How do you find the volume bounded by #x-8y=0# & the lines #x+2y# revolved about the y-axis?

1 Answer
May 21, 2018

Volume of revolution =#188/3pia^3#

Explanation:

#x-8y=0#, PASSES through the origin
#x+2y=a# has the intercepts,
# x=a, and y=a/2#
The ihtersection of the two lines happen to be,
#8y+2y=a#
#y=a/10#
#x=8a/10#

^3Considering y axis to represent the height of the cone,
Area of cross section is given by
#A=pi(r)^2#
where,
r is the x coordinate.
For x-8y=0, from y=0 to y=a/10,
x=8y
A=pi(x)^2
A=pi(8y)^2
#A=64piy^2#
Volume1 = #intAdy=int_0^(a/10) 64piy^2dy=64pi/3y^3=64pi/3((a/10)^3-0^3)#
#Volume1=64/1000pi/3a^3#
For x=2y=1 from y=a/10 to y=a/2
#x=a-2y#
A=pi(x)^2
#A=pi(a-2y)^2#
Volume2=#intAdy=int_(a/10)^(a/2) pi(a-2y)^2dy#
#=-pi/2(a-2y)^3/3=-pi/6((a-2xx(a/2)^3-(a-2xxa/10)^3)#

#=-pi/6(a-2/8a^3-a+2/1000a^3)#
#=-pi/6(-2/8a^3+2/1000a^3)#

#=2(pi/6)(1/8-1/1000)a^3#
#1/8=125/1000#
#=pi/3(125/1000-1/1000)a^3#

Volume2=#124pi/1000a^3#

Volume of revolution = Volume1+Volume2
#=64/1000(pi/3)a^3+124/1000(pi/3)a^3#
#64+124=188#
Volume of revolution =#188/3pia^3#