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Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis. y=1/x^4, y=0...Asked by Anonymous
Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis.
y=x^2,y=1; about y=7
y=x^2,y=1; about y=7
Answers
Answered by
Damon
one way
x = + y^1/2 , double answer for the negative x part
2 int[ x 2 pi (7-y)dy] from y=1 to 7
4 pi int y^.5(7-y)dy
28 pi int y^.5 dy - 4 pi int y^1.5 dy
28 pi {y^1.5/1.5} - 4 pi y^2.5/2.5
at 7
(28/1.5)pi 7^1.5 -(4/2.5)pi 7^2.5
at 1
(28/1.5)pi -(4/2.5)pi
subtract at 1 from at 7
x = + y^1/2 , double answer for the negative x part
2 int[ x 2 pi (7-y)dy] from y=1 to 7
4 pi int y^.5(7-y)dy
28 pi int y^.5 dy - 4 pi int y^1.5 dy
28 pi {y^1.5/1.5} - 4 pi y^2.5/2.5
at 7
(28/1.5)pi 7^1.5 -(4/2.5)pi 7^2.5
at 1
(28/1.5)pi -(4/2.5)pi
subtract at 1 from at 7
Answered by
Steve
Since the region is symmetric, we can just double the volume from x=0 to x=1.
using shells of thickness dy,
v = 2∫[0,1] 2πrh dy
where r=7-y and h=x=√y
v = 4π∫[0,1] (7-y)√y dy = 256π/15
using discs of thickness dx,
v = 2∫[0,1] π(R^2-r^2) dx
where R=7-y and r=6
v = 2π∫[0,1] ((7-x^2)^2-36) dx = 256π/15
using shells of thickness dy,
v = 2∫[0,1] 2πrh dy
where r=7-y and h=x=√y
v = 4π∫[0,1] (7-y)√y dy = 256π/15
using discs of thickness dx,
v = 2∫[0,1] π(R^2-r^2) dx
where R=7-y and r=6
v = 2π∫[0,1] ((7-x^2)^2-36) dx = 256π/15
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