vertex for the curve

Question

Find the volume of the solid generated when the region bounded by the curve y=(x-1)^2 and the line y=x+1 is revolved about the x – axis.

Answer 1

y=(x-1)² and y=x+1 intersect when:

x+1=(x-1)²=x²-2x+1,

x²-3x=0=x(x-3), so x=0 and 3, making the intersection points (0,1) and (3,4).

Between x=0 and x=3, the line y=x+1 is above the curve y=(x-1)². For any point x between 0 and 3, the height of the line above the x-axis is x+1, while the height of the curve is (x-1)². Note that (1,0) is the vertex of the curve, a parabola. If we take just the line and its rotation about the x-axis, the height above the x-axis (that is, the y value) is the radius of the disc produced by rotation. And its area would be πy². Give the disc a very small thickness dx and this becomes the volume πy²dx=π(x+1)²dx. So π∫₀³(x+1)²dx is the volume of the shape obtained by rotating just the line segment about the x-axis. Integration just sums the volumes of the thin discs over the interval [0,3].

Similarly, π∫₀³y²dx=π∫₀³(x-1)⁴​dx would be the volume of the shape obtained by rotating just part of the curve about the x-axis. This volume has to be "cut out" of the shape produced by rotating the line. Note that exactly at the vertex of the parabola there would be no cutout. So the volume of the remaining shape is:

π∫₀³(x+1)²dx-π∫₀³(x-1)⁴​dx=

π∫₀³[(x+1)²-(x-1)⁴]​dx=

π∫₀³[(x²+2x+1-(x⁴-4x³+6x²-4x+1)​]​dx=

π∫₀³(-x⁴+4x³-5x²+6x)​​dx=π[-x⁵/5+x⁴-5x³/3+3x²]₀³=72π/5=45.2389 cubic units (approx).