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The volume of the solid generated by revolving the region bounded by the given curve and lines y = 2x, y = 2, x = 0 about the x-axis will be
V = 8π/3
When the integration is done in the three dimensions in the three coordinates x,y and z then we will call it as a volume integration.
I set y=2x = y=2 to find where the intersect each others so I can have my boundaries for integration.
You goal is to find the area so you can integrate around that area. We're revolving around the x-axis so the area will be a circle.
V = ∫A(x)dx = ∫(πr²)dr
Since we have two different radius, we subtract them from each others.
∫(πr₂² - πr₁²)dr
∫(π(2)² - π(2x)²)dr
∫(4π - 4πx²)dr
4π∫(1 - x²)dr
integrate from 0 to 1 since that's where our boundary is.
V = 4π∫(1 - x²)dr = 8π/3
Hence the volume of the solid generated by revolving the region bounded by the given curve and lines y = 2x, y = 2, x = 0 about the x-axis will be V = 8π/3
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