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anonymous
 4 years ago
The region bounded above by the line, y = 6, below by the curve y=sqrt(x), and to the left by the yaxis. The solid is rotated along the following lines: C. The line y=6 D. The line x=36. These are two separate rotation axis and answers.
anonymous
 4 years ago
The region bounded above by the line, y = 6, below by the curve y=sqrt(x), and to the left by the yaxis. The solid is rotated along the following lines: C. The line y=6 D. The line x=36. These are two separate rotation axis and answers.

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amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1dw:1327531365366:dw

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1we need to know the point of intersection here

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Hey sorry, keyboard died. The point of intersection is (4,16).

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1really? i see it as (6,36) meself (y=6) = (y=sqrt(x)) 6 = sqrt(x) at x=36

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1327531523131:dw

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1your pic has way more parts then the question asks for

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0It's a rotation of y=6, but the shape is bounded by y=4, so the intersection of y=4 and y=sqrt(x) would be (16,4).

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1327531686984:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0So I'm thinking of using the cylindrical shell method, 2Pi(radius)(height)

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1ok, since typos happen; lets review what youve posted: The region bounded above by the line, y = 6, below by the curve y=sqrt(x), and to the left by the yaxis. The solid is rotated along the following lines: C. The line y=6 D. The line x=36

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yes, that is correct.

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1dw:1327531783399:dw

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1then this is our graph of that region

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1shell is fine, but we need to establish that this is our representation

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so you this is a graph for rotation around x=36?

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1yes, in fact its the same graph for rotate y=6 AND x=36

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1the inverse of y=sqrt(x) that we can use for this is x=y^2 that will give us the shell height for y=6 rotation

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so when rotating around a Yaxis, we should put the curve as Y too?

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1first we should simply draw out the graphs; after we have the graphs down, then we worry about where to rotate at

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1the graph itself has nothing to do with what we rotate around to begin with

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so the graph I had illustrated was inccorect

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0visualizing the graphs has been the toughest part for me!

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1dw:1327532250390:dwthis is our shell representing rotate about y=6

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1the graph you drew was something else altogther :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I HAVE NO CLUE WHY I PUT Y = 4!!

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1'sok :) so\[Area_{shell}=\int_{0}^{6}2pi\ y\ f(y)\ dy\] \[Area_{shell}=2pi\int_{0}^{6} y\ y^2\ dy\] \[Area_{shell}=2pi\int_{0}^{6} y^3\ dy\] we agree so far?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1327532422099:dw

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1i think i want to "move" our graph stuff to a better location casue I think im forgetting something; moving the graph changes no value, just positions

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Since we are rotating around y=6, wouldn't we do pi(6y)^2

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1dw:1327532555854:dw

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1ugh, yes, your right; been awhile :) x=(y6)^2

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so we could do the crosssection method

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1\[A=2pi\int_{0}^{6}y(y6)^2dy\] \[A=2pi\int_{0}^{6}y(y^212y+36)dy\] \[A=2pi\int_{0}^{6}(y^312y^2+36y)dy\] looks better to me for the shell method.

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1disc would be: \[A=pi\int_{0}^{36}(x^{1/2}6)dx\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1327532793666:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0what i posted above is for the crosssection, or it works for it..

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1\[A=pi\int_{0}^{36}(x^{1/2}6)^2dx\] forgot to r^2 the area :)

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1the disc gives me: 216 pi the shell gives me: 216 pi

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok so I'm still a little confused how we got there..

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1lets turn this on its side so it looks like an upsidedown cone; the math doesnt care how it looks :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.02pi(r)(h) 2pi\[2\pi \int\limits_{0}^{6} y(6y)\]

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1dw:1327533114400:dw

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1good, r = y and h=f(y), which is the inverse of f(x) which we moved down by 6

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1y=sqrt(x)6 y+6 = sqrt(x) (y+6)^2 = x = f(y) I used the other side of the line so I got (y6)^2 which makes no difference here

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1well, it does according to wolfram but .... fer some reason i did it right to begin with lol

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0hmm.. so what would be the equation we are trying to integrate and with what limits? 0 to 36?

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1Volume, not Area; named it wrong .... it bites getting old\[V=2\pi\int_{0}^{6}y(y+6)^2dy=216\pi\]

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1the 0 to 36 is for the disc method for this thing

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1i need to downlaod google chrome; this thing is just way too slow in IE

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0haha it runs great in chrome

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so why is the (y+6) being squared?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Thanks so much for your help, you are great!

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1college computers here clear out and reboot so any downloads are erased

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1w have to tell the integration to look at the the proper rendition of out equation for height

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ohhhhh I see!! I was looking at the limit from x=y instead of the axis of revolution.

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1the direction we are looking at for the shell measuers up from the y axis, or over ; but we have to rewrite the height line in terms of y for it to make sence in out integration using shells

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1since w is the inverse of a normal xy graph we inverse our y=sqrt(x) to match it

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1since our view is the inverse ....

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0do you mean height instead of width?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I have the correct answer on my paper now, sweet!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Could you help me do a couple more? :)

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1i can, i got the chrome downlaoded now; doesnt help the typos, but its a bit quicker lol

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0cool, should I start an entirely new question so I can give you another medal?

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1not that the medals matter, but starting a new one might help my computer to keep up the pace :)

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1we didnt get thru rotate about the x=36 yet tho, so lets do that one on another post

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Alright I'll do that!
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