Calculus volume of a solid

Volume of a solid:
y = .5x^2, y = (3/2)-x^2; about the x-axis.

Note that y = (1/2)x^2 and y = 3/2 - x^2 intersect when:
(1/2)x^2 = 3/2 - x^2 ==> (3/2)x^2 = 3/2 ==> x^2 = 1 ==> x = ±1.

So, we will be integrating from -1 to 1.

Here is the graph of the region between y = (1/2)x^2 and y = 3/2 - x^2:
http://www.wolframalpha.com/input/?i=y+%…

Since we are rotating a region bounded by two curves of the form y = f(x) about a horizontal line, we need to use the washer method.

Note that the distance between y = 0 and the lower curve, y = (1/2)x^2, is (1/2)x^2; this is the inner radius. The outer radius is the distance between y = 0 and y = 3/2 - x^2, which is just 3/2 - x^2. Therefore, the required volume is:
V = π ∫ {(3/2 - x^2)^2 - [(1/2)x^2]^2} dx (from x=-1 to 1)
= π ∫ [x^4 - 3x^2 + 9/4 - (1/4)x^4] dx (from x=-1 to 1)
= π ∫ [(3/4)x^4 - 3x^2 + 9/4] dx (from x=-1 to 1)
= π[(3/20)x^5 - x^3 + (9/4)x] (evaluated from x=-1 to 1)
= π[(3/20 - 1 + 9/4) - (-3/20 + 1 - 9/4)]
= 14π/5.

I hope this helps!