V = π∫(y^2) dx from a to b
= π∫x^4 dx from 0 to 3
= π[x^5/5] from 0 to 3
= π(243/5 - 0) = 243π/5 cubic units
Find the volume of the solid obtained by rotating the region bounded by y= x^2, y=0, and x=3 about the x-axis.
2 answers
or, using shells, you can do
V = ∫[0,9] 2πrh dy
where r = y and h = 3-x
= 2π∫[0,9] y(3-√y) dy
= 2π(3/2 y^2 - 2/5 y^(5/2)) [0,9]
= 2π(3/2 * 81 - 2/5 * 243)
= 2π(243/10)
= 243π/5
V = ∫[0,9] 2πrh dy
where r = y and h = 3-x
= 2π∫[0,9] y(3-√y) dy
= 2π(3/2 y^2 - 2/5 y^(5/2)) [0,9]
= 2π(3/2 * 81 - 2/5 * 243)
= 2π(243/10)
= 243π/5