2003 HSC Extenstion 2 Question 3 (c) (1 Viewer)

[Neborg67]

I need help in understanding how to do the 2003 maths extension 2 question 3 part (c). I generally know how to do volumes of solids of revolution, however I don't know what to do for this question because there are two possible x values or each y value and the question requires you to find the area of the annulus in terms of y. (you will need to look at the actual question to know what I mean).

I understand that somehow you have to get y in terms of x1 and x2, but i don't know how.

For those of you who have the CAMBRIDGE text book, there is a similar question with a worked example on page 182 (Example 3), however it is not clear how they manage to go from 1 stage to another. If you can help me with this question I should be able to work out the past HSC question.

The cambridge question is as follows:

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.

The worked solution is as follows: (note: x1 is x subscript 1 (same for x2))

The cross-section of each slice is an annulus with radii x1, x2, where x2 > x1 and x1, x2 are the roots of y = x(4 - x) considered as a quadratic equation in x.
The annulus has area
A = pi(x2 + x1)(x2 - x1).
The slice has volume dV = Ady.

y = x(4 - x)
x^2 - 4x + y = 0
therefore, x2 + x1 = 4, x1x2 = y (i dont see how they got to this stage)
(x2 - x1)^2 = (x2 + x1)^2 - 4x1x2
therefore, (x2 - x1) = squareroot(16 - 4y)
dV = pi(x2 + x1)(x2 - x1)dy
therefore, dV = 8pi (multiplied by) (squareroot(4 - y))dy

The rest just follows by integegrating the above to find a total volume of (128pi)/3

please help in understanding how they get to the third stage
(i.e. where x2 + x1 = 4, x1x2 = y)

 

[haboozin]

y = x(4 - x)
x^2 - 4x + y = 0
therefore, x2 + x1 = 4, x1x2 = y (i dont see how they got to this stage)

sum of roots, product of roots

 

[KFunk]

The cross-section of each slice is an annulus with radii x1, x2, where x2 > x1 and x1, x2 are the roots of y = x(4 - x) considered as a quadratic equation in x.
The annulus has area
A = pi(x2 + x1)(x2 - x1).
The slice has volume dV = Ady.

y = x(4 - x)
x^2 - 4x + y = 0
therefore, x2 + x1 = 4, x1x2 = y

please help in understanding how they get to the third stage
(i.e. where x2 + x1 = 4, x1x2 = y)

What they're saying is that x1 and x2 are the roots of the equation x^2 - 4x + y = 0. This makes sense since for (almost) every y value you have two solutions for x.

Using sum of roots: x1 + x2 = 4

Using product of roots: x1x2 = y

 

[Neborg67]

Ahhhhhhh, of course (stupid me). Thanks.

(I take it when you say almost, you mean where x = 2)

 

[KFunk]

I need help in understanding how to do the 2003 maths extension 2 question 3 part (c). I generally know how to do volumes of solids of revolution, however I don't know what to do for this question because there are two possible x values or each y value and the question requires you to find the area of the annulus in terms of y. (you will need to look at the actual question to know what I mean).

The equation of the curve is y = -3 + 4x - x². Rearranging you get:

x² - 4x + (y+3) = 0

... then treat it like a quadratic with the y acting as a constant as per the example you gave and use the quadratic equation to get:

x = 2 ± √(16 -4[y+3])/2

= 2 ± √(1 - y)

here you use a bit of logic. In this question 0 ≤ y ≤ 1 (these will be your limits of integration). Hence between these two values √(1-y) ≥ 0.

Hence 2 + √(1-y) ≥ 2 - √(1-y) which means that the left hand term is the point closest to the axis of rotation and the right hand term is the one furthest away i.e. they are the inner and outer radii of the annulus, respectively (I hope this helps out a bit for the two values thing).

 

[Neborg67]

The equation of the curve is y = -3 + 4x - x². Rearranging you get:

x² - 4x + (y+3) = 0

... then treat it like a quadratic with the y acting as a constant as per the example you gave and use the quadratic equation to get:

x = 2 ± √(16 -4[y+3])/2

= 2 ± √(1 - y)

here you use a bit of logic. In this question 0 ≤ y ≤ 1 (these will be your limits of integration). Hence between these two values √(1-y) ≥ 0.

Hence 2 + √(1-y) ≥ 2 - √(1-y) which means that the left hand term is the point closest to the axis of rotation and the right hand term is the one furthest away i.e. they are the inner and outer radii of the annulus, respectively (I hope this helps out a bit for the two values thing).

Cool i think i get it now. I just didn't figure that the some of roots and products of roots thing would work seeing that y is a variable, but i guess it does. (I originally thought it only worked for when y = 0). Funny, I don't think i ever learnt that. Maybe I forgot... pre HSC stress.

Once again thanks for your help!