How do you find the volume of the solid #y=sqrt(9-x^2)# revolved about the x-axis?
1 Answer
Feb 12, 2017
Volume
Explanation:
graph{(y-sqrt(9-x^2))=0 [-6, 6, -2, 4]}
The Volume of Revolution about
# V= int_(x=a)^(x=b) \ pi y^2 \ dx #
So for for this problem, Noting that
# V= 2int_0^3 \ pi (sqrt(9-x^2))^2 \ dx #
# \ \ \= 2pi \ int_0^3 \ (9-x^2) \ dx #
# \ \ \= 2pi \ [9x-x^3/3]_0^3#
# \ \ \= 2pi \ {(27-27/3) - (0-0)}#
# \ \ \= 36pi#