posted by rachel .
the area under the curve y= square root of 9-x^2 with coordinates x=-3 and 3 , find the volume rotated aroudn the x-axis
Hey, that's just the top half of the circle
x^2 + y^2 = 9 rotated, giving us the whole sphere
radius is 3,so the volume is (4/3)π(3^3) = 36π
But I guess you want it done using integration....
volume π∫(9-x^2) dx from -3 to 3
= π[9x - (1/3)x^3] from -3 to 3
= π(27 - 9 - (-27 + 9))