Use the method of cylindrical shells to find the volume V of the solid obtained by rotating the region bounded by the given curves about the x-axis.
x = 1 + (y − 6)2, x = 10

- anonymous

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- amistre64

i remember these :)

- anonymous

first, you should graph the given curves and find the points of intersection

- anonymous

Here's the graph

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- anonymous

I calculated that the intersections are at (10,3) and (10,9)

- amistre64

i wonder if moving it is a good idea ... prolly not at the moment tho
x = 1 + (y − 6)2, x = 10
x-10 = 1 + (y − 6)2, x = 10 - 10
x = 11 + (y − 6)2, x = 0
-x = 11 + (y − 6)2, -x = 0
x = -11 - (y − 6)2, x = 0

- anonymous

Okay, what did you just do there ? lol

- anonymous

i think it's better not to move it...

- amistre64

changing x for y at the moment only orients it normally
y = -11 - (x − 6)^2, y = 0
but yeah ... lets not try to move it this time :)

- amistre64

define the function for height, and radius using the given graph as a guide

- anonymous

|dw:1377638073740:dw|

- anonymous

oops...

- amistre64

yeah ... shells :)

- anonymous

the thickness is dy.

- anonymous

How did you get (y+dy)^2 - y^2 as your radius ?

- anonymous

I understand 10 - (1 +(y-6)^2) being your height

- amistre64

r = y = 0 to 9
h = 1-(y-6)^2 = 0 when y=3 and 9
1-(3-6)^2 = 1-9 = -8
1-(9-6)^2 = 1-9 = -8
so adding 8 seems appropriate to me
h = 1-(y-6)^2 + 8
or is that in error?

- amistre64

r = 3 to 9 ... not 0 to 9 :/

- anonymous

|dw:1377638369210:dw|

- anonymous

i think you use 2ydy as the square of the radius.

- amistre64

|dw:1377638490040:dw|
\[\int 2\pi rh\]
define r and h in terms of y

- anonymous

yep... you get this integral:|dw:1377638607253:dw|

- anonymous

I originally put down r = y + 3.. was that wrong ?

- amistre64

r = y+3 is wrong

- anonymous

have a look

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- anonymous

Okay so its just y then

- amistre64

yep: r = y, from 3 to 9

- amistre64

and h = x, but adjusted for the odd orientation

- anonymous

i did it right the first time... only difference is that dy^2 goes to 0 much much faster than dy and that's why it drops from the integral.

- anonymous

(y+dy)^2 - y^2 = 2ydy+dy^2 -> 2ydy

- anonymous

@beeca_boo73 do you understand the problem now?

- anonymous

Yeah, I got it, I think. I was just working it out. :b
So in the end I got a really big number.. lol
6912(pi)

- anonymous

i didn't compute but trust you did it correctly.

- anonymous

Nope, webassign says its still wrong. ):

- anonymous

hold on...

- anonymous

\[2\pi \int\limits_{3}^{9}y(9-(y-6)^{2})dy\] Is where I should have gone from, right ?

- anonymous

Thats from 3 to 9 by the way, don't know why it looks so funky

- amistre64

looks good from here

- amistre64

when y=3, h=0
when y=6, h=9
when y=9, h=0
it fits

- anonymous

Yeah, okay, but when I computed it it apparently came out wrong. Maybe I messed up on arithmetic somewhere

- anonymous

So from there I went to\[2\pi \int\limits_{3}^{9}y(9-(y ^{2}-12y+36))dy\]further simplified and finally got\[2\pi \int\limits_{3}^{9}(y ^{3}+12y-27y)dy\] before I integrated

- anonymous

-y^3

- anonymous

i get 2808*pi

- anonymous

oops... double that.

- anonymous

Oh, oops. Yeah, -y^3. And in the end I got 432(pi) and webassign says its correcy so yay

- anonymous

*correct

- anonymous

ok... my calcs must be off.

- amistre64

i was trying to latex it and broke the chrome :/

- anonymous

I suppose. 0.0
After integrating I got \[V= 2\pi[-y ^{4}/4+4y ^{3}-27y ^{2}/2] from 3-9\]

- anonymous

Anyways, thanks for the help guys ! I appreciate it (:

- anonymous

me too.

- anonymous

sorry, i messed up summing my terms! i got 432*pi as well.

- anonymous

Yay, we helped each other ! lol Little miscalculations are what mess me up like 50% of the time. XD

- anonymous

you're not the only one!

- anonymous

Haha yeah. Thanks again !

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