1.Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Then find the area of the region. y=6x^2, y=x^2+2
2. Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis.
about the yaxis.
y=x^2, y=1, x=0, x=1
2. Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis.
about the yaxis.
y=x^2, y=1, x=0, x=1

1. both functions are even with respect to x. By leting 6x^2 = x^2 +2, we can get x = +/ sqrt(2/5) as the intersecting point
we can integrate x^2+2  6x^2 from 0 to sqrt(2/5) and double this to get the area
which is 2x  5/3 x^3 = 2*sqrt(2/5)  5/3*2/5*sqrt(2/5) = 2/3 * sqrt(2/5)
so the area is 4/3 sqrt(2/5)
2. the volume is calculated by integrating
pi x^2 dy for y from 0 to 1
or pi y dy = pi/2 y^2 = pi/2
we can integrate x^2+2  6x^2 from 0 to sqrt(2/5) and double this to get the area
which is 2x  5/3 x^3 = 2*sqrt(2/5)  5/3*2/5*sqrt(2/5) = 2/3 * sqrt(2/5)
so the area is 4/3 sqrt(2/5)
2. the volume is calculated by integrating
pi x^2 dy for y from 0 to 1
or pi y dy = pi/2 y^2 = pi/2

1/
intersections of 2 graphs
6x^2 = x^2 + 2 > x = sqrt(2/5) , sqrt(2/5)
area = int of (2  5x^2 )dx from x = sqrt(2/5) to sqrt(2/5)
2/
V = pi int of y dy , from y = 0 to 1
intersections of 2 graphs
6x^2 = x^2 + 2 > x = sqrt(2/5) , sqrt(2/5)
area = int of (2  5x^2 )dx from x = sqrt(2/5) to sqrt(2/5)
2/
V = pi int of y dy , from y = 0 to 1