Hsc 2002 (1 Viewer)

[Premus]

Hey

can you guys help me do question 4 , Part a ) ii , iii from the 2002 HSC exam?

Thanks!!

 

[:: ck ::]

post up the q plz..

 

[Rorix]

What? You mean you don't have all the exam papers from the last 10 years at your fingertips! How do you plan on succeeding ?

OK, I went to all the effort of going to the BoS site..

ii) radius = 1+x
height = 3 - x^2 - x - x^2

you should be able to get it from there..

 

[CM_Tutor]

(i) P is at (1, 2)

(ii) δV = 2π(x + 1)(3 - x - 2x²)δx

So, V_solid = lim_δx-->0Σ (x=-1 to x=1) δV
= 2π * ∫ (-1 --> 1) (x + 1)(3 - x - 2x²) dx
= 2π * ∫ (-1 --> 1) 3 + 2x - 3x² - 2x³ dx

(iii) Evaluate the integral, which is straight forward, so you can do it yourself. I get 4π cu units, but that's without writing it out, so I could have made an error.

 

[Premus]

Hi

I dont understand how you guys worked out the radius and height???
i dont see how you got the radius and height from the two different curves? where did u start from??

i could evaluate the integral ....but i got 8 pi cu units.

THanks for ur help!!

 

[McLake]

Oooh, the 2002 paper. That brings back memories ...

The radius equals the x co-ordinate value + the distance from the x-axis to the bounding line (which is x = -1). So that = |x| + |-1| = x + 1
The height 3 - x^2 - (x + x^2) which is 3 - x^2 - x - x^2 which is 3 - x - 2x^2

So radius is (x + 1) and height is (3 - x - 2x^2)

Then follow CM Tutor's steps.

I get 8π too. Sorry CM.

 

[Premus]

Thanks a lot!

 

[CM_Tutor]

I get 8π too. Sorry CM.

So do I - it just goes to show that you shouldn't try to do 4u in your head - I just forgot the final factor of 2 when I changed it to
V = 2π * 3 * ∫ (0 --> 1) 1 - x² dx * 2 = 2π * 3 * (2 / 3) * 2 = 8π cu units.