Asked by Fred
Find the volume generated when the area bounded by the curve y^2 = 16x, from x = 0 to x = 4 is rotated around the x-axis.
Answers
Answered by
Steve
using discs:
v = ∫[0,4] πr^2 dx
where r=y
v = ∫[0,4] πy^2 dx
= π∫[0,4] 16x dx
= 16π∫[0,4] x dx
= 16π(1/2 x^2)[0,4]
= 16π(8-0)
= 128π
using shells:
v = ∫[0,8] 2πrh dy
where r = y and h = 4-x = 4-y^2/16
v = 2π∫[0,8] y(4 - y^2/16) dy
= 2π (2y^2 - y^4/64)[0,8]
= 2π(128 - 64)
= 128π
v = ∫[0,4] πr^2 dx
where r=y
v = ∫[0,4] πy^2 dx
= π∫[0,4] 16x dx
= 16π∫[0,4] x dx
= 16π(1/2 x^2)[0,4]
= 16π(8-0)
= 128π
using shells:
v = ∫[0,8] 2πrh dy
where r = y and h = 4-x = 4-y^2/16
v = 2π∫[0,8] y(4 - y^2/16) dy
= 2π (2y^2 - y^4/64)[0,8]
= 2π(128 - 64)
= 128π
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