## adnanchowdhury Group Title Integration question: http://d.pr/RTc6 2 years ago 2 years ago

How do you solve it?

how do i integrate: $π \int\limits_{0}^{6} (y ^{2}/12 +1)^{2} dy$

3. bahrom7893 Group Title

Rotate the rectangle first, then rotate the parabola itself, find individual volumes then subtract the volume of the solid generated by the parabola from the solid generated by the rectangle

4. bahrom7893 Group Title

and for that integral, just multiply it out.

5. bahrom7893 Group Title

And then integrate term by term that will make your life easy.

6. bahrom7893 Group Title

actually finding the vol of the solid generated by the rectangle is easy.

I've been taught to integrate like this when the limits are on the y axis: $\int\limits_{a}^{b} π x ^{2} dy$

8. bahrom7893 Group Title

It generates a cylinder with radius 6 and height 4

9. bahrom7893 Group Title

Yea but you don't really have to do that for the solid generated by the rectangle. If you rotate it around it generates a cylinder.

What rectangle are you referring to?

11. bahrom7893 Group Title

Imagine a line passing through (4,6) and (4,0)

12. bahrom7893 Group Title

that will form a rectangle right? with the line y = 6, x = 4, and x and y axes

yes, i see. But what do you after that?

14. TuringTest Group Title

you can do this with disk or shell method. Do you know both?|dw:1329324049040:dw|

15. bahrom7893 Group Title

So find the volume of the solid generated by that rectangle - it will be a cylinder

I haven't heard of the 'disk' or 'shell' method?

What do i do after working out the vol. of solid rectangle?

18. bahrom7893 Group Title

Find the volume generated by revolving the following area and then subtract it from the vol of the cylinder.

19. bahrom7893 Group Title

|dw:1328460177913:dw|

How would you integrate this? http://d.pr/2Eo2

21. bahrom7893 Group Title

Just expand and then integrate term by term, it's not hard.

without expanding? Like chain rule?

23. bahrom7893 Group Title

I'm not sure if you can. Does the question force you to integrate without expanding?

no but that is one technique i have to learn.

25. TuringTest Group Title

your formula is for a revolution about y, so it's not right

26. bahrom7893 Group Title

no that's a diff question

27. myininaya Group Title

yeah bahrom is right i'm pretty sure the only way to integrate that is by expanding it

28. myininaya Group Title

that is the only way i see anyways

29. bahrom7893 Group Title

I mean it's not u-sub, and I don't recognize any inverse formulas.. inverse tan won't work

30. bahrom7893 Group Title

and integration by parts is gonna give u a horrible mess

31. TuringTest Group Title

...but this integral is not supposed to represent the area in the above problem? separate Q you say?

32. myininaya Group Title

it is not so nasty to expand this it is the easiest method to expand

33. bahrom7893 Group Title

Yea turing, it's separate

34. bahrom7893 Group Title

I think expanding is the only way.

Basicallly, I ended up with this integral, by rearranging the given equation to make x the subject. Then substituting into this: http://d.pr/PR8z

The final answer is suppose to be 90π.

37. TuringTest Group Title

so it is supposed to be the same problem

38. TuringTest Group Title

...which means the integral is wrong

39. bahrom7893 Group Title

No, there are two diff problems here as far as i can tell, the integral is a completely diff question

40. TuringTest Group Title

No ,they just said they got the integral by subbing in for y, but they are integrating on the wrong axis with the wrong method

That is the formula i am given in the book for integrating on the y-axis. So, i can't be integrating on the wrong axis.

42. TuringTest Group Title

$2\pi\int_{0}^{6}y(\frac{y^2}{12}+1)dy$ is shell method, which is how I would do it

43. myininaya Group Title

$\pi \int\limits_{0}^{6}(\frac{y^4}{144}+\frac{y^2}{6}+1) dy=\pi(\frac{y^5}{5 \cdot 144}+\frac{y^3}{3 \cdot 6}+y)|_0^6$ $=\pi(\frac{6^5}{5 \cdot 144}+\frac{6^3}{3 \cdot 6}+6)=\pi(\frac{6^4 \cdot 6}{5 \cdot 144}+\frac{6^2}{3}+6)$ $=\pi(\frac{6^4 }{5 \cdot 24}+\frac{ 6 \cdot 6}{3}+6)=\pi(\frac{6^4}{5 \cdot 24} +6(2)+6)$ $=\pi(\frac{6 \cdot 6^3}{5 \cdot 24}+12+6)=\pi(\frac{6^3}{5 \cdot 4}+18)$ $=\pi(\frac{6 \cdot 6^2}{5 \cdot 4}+18)=\pi(\frac{3 \cdot 6^2}{5 \cdot 2}+18)$ $=\pi(\frac{3 \cdot 36}{10}+18)=\pi(\frac{12 \cdot 9}{10}+18)=\pi(\frac{108}{10}+18)$ so this is the same problem or different? lol i don't know anymore by the way getting the answer is hard without a calculator as you can see

44. TuringTest Group Title

shell method:|dw:1329325465495:dw|Look at the radius of each cylinder( y) and the height of each cylinder (x(y)) and we get$2\pi\int_{0}^{6}rhdr=2\pi\int_{0}^{6}y(\frac{y^2}{12}+1)=90\pi$

45. bahrom7893 Group Title

46. TuringTest Group Title

I am integrating along y because I used shell method, disk method would be a pain here because we would need two integrals

@myininaya did it right. I think you used the same method as me, but you expanded x^2 in the beginning? @bahrom7893 yes

I'm still quite confused about the shell method.

49. TuringTest Group Title

myin did the integral right, but it is not a solid about the x-axis

Yes it is. ?

51. TuringTest Group Title

no, hers are disks about the y-axis

52. TuringTest Group Title

|dw:1329325925783:dw|like this but centered on the y-axis

53. TuringTest Group Title

area of each disk is pi*r^2, but if r is x(y) we are going around the y-axis

Oh I see.

This is rotation 360 degree about y-axis?

56. TuringTest Group Title

if you want shell method for your integral it can be done, but you will need to split the integral in two, or subtract the second area as bahrom did disks of radius y(x)=6 between 0<x<1 rings of inner radius y(x)=sqrt(12x-1) and outer radius y=6 between 1<x<6

Ok, let me try.

Yes! I got it. I did what bahrom7893 told me to do. Find volume of cylinder made by rectange on this graph. Then subtact it by the volume between where the line cross x-axis and where the rectangle ends (at x =4).

59. TuringTest Group Title

that'll work :D

60. TuringTest Group Title

Shell method$2\pi\int_{0}^{6}y(\frac{y^2}{12}+1)dy$Disk method (my way):$\pi\int_{0}^{1}6^2dx+\pi\int_{1}^{6}6^2-(\sqrt{12(x-1)})^2dx$Disk method bahrom's way$\pi\int_{0}^{6}6^2dx-\pi\int_{1}^{6}(\sqrt{12(x-1)})^2dx$all should give the same answer. Try to think about the difference in the concept of each formula if you can, that will broaden your understanding.