y^2-x^2=1 or
x^2 - y^2 = -1 <----- vertical hyperbola, vertices at (0, ± 1)
y=2 intersect the upper part of the hyperbola at (-√3,2) and (√3,2)
because of the symmetry we could just go from 0 to √3, then double our answer.
Volume = 2π∫ (2^2 - y^2) dx from x = 0 to √3
= 2π∫(4 - 1 - x^2) dx from 0 to √3
= 2π[3x - x^3/3] from 0 to √3
= 2π(3√3 - 3√3/3 - 0)
= 4π√3
check my arithmetic , should have written it out on paper.
Find the volume of a solid obtained by revolving the region bounded by y^2-x^2=1 and the y=2 about the x-axis.
1 answer
1 answer