Kilbaha!

[kenhung123]

This question is very difficult. I cannot do parts d and e

[Slumdawg]

What year is this from?

[kenhung123]

2006

[8039]

Isn't it directly telling you how to find the volume?

[kenhung123]

How did you do it?

[Slumdawg]

You need your terminals to be 1 and 26 cos it's dy not dx. Then you sub in 4(y-1) in the x position and do the normal integral stuff to find volume.. You need to use the equation they give you..

at the end your final equation should be V = 4pi[ (y^2)/2 - y] ²⁶₁...(so you just sub in 1 into y spot of the equation then minus the equation with 26 subbed in)  giving you -3926.99 which you make positive and round, to give you 3926 cubic units..

Hope that helps somewhat.

[8039]

How did you do it?

Well previous question asked you to find y in terms of x which is the first clue. Plug that into the given equation (negative or positive doesn't matter). Next we just have to figure out which terminals to use. Basically the x-value is the radius (half diameter) and the corresponding y-value is the ending terminal. So when radius is 0 the y value is 1, when x is the maximum radius of 10 the y value is 26. There's your a and b values.

For part d it has a radius of 3 therefore x=3, plug that into the equation you got in part a and find that the corresponding y-value is 3.25. The new terminals are simply 1-3.25


Btw this is all a very simple integration question. The reason for flipping it to it's side was just to use the y values as you would x values.

[kenhung123]

Thanks man

[kenhung123]

OK this is another one I struggled with. I just did getting A. ANyone know why its wrong?

[Slumdawg]

Is the answer B? If it is I'll explain it Probably isn't though haha.

[wildareal]

Yes this paper was very hard. The Jamaican Professors at Kilbaha in Kew are crazy!

[Slumdawg]

Well if the answer is B. Here's how I did it:

Firstly I put the two graphs in Y1 then saw where it was bounded just to figure out what I needed to do. I realised you need to get the integral of the abs(6x-11) graph from the points which the other graph intersects it. These points were 1.57 and 2.2 found using F5:intersection.

Then I went back to HOME screen. Typed integral (abs(6x-11),x,1.57,2.2) enter. Which gave .61136667 so that matches option B.

Hope that helps! Let me know if you're confused by the explanation or need more clarification.

[8039]

Is the answer B? If it is I'll explain it Probably isn't though haha.

I think it's C?

[Slumdawg]

Is the answer B? If it is I'll explain it Probably isn't though haha.

I think it's C?

I just checked the answers. It is infact B (this must be my lucky day with methods). Look at the above explanation if you're confused

[8039]

Oops I meant A. Which turns out is wrong anyway. Good work

Edit: hmm my method was similar in finding 1.57 to 2.2, then I went |x| minus |6x-11| and got an entirely different answer :/

Edit: More hmm, kilbaha says area from 1.57 to 2.2 for |x| is 0.576... my calculator says 1.18755  :idiot2: