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[FD3S-R]

Cambridge. Exercise 6.1
q7

Use the method of slicing to find the volume of the solid obtained by rotating the region:
enclosed within the circle (x-1)^2 + y^2 = 1 about the y-axis

more stuck with the integration, prob aint too hard or easy im just missing something

 

[Dumsum]

(x-1)^2 + y^2 = 1
x^2 - 2x + 1 + y^2 = 1
x^2 - 2x + y^2 = 0

x₂ + x₁ = 2, x₂x₁ = y^2

(x₂ - x₁)^2 = (x₂ + x₁)^2 - 4x₂x₁
= 4 - 4y^2

x₂ - x₁ = 2sqrt(1-y^2)

δV = 4pi.sqrt(1-y^2).δy

V = 4pi Int{-1 -> 1} sqrt(1-y^2).dy

let y = sin@
dy = cos@.d@

when y = -1, @ = -pi/2
when y = 1, @ = pi/2

V = 4pi Int{-pi/2 -> pi/2} sqrt(1-sin^2 @).cos@.d@
= 4pi Int{-pi/2 -> pi/2} cos^2 @ . d@
= 4pi (@/2 + (sin2@)/4) {-pi/2 -> pi/2}
= 4pi (pi/4 + pi/4)
= 2pi^2 cubic units.

Trig substitutions will usually work wonders when there are square roots and whatnot.

 

[FD3S-R]

sorry redid that question was real easy made stupid mistakes

anyway this problem always occurs with me with questions that use isoceles triangles

5. Cambridge exercise diagnostic test
The base of a particular solid is the circle x^2 + y^2 - 9. Find the volume of the solid if every cross-section perpendicular to the x-axis is an isosceles right angled triangle with the hypotenuse in the base of the solid.


anyhelp deeply appreciated

 

[KFunk]

Here you can use the area of the triangle = 1/2absinθ or just half base times height since it is a right angled triangle.

We have to formulate the area of each triangle in terms of the hypotenuse (since the hypotenuse falls within the bounds of the circle).
Area of triangle = 1/2a.a = 1/2a² (where a is the length of the two same length sides of the isoc. Δ )

Using Pythagoras a² + a² = b² ---> a=b/√2
hence Area Δ = b²/4

We then want the value of b as a function of x so convert x² + y² = 9 into x = √(9 - x²)
since b = 2.x (the above gives single x values of a semi circle -3≤x≤3 which is symmetrical about the x-axis hence we can multiply by2)
Area Δ in terms of x = 4/4(√(9 - x²))²
= 9 - x²

Now that we have the area of each triangular section for any given value of x we can integrate between -3 and 3 (in this case just integrate between 0 and 3 and multiply by 2 --> i.e use the symmetry of a circle to make your job easier)

(between 0 and 3) 2 ∫ 9 - x² dx
= 2[ 9x - 1/3.x³]
= 2[27 - 9]
= 2.18 = 36 units³

I Hope that helps you.