We sketch the upper and lower bounding curves:
Area bounded by `y = 2x^2` (the bottom curve), `y = x+1` (the line above), and `x = 0`, showing a typical rectangle.
The lower limit of integration is `x = 0` (since the question says `x ≥ 0`).
Next, we need to find where the curves intersect so we know the upper limit of integration.
Equating the 2 expressions and solving:
2x2 = x + 1
2x2 − x − 1 = 0
(2x + 1)(x − 1) = 0
x = 1 (since we only need to consider x ≥ 0. This is consistent with what we see in the graph above.)
So with y2 = x + 1 and y1 = 2x2, the volume required is given by:
`text[Volume]=pi int_0^1 [(x+1)^2-(2x^2)^2]dx`
`=pi int_0^1 [(x^2+2x+1)-(4x^4)] dx`
`=pi[(1/3 + 1+1 - 4/5)-(0)]`
Here's an illustration of the volume we have found. A typical "washer" with outer radius y2 = x + 1 and inner radius y1 = 2x2 is shown.
The cup resulting from rotating the area bounded by `y = 2x^2`, `y = x+1`, and `x = 0` about the `x`-axis.