How do you find the volume of the solid generated by revolving the region bounded by the curves $y = 2 x^{2}$ $y=2x^2$ ; y=0; x=2 rotated about the x-axis?

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Answer 1 · 2015-10-14T01:23:02

See the explanation below.

Here is the region and a thin slice taken perpendicular to the axis of rotation. (So we'll be using disks.)

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The representative disk (at $x$ $x$ ) has volume $π r^{2} \cdot thickness$ $pir^2*"thickness"$

In this case, radius $r = 2 x^{2}$ $r = 2x^2$ and thickness = $d x$ $dx$ .

$x$ $x$ varies from $0$ $0$ to $2$ $2$ , so we need

$\int_{0}^{2} π {(2 x^{2})}^{2} d x = 4 π \int_{0}^{2} x^{4} d x$ $int_0^2 pi(2x^2)^2 dx = 4pi int_0^2 x^4 dx$ which is straightforward to evaluate.

How do you find the volume of the solid generated by revolving the region bounded by the curves y=2x2y=2x^2; y=0; x=2 rotated about the x-axis?