First, solve for x as a function of y because we will be using the disc method for revolving a function about the y-axis.
xy2=4(2-x)
xy2=8-4x
xy2+4x=8
x(y2+4)=8
x= 8/(y2+4)
Note: Wyzant does not allow to right an improper integral from negative infinity to positive infinity so I will simply use ∫ symbol when I mean an improper integral from negative infinity to positive infinity.
Next use the disc method for revolving a function about the y-axis, we have the Volume is given by
V
=π∫x2dy
=π∫(8/(y2+4))2dy
=π∫64/(y2+4)2dy
=64π∫1/(y2+4)2dy.
Next note that integrated is an even function which means it is symmetric across the x-axis or equivalently x(-a)=x(a) for any number a, for example, x=y2 is an even function of y. The integral of an even function from −A to +A is twice the integral from 0 to +A. Feel free to reach out to me or look this up online if you are still confused. Then we have
V
=64π∫1/(y2+4)2dy
=64π (2∫0∞ 1/(y2+4)2dy)
=128π ∫0∞ 1/(y2+4)2dy.
Note that if we let y=2 tan(u) the denominator will turn into a single term by a Pythagorean identity. Then
dy/du=2sec2(u) so dy=2sec2(u)du. Next to find the bounds after making our substitution we have
y(0)=2tan(0)=2(0)=0 and Iim b→∞ of 2tan(b)=π/2. Therefore
V
=128π ∫0∞ 1/(y2+4)2dy.
=128π ∫0π/2 1/((2 tan(u))2+4)2(2sec2(u)du)
=128π ∫0π/2 1/(4 tan2(u)+4)2(2sec2(u)du)
=128π ∫0π/2 (2sec2(u))/(4 tan2(u)+4)2du
=128π ∫0π/2 (2sec2(u))/(4 (tan2(u)+1) )2du
=128π ∫0π/2 (2sec2(u))/(4 (sec2(u)) )2du
=128π ∫0π/2 (2sec2(u))/(16sec4(u)) du
=128π ∫0π/2 1/(8sec2(u)) du
=(128/8)π ∫0π/2 1/sec2(u) du
=16π ∫0π/2 1/sec2(u) du
=16π ∫0π/2 cos2 (u) du
=16π ∫0π/2 cos2 (u) du
=16π ∫0π/2 (1+cos(2u))/2 du ,(Using the power reduction formla)
=(16/2)π ∫0π/2 (1+cos(2u)) du
=8π ∫0π/2 (1+cos(2u)) du
=8π(u+0.5sin(2u)) |0π/2
=8π( ((π/2)+0.5sin(2(π/2)) - ((0)+0.5sin(2(0)) )
=8π( ((π/2)+0.5sin(π) - 0.5sin(0) )
=8π( ((π/2)+0.5(0)- 0.5(0) )
=8π( π/2)
=4π2
