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?

1 Answer
Mar 15, 2018

V=8pi volume units

Explanation:

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