2}y\left(\frac1y-\frac y2\right)\,\mathrm dy\right)[/tex]

Either integral gives a volume of

[tex]\dfrac{4(\sqrt2-1)\pi}3[/tex]

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