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How do you find the volume of the solid generated when the regions bounded by the graphs of the given equations #y = 2/sqrtx, x=1, x=5# and the #x#-axis are rotated about the #x#-axis?

Answer 1

Isabella Ackerman

#V=4ln5pi#

The volume of revolution, #V#, given by a curve #y=f(x)# rotated about the #x#-axis between #b# and #a# is given by:

#V=piint_a^by^2# #dx#

Here #y=2/sqrtx rArry^2=4/x#, #b=5# and #a=1#

#thereforeV=piint_1^5 4/x# #dx=pi[4ln|x|]_1^5-4ln5pi-0=4ln5pi#

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Answer 2

Cameron Alexander

To find the volume of the solid generated by rotating the region bounded by the graphs of the equations (y = \frac{2}{\sqrt{x}}), (x = 1), (x = 5), and the x-axis about the x-axis, you would use the method of cylindrical shells.

The volume can be calculated using the integral:

[ V = 2\pi \int_{1}^{5} x \left(\frac{2}{\sqrt{x}}\right) dx ]

Solving this integral yields the volume of the solid generated by the rotation.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.