Let R be the region in the first and second quadrants bounded above by the graph of #y=20/(1+x^2)# and below by the horizontal line y=2, how do you find volume of the solid generated when R is rotated about the x-axis?

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Answer 1 · 2016-07-04T19:20:47

# = pi(96 +400 arctan (3))#

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the small strip width dx as indicated will have volume when revolved about x axis given by

#dV = pi( y_2^2 - pi y_1^2) dx#

where #y_2 = 20/(1+x^2)# and #y_1 = 2#

so

#dV = pi( y_2^2 - y_1^2) dx#

#= pi( (20/(1+x^2))^2 - 2^2) dx#

#V = pi int_{-3}^3 dx qquad (20/(1+x^2))^2 - 4 pi int_{-3}^3 dx qquad star#

if we do the LHS of #star# first as an indefinite integral just to keep it a bit cleaner:

#V = pi int dx qquad (20/(1+x^2))^2 #

#x = tan psi, dx = sec^2 psi d psi#

#V =400 pi int dpsi qquad sec^2 psi(1/(1+tan^2 psi^2))^2 #

# =400 pi int dpsi qquad cos^2 psi #

from the double angle formula #cos 2 psi = 2 cos^2 psi - 1#
# =200 pi int dpsi qquad cos 2 psi + 1 #

# =200 pi ( 1/2sin 2 psi + psi + C )#

#sin 2 psi = 2 cos psi sin psi#

# =200 pi ( sin psi cos psi + psi + C )#

see this for converting back
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# =200 pi ( x/sqrt(1+x^2)*1/sqrt(1+x^2) + arctan x + C )#

# =200 pi ( x/(1+x^2) + arctan x + C )#

the RHS of #star# amounts to #-4pix#, so reinstating the integration interval we have

#V = pi [200 ( x/(1+x^2) + arctan x) -4x]_{-3}^3 #

# = pi(96 +400 arctan (3))#