The region bounded by the given curves is rotated about the specified axis. Find the volume of the resulting solid by any method.
y=5, y=x+(4/x); about x=-1
I know that the answer is 2π(12-4ln4) but I don't know how to get to that solution. Been working on it for 20 minutes with no luck. Please help!
y=5, y=x+(4/x); about x=-1
I know that the answer is 2π(12-4ln4) but I don't know how to get to that solution. Been working on it for 20 minutes with no luck. Please help!
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The enclosed region is a sort of semicircular region that intersects y= 5 at (1,5) and (4,5).
Using shell, integrate 2pi*R*H ,dx, for x from 1 to 4
R= x+1, since the rotaion axis is x=-1
H= distance from top to bottom= 5-(y)= 5-( x+4/x)
INT [1,4] [ 2pi(x+1)(5-x-4/x) ] dx
= 2pi* INT [ -x^2+ 4x +1 -4/x] dx
= 2pi[( -x^3)/3 + 2x^2 +x -4lnx] | [1,4]
= 2pi[( -64/3 +32 +4-4ln4)- (-1/3 + 2+ 1-0)]
= 2pi( 12- 4ln4)
Hoping this helps!
Using shell, integrate 2pi*R*H ,dx, for x from 1 to 4
R= x+1, since the rotaion axis is x=-1
H= distance from top to bottom= 5-(y)= 5-( x+4/x)
INT [1,4] [ 2pi(x+1)(5-x-4/x) ] dx
= 2pi* INT [ -x^2+ 4x +1 -4/x] dx
= 2pi[( -x^3)/3 + 2x^2 +x -4lnx] | [1,4]
= 2pi[( -64/3 +32 +4-4ln4)- (-1/3 + 2+ 1-0)]
= 2pi( 12- 4ln4)
Hoping this helps!