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