How do you find the volume bounded by x8y=0 & the lines x+2y revolved about the y-axis?

1 Answer
May 21, 2018

Volume of revolution =1883πa3

Explanation:

x8y=0, PASSES through the origin
x+2y=a has the intercepts,
x=a,andy=a2
The ihtersection of the two lines happen to be,
8y+2y=a
y=a10
x=8a10

^3Considering y axis to represent the height of the cone,
Area of cross section is given by
A=π(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=64πy2
Volume1 = Ady=a10064πy2dy=64π3y3=64π3((a10)303)
Volume1=641000π3a3
For x=2y=1 from y=a/10 to y=a/2
x=a2y
A=pi(x)^2
A=π(a2y)2
Volume2=Ady=a2a10π(a2y)2dy
=π2(a2y)33=π6((a2×(a2)3(a2×a10)3)

=π6(a28a3a+21000a3)
=π6(28a3+21000a3)

=2(π6)(1811000)a3
18=1251000
=π3(125100011000)a3

Volume2=124π1000a3

Volume of revolution = Volume1+Volume2
=641000(π3)a3+1241000(π3)a3
64+124=188
Volume of revolution =1883πa3