anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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amistre64
  • amistre64
i remember these :)
anonymous
  • anonymous
first, you should graph the given curves and find the points of intersection
anonymous
  • anonymous
Here's the graph
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anonymous
  • anonymous
I calculated that the intersections are at (10,3) and (10,9)
amistre64
  • 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
  • anonymous
Okay, what did you just do there ? lol
anonymous
  • anonymous
i think it's better not to move it...
amistre64
  • 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
  • amistre64
define the function for height, and radius using the given graph as a guide
anonymous
  • anonymous
|dw:1377638073740:dw|
anonymous
  • anonymous
oops...
amistre64
  • amistre64
yeah ... shells :)
anonymous
  • anonymous
the thickness is dy.
anonymous
  • anonymous
How did you get (y+dy)^2 - y^2 as your radius ?
anonymous
  • anonymous
I understand 10 - (1 +(y-6)^2) being your height
amistre64
  • 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
  • amistre64
r = 3 to 9 ... not 0 to 9 :/
anonymous
  • anonymous
|dw:1377638369210:dw|
anonymous
  • anonymous
i think you use 2ydy as the square of the radius.
amistre64
  • amistre64
|dw:1377638490040:dw| \[\int 2\pi rh\] define r and h in terms of y
anonymous
  • anonymous
yep... you get this integral:|dw:1377638607253:dw|
anonymous
  • anonymous
I originally put down r = y + 3.. was that wrong ?
amistre64
  • amistre64
r = y+3 is wrong
anonymous
  • anonymous
have a look
anonymous
  • anonymous
Okay so its just y then
amistre64
  • amistre64
yep: r = y, from 3 to 9
amistre64
  • amistre64
and h = x, but adjusted for the odd orientation
anonymous
  • 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
  • anonymous
(y+dy)^2 - y^2 = 2ydy+dy^2 -> 2ydy
anonymous
  • anonymous
@beeca_boo73 do you understand the problem now?
anonymous
  • 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
  • anonymous
i didn't compute but trust you did it correctly.
anonymous
  • anonymous
Nope, webassign says its still wrong. ):
anonymous
  • anonymous
hold on...
anonymous
  • anonymous
\[2\pi \int\limits_{3}^{9}y(9-(y-6)^{2})dy\] Is where I should have gone from, right ?
anonymous
  • anonymous
Thats from 3 to 9 by the way, don't know why it looks so funky
amistre64
  • amistre64
looks good from here
amistre64
  • amistre64
when y=3, h=0 when y=6, h=9 when y=9, h=0 it fits
anonymous
  • anonymous
Yeah, okay, but when I computed it it apparently came out wrong. Maybe I messed up on arithmetic somewhere
anonymous
  • 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
  • anonymous
-y^3
anonymous
  • anonymous
i get 2808*pi
anonymous
  • anonymous
oops... double that.
anonymous
  • anonymous
Oh, oops. Yeah, -y^3. And in the end I got 432(pi) and webassign says its correcy so yay
anonymous
  • anonymous
*correct
anonymous
  • anonymous
ok... my calcs must be off.
amistre64
  • amistre64
i was trying to latex it and broke the chrome :/
anonymous
  • anonymous
I suppose. 0.0 After integrating I got \[V= 2\pi[-y ^{4}/4+4y ^{3}-27y ^{2}/2] from 3-9\]
anonymous
  • anonymous
Anyways, thanks for the help guys ! I appreciate it (:
anonymous
  • anonymous
me too.
anonymous
  • anonymous
sorry, i messed up summing my terms! i got 432*pi as well.
anonymous
  • anonymous
Yay, we helped each other ! lol Little miscalculations are what mess me up like 50% of the time. XD
anonymous
  • anonymous
you're not the only one!
anonymous
  • anonymous
Haha yeah. Thanks again !

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