[jay_shark]

h(x)=x^2 and f(x)=+-sqrtx and notice that h(x) and f(x) intersect at (0,0) and (1,1)

g(x)= h(x)-f(x)
g(x)= x^2-sqrtx

So we want the integral(or area under the curve) of g(x) evaluated at x=0 and x=1 .

G(x) = 1/3*x^3 -2/3*x^(3/2) | x=0 and x=1
G(1) - G(0) = -1/3 - 0 = -1/3

So the area bounded by the two equations is 1/3 since it cannot be a negative number .

2) Use this formula :

V= pi* integral of f(x)^2dx |x=a to x=b

V=pi* integral of (3x-x^2)^2dx |x=0 to x=3

To solve for a and b I solved the equation 3x-x^2 =0
x(3-x)=0 , x=0 or x=3 .

V=pi* int [9x^2 -6x^3 + x^4)dx | x=0 to x=3
V=pi*[3x^3 - 6/4*x^4 + 1/5*x^5] |x=0 to x=3

Now just solve for V(3) - V(1) and we're done .