Question

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

Answer 1

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`