A good way to solve such questions is understand the geometry
when the parabola is rotated around x axis , each point on the parabola will rotate in a circle. Thus the radius of every point on the parabola will be given by
r= 2-x for rotation. (Think about it, the point of origin will go till 4,0 when rotated, and so are other points according to 2-x)
The height of every point during rotation would be 2y
It can be seen that when the rotation would happen each point will move like a cylinder in 3d, whose area will be 2pirh
V = Adx
Integral V = 2pi*(2-x)* 2ydx
Integral V = 4pi*(2-x)*_/8x dx
Integral V = 8_/2pi [2_/x - x^(3/2)]
V = 8_/2pi [2/3* x^(3/2) - 2/5* x^(5/2)]
adding limits of x from 0 to 2
V = 16pi [8/3 - 8/5]
V = 256pi/15
The volume of solid generated by revolving the region bounded by y^2 = 8x and x=2 about the line x=2 is api/b. What is a+b
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@abhishek.exl can you solve this one? This is difficult for me
Can someone solve this?
can someone solve this?
can someone solve this?
A good way to solve such questions is understand the geometry
when the parabola is rotated around x axis , each point on the parabola will rotate in a circle. Thus the radius of every point on the parabola will be given by
r= 2-x for rotation. (Think about it, the point of origin will go till 4,0 when rotated, and so are other points according to 2-x)
The height of every point during rotation would be 2y
It can be seen that when the rotation would happen each point will move like a cylinder in 3d, whose area will be 2pirh
V = Adx
Integral V = 2pi*(2-x)* 2ydx
Integral V = 4pi*(2-x)*_/8x dx
Integral V = 8_/2pi [2_/x - x^(3/2)]
V = 8_/2pi [2/3* x^(3/2) - 2/5* x^(5/2)]
adding limits of x from 0 to 2
V = 16pi [8/3 - 8/5]
V = 256pi/15
@abhishek.pgp17158 can you explain this a little more. Have one doubt
@randominbox9 see this
@abhishek.pgp17158 can you please explain the part where you did the integration?
@randominbox9 see this
I solved this. This was a nice question.
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