How do you find the volume of the solid generated by revolving the region bounded by the graphs of the equations #y=sqrtx#, y=0, and x=4 about the y-axis?

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Answer 1 · 2018-03-15T09:38:58

V=#8pi# volume units

Essentially the problem you have is:

V=#piint_0^4 ((sqrtx))^2 dx#

Remember, the volume of a solid is given by:

V=#piint (f(x))^2 dx#

Thus, our original Intergral corresponds:

V=#piint_0^4(x) dx#

Which is in turn equal to:

V=#pi [ x^2/(2)]# between x=0 as our lower limit and x=4 as our upper limit.

Using The fundamental theorem of Calculus we substitute our limits into our integrated expression as subtract the lower limit from the upper limit.

V=#pi[16/2-0]#

V=#8pi# volume units