Asked by Edward
Find volume of solid generated by revolving around region y=x^(1÷2) and y=2,x=0 about 1)x=0 2)y=2
Answers
Answered by
Steve
Using discs (washers), they stack up with thickness dy or dx, as needed.
(a) v = ∫[0,2] π(R^2-r^2) dy
where R=4 and r=x=y^2
= ∫[0,2] π(16-y^4) dy = 128π/5
(b) v = ∫[0,4] π(R^2-r^2) dx
where R=2 and r=2-y=2-√x
v = ∫[0,4] π(4-(2-√x)^2) dx = 40π/3
Or, you can use nested cylinders (shells). Then you have
(a) v = ∫[0,4] 2πrh dx
where r=x and h=y=√x
v = ∫[0,4] 2πx√x dx = 128π/5
(b) v = ∫[0,2] 2πrh dy
where r=2-y and h=4-x=4-y^2
v = ∫[0,2] 2π(2-y)(4-y^2) dy = 40π/3
(a) v = ∫[0,2] π(R^2-r^2) dy
where R=4 and r=x=y^2
= ∫[0,2] π(16-y^4) dy = 128π/5
(b) v = ∫[0,4] π(R^2-r^2) dx
where R=2 and r=2-y=2-√x
v = ∫[0,4] π(4-(2-√x)^2) dx = 40π/3
Or, you can use nested cylinders (shells). Then you have
(a) v = ∫[0,4] 2πrh dx
where r=x and h=y=√x
v = ∫[0,4] 2πx√x dx = 128π/5
(b) v = ∫[0,2] 2πrh dy
where r=2-y and h=4-x=4-y^2
v = ∫[0,2] 2π(2-y)(4-y^2) dy = 40π/3
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