How do you find the volume of the solid obtained by rotating the region bounded by the curves #y= x^2 - 4 # and #y= 3x# and #x=0# about the y axis?

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Answer 1 · 2015-09-28T02:54:58

#136pi#

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Let's find intersection:

#x=y/3 => y=(y/3)^2-4#

#y^2/9-y-4=0#

#y^2-9y-36=0#
#y^2-12y+3y-36=y(y-12)+3(y-12)=0#
#(y-12)(y+3)=0 <=> y=12 vv y=-3#

#y=x^2-4 => x=sqrt(y+4)#
#y=3x => x=y/3#

#V=pi (int_-4^12 (sqrt(y+4))^2dy - int_0^12 (y/3)^2dy)#

#V=pi (int_-4^12 (y+4)dy - int_0^12 y^2/9dy)#

#V_1=pi [y^2/2+4y]|_-4^12#

#V_1=pi (144+48-8+16)=200pi#

#V_2=pi [y^3/27]|_0^12#

#V_2=64pi#

#V=200pi-64pi = 136pi#