Can any math geniuses figure this problem out?

Question

What is the volume, in cubic length units, formed by revolving around the x-axis the area bounded by the curve y = (9 - x^2 )^1/2 and the-x axis.

Answer 1

The radius of the revolution is y.
y = (9 - x^2)^(1/2)
The cross section area of the revolved disk is A.
A = pi y^2 = pi (9 - x^2)
The differential volume element is dV.
dV = A dx = 9 pi dx - pi x^2 dx
Adding up the volume elements to get V, the volume of the solid of revolution.
V = integral[a,b]{ 9 pi dx } - integral[a,b]{ pi x^2 dx }
V = { 9 pi x - (pi/3) x^3 } |(a,b)
V = 9 pi (b-a) + (pi/3)(a^3 - b^3)
Assumed: a,b are the roots of y(x).
a = -3
b = +3
V = 9 pi (3+3) + (pi/3)[-27 - 27]
V = 54 pi - (1/3) 54 pi
V = (2/3)(54 pi)
V = 36 pi

Answer 2

can you see what the graph of the curve normally looks like? thats the important part. if youre not sure, square both sides of the equation and add x^2 to both sides. it looks like this:

x^2 + y^2 = 9

which is the equation of a circle with its center at (0,0) and a radius of 3

but . . . its only the top half of the circle, because the equation you were given is only the positive part of the square root, not +/- the square root.

rotate the top half of a circle in 3-D and what do you get? the top half of a sphere! so all you need to solve this problem is the equation for the volume of a sphere:

v = 4/3 * pi * r^2

plug in 3 for r, and dont forget to divide by 2 at the end because you only want the top half.

v = 4/3 * pi * 3^2
v = 4/3 * pi * 9
v = 12pi --> 6pi (or about 18.8) units^3

the end! hope that helps :)