volume enquiry (1 Viewer)

[sasquatch]

For this question:

{(x,y): 0<=x<=2, 0<=y<=2x-x²} about the y-axis

The method the book introduces in its example, does it by having two radii things, x₁ and x₂.

My first solution is:

∆V ≈ (πx₂² - πx₁²)∆y
= π(x₂² - x₁²)∆y
= π(x₂ - x₁)(x₂ + x₁)∆y

y = 2x - x²
x² - 2x + y = 0

x₁ + x₂ = 2
x₁x₂ = y

(Can someone answer why the sum and product of the roots are used, im just copying the working without understanding )

(x<sub>2[/sub - x1)² = (x2² -2x2x1+ x1²
= (x2[/sub + x1)² -4x2x1
= 4 - 4y
= 4(1-y)
(x2[/sub - x1) = 2(1-y)^1/2

so
∆V ≈ 4π(1-y)^1/2∆y

V = 4π * int(0,1) (1-y)^1/2dy
= 8π/3 units cubed

I think i have an alternate, can someone check it please:

ri = x
ro = 2-x

∆V ≈ (π(2-x)² - πx²)∆y
≈ π(4-4x + x² - x²)∆y
≈ π(4-4x)∆y
≈ 4π(1-x)∆y

x² - 2x + y = 0

x = [2 ± root(4-4y)] / 2
= [2 ± 2root(1-y)] / 2
= 1 ± root(1-y)

so

∆V≈ 4π(1-[1 ± root(1-y)])∆y
≈ 4π(1-1 ∓ root(1-y))∆y
≈ 4π(∓ root(1-y))∆y
≈ 4π(root(1-y))∆y (as volume is positive)

V = 4π * int(0,1) (1-y)^1/2dy
= 8π/3 units cubed

So yeah is my method ok? And also could somebody explain that sum/product of roots thingo.. Thanks!</sub>

 

[onebytwo]

when you apply the quad formula to the equation and you get you two x values take the bigger one as x2 and the smaller as x1 and put this into the
delta v = pi(x2 - x1)(x2 + x1), it should work out normally (or as ive learnt it anyway)
as for the sums and product thing, ive never come across it

 

[Riviet]

x₁ + x₂ = 2
x₁x₂ = y

(Can someone answer why the sum and product of the roots are used, im just copying the working without understanding )

From ∆V = π(x₂ - x₁)(x₂ + x₁)∆y,
We can clearly see why the sum of the roots are used as you can replace it with 2. As for the product of the two x's, the book has tried to get an expression for x₂-x₁ in terms of the sum and product of roots. The advantage of this is you can simply substitute these into ∆V and get an expression that will be ready to integrate straight away:

(x₂ - x₁)² = x₂² -2x₂x₁+ x₁²
= (x₂ + x₁)² -4x₂x₁
= 4 - 4y since x₂+x₁=2 and x₂x₁=y
= 4(1-y)
.'. (x₂ - x₁) = 2(1-y)^1/2 by taking the square root of both sides.

 

[sasquatch]

Um i kinda think you misinterperated my question, but thats ok cuz i figured it out anyway... hehe thanks..