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:

2

x^{2}=x+ 12

x^{2}−x− 1 = 0(2

x+ 1)(x− 1) = 0

x= 1 (since we only need to consider≥ 0. This is consistent with what we see in the graph above.)x

So with *y*_{2} = *x* + 1 and *y*_{1} = 2*x*^{2}, 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 [(x^3)/(3)+x^2+x-(4x^5)/(5)]_0^1`

`=pi[(1/3 + 1+1 - 4/5)-(0)]`

`=pi[(5+30-12)/(15)]`

`=(23pi)/(15)`

`=4.817\ text[cm]^3`

Here's an illustration of the volume we have found. A typical "washer" with outer radius *y*_{2} = *x* + 1 and inner radius *y*_{1} = 2*x*^{2} is shown.

The cup resulting from rotating the area bounded by `y = 2x^2`, `y = x+1`, and `x = 0` about the `x`-axis.

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