## anonymous 4 years ago Calculus II - Volume By Shells Hello, having some problems with a volume by shells problem. I believe my issue is the setup. Question: Use the shells method to find the volume of a solid generated by revolving the region bounded by the lines y=x+6 and y=x^2 about the following lines; a) The line x=3 b) The line x=-2 c) The x-axis d) The line y=9 So - I convert; y = x+6 x = y-6 y = x^2 x = sqrt(y) Rest coming in an equation post -

1. anonymous

$2\pi \int\limits_{0}^{9} (3-x)(\sqrt(y) - (y-6))$ Is this the correct setup?

2. TuringTest

for a ?

3. anonymous

Yeah, sorry.

4. TuringTest

well I would set it up the other way, in terms of x...

5. anonymous

That was a typo, I meant to have (3-y)(sqrt(y) etc, etc...

6. TuringTest

|dw:1327780091605:dw|$2\pi\int_{-2}^{3}(3-x)[(x+6)-x^2]xdx$is the way I see it easiest...

7. anonymous

what letter are you working on now?

8. anonymous

A (Turings post above)

9. TuringTest

doing shell method here in terms of y seems strange to me, because the area is bonded by y=x^2 on both sides between 0<y<2

10. TuringTest

bounded*

11. TuringTest

sorry I had an extra x earlier$2\pi\int_{-2}^{3}(3-x)[(x+6)-x^2]dx$

12. anonymous

Yeah, that sure does look a lot easier. Every example I have seen if you are rotating around X (axis or line) then you get the equation in terms of Y. That is the reason I convert them... I have a feeling I'm telling myself incorrect rules.

13. anonymous

Going to try working it out, brb!

14. TuringTest

|dw:1327780376562:dw|look at the picture and you can see that the cylinders have radius r=(3-x) and height h=x+6-x^2 I see no reason to integrate with respect to y.

15. TuringTest

|dw:1327780500851:dw|

16. anonymous

Yeah, you're right. I really need to get better at drawing/interpreting a graph on every problem. I've been sort of lazy with them up to this point.

17. TuringTest

Practice, practice :D

18. TuringTest

as for part b and c you will just change the radius... part d I have to look at though.

19. anonymous

Bam! Got it! Thanks a lot for the help! I'm going to try to do the second one by myself, although I already know it will be insanely easy since we already have the setup.

20. anonymous

Yeah. Cool, I'll get b/c and then work on d and hopefully be back with the right answer :) Thanks again, mate!

21. anonymous

Hrm, rotating about the X axis everything is the same, but the radius is X, right? I got B no problem, but I'm having problems with C.

22. TuringTest

Yes to your last question^ I think we have to split the integral in part d up because as I mentioned, the area is bound by y=x^2 on both sides between y=0 and y=4, so$h=2\sqrt y,0\le y\le4$$h=\sqrt y-(y-6),4\le y\le9$$r=9-y$|dw:1327780922685:dw|$\int_{0}^{4}(9-y)(2\sqrt y)dy+\int_{4}^{9}(9-y)[\sqrt y-(y-6)]dy$

23. TuringTest

oh sorry, "no" to your last question archetype you could do the same with x for the radius if we were going about the y-axis

24. anonymous

Ah, thanks Turing. That's what i was thinking, but then I saw you say yes and I was, well.. confused. So for C its; $2\pi \int\limits_{0}^{9} y(\sqrt(y) - (y-6))$

25. TuringTest

@phi what do you mean the part to the left is double counted ? you mean the negative x portion?

26. TuringTest

@archetype but between$0\le y\le4$the height will not be what you have

27. TuringTest

ok, but do what radius do you mean? along the y axis the radius has no 'left' portion, just positive (up)|dw:1327781599409:dw|If you are referring to an earlier problem I think we have confirmed the answer already, right archetype?

28. TuringTest

that is supposed to be about y=9 above*

29. TuringTest

that should be symmetric, but the drawing sucks|dw:1327781872267:dw|

30. anonymous

I've got an answer for A and B (both 625Pi/6). I do not yet have C or D.

31. phi

sorry I was doing part C @Tur you are doing D correct

32. anonymous

When looking at the graphs do you ALWAYS use the original equation? For some reason I'm looking at the graph of x=y-6 and x=sqrt(y) to solve part C, but something is telling me I'm looking at the wrong graph.

33. TuringTest

cool, ok so for C the only thing we are missing is that fact that the height formula for each shell depends which interval you are in$h=2\sqrt y$for the interval$0\le y\le4$and$h=\sqrt y-(y-6)$for the interval $4\le y\le9$look at the picture carefully to see this

34. TuringTest

|dw:1327782089205:dw|see how the line y=4 divides our region in two? the height below it is 2sqrty, above it is sqrty-(y-6)

35. anonymous

Awesome! I graphed the two things out myself and was going to ask if it was 2sqrt(y) because of the other line intersecting at that point.

36. TuringTest

good so the integral will be divided by the interval as well:$2\pi\int_{0}^{4}y(2\sqrt y)dy+2\pi\int_{4}^{9}y[\sqrt y-(y-6)]dy$

37. anonymous

I find this stuff so insanely difficult, haha. I guess like you said, practice, practice, practice. This is one of those days I wish I was a savant. Or just good at math. I'd take good at math.

38. TuringTest

I'd like to think I'm good at math, but really it's just a matter of familiarity with the ways to visualize things

39. anonymous

Yeah - and I think that is a lot of my problem. Not much else in school requires visualization like Volumes of Solids. At least nothing that I have taken.

40. anonymous

Drawing that graph and putting the picture in like you did above really helped me.

41. TuringTest

these are really good notes http://tutorial.math.lamar.edu/Classes/CalcI/VolumeWithCylinder.aspx they even teach you how to do things other than just revolutions here if you read more than just this section

42. anonymous

Cool! Thanks! Well I really appreciate all of the help. Going to try to get my final answer on these guys.

43. TuringTest

no prob good luck!

44. anonymous

Sweeet, got them both. And I feel like I came away with a far better understanding of the setup. Thanks again, man, it was really awesome of you to help me out (mainly since we've been working together for over an hour).

45. TuringTest