How do you find the volume bounded by #x^2y^2+16y^2=6# and the x & y axes, the line x=4 revolved about the x-axis?

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Answer 1 · 2016-11-16T14:34:00

#3/8pi^2#

= 3.7011 cubic units, nearly.

The graph of #x^2y^2+16y^2=6# is inserted.

In #Q_1#, The curve meets x = 0 ( y-axis ) at #(0, 1/2sqrt(3/2))# and

the parallel x = 4 at #( 4, sqrt3/4 ).#

The volume is

#V = pi int y^2 dx#, from x = 0 to x = 4

#=pi int 6/(x^2+4^2) dx#, between the limits

#=6pi(1/4)[tan^(-1)(x/4)]#, between the limits

#=3/2pi[tan^(-1)1-tan^(-1)0]#

#3/2pi(pi/4)#

#=3/8pi^2#

= 3.7011 cubic units, nearly.-

graph{x^2y^2+16y^2-6=0 [-10, 10, -5, 5]}