## ijlal 5 years ago http://www.xtremepapers.com/CIE/International%20A%20And%20AS%20Level/9709%20-%20Mathematics/9709_s10_qp_13.pdf help me in Question no.9 pls reply quick thank you :)

1. TuringTest

First we need the points A and B, know how to find that?

2. ijlal

yes i know that

3. ijlal

A has (1,5) and B(4,5)

4. ijlal

M(2,4)

5. TuringTest

right, so our integration will be from 1 to 4 what is the outer radius of each disk we will use in our method? what is the radius of the inner disk?

6. ijlal

i dont know how a cylinder is formed in this curve

7. TuringTest

we are making rings out of each section of area|dw:1327072712405:dw|the outer radius of each ring will be$r_0=5$the inner radius will be$r_i=x+\frac{4}{x}$the area of each disk wile therefor be$\pi r_0^2-\pi r_i^2\to\pi(r_0^2-r_i^2)$

8. TuringTest

|dw:1327073020023:dw|here is a cross-section of each ring|dw:1327073083149:dw|therefor adding up these rings we will integrate the areas of the rings from x=1 to x=4$\pi\int_{1}^{4}r_0^2+r_i^2dx=\pi\int_{1}^{4}5^2-(x+\frac{4}{x})^2dx$make sense?

9. TuringTest

*slight typo above, should be minus in the first integral (because we are subtracting the area of the inner disk from that of the outer)$\pi\int_{1}^{4}r_0^2-r_i^2dx=\pi\int_{1}^{4}5^2-(x+\frac{4}{x})^2dx$

10. ijlal

Yes the Volume of the ring is i guess 57pie

11. TuringTest

I guess, I haven't checked. I'll leave the integration for you to do. But your final answer is a volume of a shape, each section is a ring. (Just pointing out the discrepancy in calling the whole thing a ring)

12. ijlal

the answer of the question is 18pie i dont know how the cylinders height is 4units can you tell me that i will be done with my question done

13. TuringTest

which cylinder has a height of 4 units?

14. ijlal

there is only one cylinder formed

15. TuringTest

$\pi\int_{1}^{4}r_0^2+r_i^2dx=\pi\int_{1}^{4}5^2-(x+\frac{4}{x})^2dx=\pi\int_{1}^{4}25-x^2-8-\frac{16}{x^2}dx$$=\pi(17x-\frac{x^3}{3}+\frac{16}{x})|_{1}^{4}$$=\pi[(68-\frac{16}{3}+4)-(17-\frac{1}{3}+16)]$whatever that simplifies to... The whole shape will have a length of 3 units (from x=1 to x=4) if that is what you mean....

16. TuringTest

I got$\pi\int_{1}^{4}r_0^2-r_i^2dx=\pi\int_{1}^{4}5^2-(x+\frac{4}{x})^2dx=\pi\int_{1}^{4}25-x^2-8-\frac{16}{x^2}dx$$=\pi(17x-\frac{x^3}{3}+\frac{16}{x})|_{1}^{4}$$=\pi[(68-\frac{16}{3}+4)-(17-\frac{1}{3}+16)]=32\pi$and you say that you know the answer to be 18pi ?

17. ijlal

yes the answer is 18 pi it is given in the mark scheme of this question

18. ijlal

and this solution simplifies to 34pi not 32pi

19. TuringTest

Oh, I just made some silly arithmetic mistake apparently (this is why I left the integral for you to do) As you can see by wolfram it is set up right$\pi\int_{1}^{4}r_0^2-r_i^2dx=\pi\int_{1}^{4}5^2-(x+\frac{4}{x})^2dx$ http://www.wolframalpha.com/input/?i=int%20from%201%20to%204%20(25-(x%2B4%2Fx)%5E2)dx&t=crmtb01 so you figure out where I messed up...

20. ijlal

the solution of this integral gives 90pi

21. TuringTest

no, look at the link I posted. Wolfram does no lie: http://www.wolframalpha.com/input/?i=pi*int+from+1+to+4+%285%5E2-%28x%2B4%2Fx%29%5E2%29dx there it is explicitly, clearly the answer to the integral I wrote is 18pi

22. ijlal

oh sorry i neglected the minus sign by mistake :P thank you :)

23. TuringTest

very welcome :) I apparently made a similar mistake trying to do it by eye, so it happens.