volumes (1 Viewer)

[want2beSMART]

1. {(x,y) : 0<=x<=2, 0<=y<= 2x-x^2} about th y-axis

2. the region bounded by the curve y= x(4-x) and the x-axis is rotated about the y-axis. find the volume of the solid of revolution by taking slices perpendicular to the y-axis

cheers

 

[FinalFantasy]

1. {(x,y) : 0<=x<=2, 0<=y<= 2x-x^2} about th y-axis

2. the region bounded by the curve y= x(4-x) and the x-axis is rotated about the y-axis. find the volume of the solid of revolution by taking slices perpendicular to the y-axis

cheers

1.
y=2x-x², x²-2x+y=0 ---> x=[2+-sqrt(4-4y)]\2
.: x=1+-sqrt(1-y)

delta-V=pi[(1+sqrt(1-y))²-(1-sqrt(1-y))²)delta-y
=4pi*sqrt(1-y) delta-y
.: v=4pi int. sqrt(1-y) dy from 0 to 1
=8pi\3 units ³

 

[FinalFantasy]

1. {(x,y) : 0<=x<=2, 0<=y<= 2x-x^2} about th y-axis

2. the region bounded by the curve y= x(4-x) and the x-axis is rotated about the y-axis. find the volume of the solid of revolution by taking slices perpendicular to the y-axis

cheers

2.y=x(4-x)=4x-x²
x²-4x+y=0
x=[4+-sqrt(16-4y)]\2
simplify, get radius like da above q. and do it

 

[gonz]

cant u just do it like this instead of finding the roots ...

@V=pi((x+@x)^2-x^2).(2x-x^2)
=pi(4x^2-x^4)@x <-----. then just form your integral with with boundries 2 and 0 and your done.

 

[want2beSMART]

i dont know how you got

x²-2x+y=0 ---> x=[2+-sqrt(4-4y)]\2

 

[KFunk]

He used the quadratic formula to get an equation in y, since you're integrating along the y-axis.

x²-2x+y=0

x = (-b &plusmn;&radic;[b² - 4ac])/2a where a= 1 , b = -2 , c= y



EDIT: The alternative method is to complete the square:

-y = x² - 2x
-y = x² - 2x + 1 - 1
-y = (x - 1)² - 1
1 - y = (x - 1)²

hence x = 1 &plusmn; &radic;(1 - y) ..... which is the same thing

 

[want2beSMART]

o0o0o0o i dont know this...