Question
Find the volume of the solid generated by revolving the following region about the given axis. The region in the first quadrant bounded by the curve y=x^2, below by the x-axis, and on the right by the line x=1, about the line x=-2
Answers
Using shells,
v = Int(2πrh dx)[0,1]
where
r = x+2
h = y = x^2
2π*Int((x+2)x^2 dx)[0,1]
2π*Int(x^3 + 2x^2 dx)[0,1]
= 2π(1/4 x^4 + 2/3 x^3)[0,1]
= 2π(1/4 + 2/3) = 11π/6
v = Int(2πrh dx)[0,1]
where
r = x+2
h = y = x^2
2π*Int((x+2)x^2 dx)[0,1]
2π*Int(x^3 + 2x^2 dx)[0,1]
= 2π(1/4 x^4 + 2/3 x^3)[0,1]
= 2π(1/4 + 2/3) = 11π/6
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