Question

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?

Answer 1

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