How do you use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region #y=x^(1/2)#, #y=0#, and #x=4# rotated about the x axis?

Question
Answer 1

You get: #y = x^2# #x = 0# #y = 4#

Now your domain is your range and your interval is #[0,2]#.

#V = 2piintxf(x)dx#

#f(x) = x^2 in [0,2]#, bounded from above by #y = 4#, thus #f(x) = 4-x^2#

#V = 2piint_0^2 x*(4-x^2)dx#

#= 2pi[2x^2 - x^4/4]|_(0)^(2)#

#= 2pi[(2(2)^2 - (2)^4/4) - cancel((2(0)^2 - (0)^4/4))]#

#= color(blue)(8pi "u") ~~ 25.1327 "u"#

Note: if you evaluated this using #f(x) = x^2# instead, you would coincidentally get the same answer, but for the wrong work, so if you did that on a test, you wouldn't get full credit.

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