## anonymous one year ago Volume of a graph (Washer)

1. anonymous

Find the volume of $y=\sqrt{x}$ about x=4

2. anonymous

I started with $x=y^2$ so would that be the outer radius?

3. SolomonZelman

|dw:1441595115874:dw|roughly

4. SolomonZelman

x=y² is not the same as y=√x because it will have a twice larger volume

5. anonymous

I'm sorry I forgot to mention it is bounded by y=0 so just quadrant 1

6. SolomonZelman

oh, and then if it is rotated about x=4, then I will assume that this is where the √x region ends at?

7. anonymous

Yes

8. SolomonZelman

Oh, ok, so you know that your limits of integration are from 0 to 2. $$\large\color{black}{ \displaystyle \int_{0}^{2} \pi\left(y^2\right)dy }$$

9. SolomonZelman

So, you can tell that your radius is y², and it is from y=0 to y=2. THis is what I would do.

10. SolomonZelman

I missed it should e y^4

11. SolomonZelman

$$\large\color{black}{ \displaystyle \int_{0}^{2} \pi\left(y^2\right)^2dy }$$ because radius squared. Sorry

12. SolomonZelman

So that is just an integral of $$\pi$$y$$^4$$ from y=0 to y=2.

13. anonymous

Wow I tried this the first time and I guess I integrated incorrectly so I was so confused!

14. anonymous

Thank you

15. SolomonZelman

you can put integral into wolfram. what is important is to get a good practice of making a setup of the integral for volume. integration you know already....

16. anonymous

Absolutely

17. SolomonZelman

32π/5 is what i got. (want to know how to make a π • ÷ × √ with no latex?)

18. anonymous

Sure!

19. DanJS

here are some probs with the solns, pg3 has a nice little summary

20. anonymous

That will definitely be useful, thank you

21. SolomonZelman

I am glitching a bit

22. SolomonZelman

Ok, so here is going to be a short guide. The algorithm is: $$1)$$ Click ALT and hold it $$2)$$ Click the "Number Code" (on the numberpad on the right of the keyboard- if you got one) $$3)$$ Release the ALT ------------------------------ Number code Result 0, 2, 1, 5 × 2, 5, 1 √ 7 • 2, 4, 6 ÷ 2, 2, 7 π

23. SolomonZelman

there are some others too.... but these are useful examples

24. anonymous

So if it was x=6 would the outer radius be 2+y^2 and the inner radius be 2?

25. SolomonZelman

if it was x=6, |dw:1441596092489:dw|

26. anonymous

27. SolomonZelman

Do you mean, if it was a region of y^2 bound by y=0 and x=4, but rotated about x=6?

28. anonymous

Yes

29. SolomonZelman

30. SolomonZelman

31. anonymous

I thought I had to subtract the inner area?

32. SolomonZelman

yeah, my bad, I am overheating let me think

33. anonymous

|dw:1441596258832:dw|

34. anonymous

Haha no problem

35. SolomonZelman

|dw:1441596426856:dw|

36. anonymous

For this problem I can only use washer

37. SolomonZelman

Oh, you can do it with respect to x, and do f(x)-g(x)

38. SolomonZelman

But, shell is also good.

39. SolomonZelman

it is a matter of preference

40. anonymous

Would f(x) be y^2+2 and g(x) be 2?

41. anonymous

$\int\limits_{0}^{2} (2+y^2)^2-(2)^2$ ?

42. SolomonZelman

lets rvw the washer with x's

43. SolomonZelman

|dw:1441596741353:dw|

44. SolomonZelman

of that integral should say π INTEGRAL f(x)²-g(x)² dx

45. SolomonZelman

from a to b

46. anonymous

Yes I get that

47. anonymous

I just don't understand what g(x) would be in this situation

48. SolomonZelman

and with y, you get the same thing|dw:1441596949137:dw|

49. SolomonZelman

the area of the cylinder with h=2, r=2 is 8π

50. SolomonZelman

Or you can say that g is 2.

51. SolomonZelman

if instead of 4, you had some z(x) boundary for the region, then the radius for g would be 6-z(x) (of course from y=0 to y=2)

52. anonymous

Okay I think I understand this better now thank you

53. SolomonZelman

Anytime... thank you for refreshing me on these rotations:)

54. SolomonZelman

I mean, I really should tell you that Shell method rocks in so many cases, so try to use that as well. in any case, good luck!

55. anonymous

Actually, did I plug this in incorrectly? http://www.wolframalpha.com/input/?i=integrate+from+0+to+2+pi%28%282%2By%5E2%29%5E2-4%29

56. anonymous

57. SolomonZelman

for the problem where you want to know about the redion of y62 bound by y=0 and x=4, rotated about x=6....?

58. SolomonZelman

region*

59. SolomonZelman

y^2 (not y62)

60. anonymous

yes

61. SolomonZelman

|dw:1441597714083:dw|

62. SolomonZelman

I actually do not get why the answer isn't what you got in wolfram.

63. anonymous

hmmm

64. SolomonZelman

you can take the whole volume of radius y^2+2 with limits of y=0 to y=2, and subtract 8π cylinder in the middle, and you get precisely the same.

65. SolomonZelman

I have to go, it is almost 12am in my location. maybe I was answering a wrong question idk, but for what I asked, it should be 256π/15

66. SolomonZelman

Maybe my brain just shot down xD

67. anonymous

Hahaha

68. anonymous

But I really appreciated your time so thank you

69. SolomonZelman

I will look at it when I have time. Whatever I can do with my little knowledge:) gtg c(u)

70. anonymous

Bye!