5. Find the x coordinate of the centre of mass of the region under the function A(x) on the interval given. A(x) = πR²(x), where R(x) = 4x, on x = 0 to x = 2. (This corresponds to the centre of mass of a cone.)

Find a curve such that when the portion of the curve from x = a to x = b is rotated around the x-axis (for any a and b), the volume of the solid of revolution is equal to its surface area. You may assume dy/dx≥ 0

-2. Let S be the solid obtained by rotating the region bounded by the curves y = x + (9/x), y= 10 , about x = dx , where a = and 6 = The volume V = Therefore the volume V =

19