Member of the Month

KilaTsunami
October, 2024

Online

Welcome to the Newschoolers forums! You may read the forums as a guest, however you must be a registered member to post. Register to become a member today!

Calc help

profile picture
Sklar
Posts: 8864 - Karma: 31,612
i missed this lesson, skiing that day, and am wondering if im doing this right

the question is:
Find the volume of the solid that results when the region enclosed by the given curves is revolved around the given axis.
y=√x, y=0, x=0, x=4; about y=2

i wrote it as pi times the integral, from 0 to 4, of 2^2-(√x)^2
and got 8pi

can anyone confirm, or refute that? or offer some help

thanks for good info
Mar 12 2011 10:14PM
0 0
profile picture
000000000000
Posts: 843 - Karma: 1,552

really looking forward to calc next year...

If you put this on yahoo answers, you'll get an answer pretty quick

Mar 12 2011 10:24PM
0 0
profile picture
californiagrown
Posts: 9794 - Karma: 2,014
dis is calc II if my memory is correct. the brightside is that calcIII is cake.


and i have the notes, but im not gonna be doing any math right now.
Mar 12 2011 10:25PM
0 0
profile picture
hurrisun
Posts: 6284 - Karma: 49
if you just derive it i know the anwser in my calc experience deriving is the anwser to most questons
Mar 12 2011 10:27PM
0 0
profile picture
Sklar
Posts: 8864 - Karma: 31,612
AP calc AB
Mar 12 2011 10:30PM
0 0
profile picture
Gnarshred
Posts: 5670 - Karma: 18
That sounds right yeah, but thats just doing it in my head, but I think the method is right.
Mar 12 2011 10:42PM
0 0
profile picture
Sklar
Posts: 8864 - Karma: 31,612
.
Mar 13 2011 7:15PM
0 0
profile picture
Bar_Down
Posts: 5896 - Karma: 395
its not right, ill tell you that much. too lazy to look at my notes and i cant remember exactly how to do it because i barely passed that class by not doing shit and just cramming the night before the exam, but i know its wrong first of all because pi is equal to 180 degres, so if you integrate to find the area under the curve and then multiply that by pi, in theory, if this method is correct to start with, it would only give you the area for 180 degres instead of a full revolution around the axis
Mar 13 2011 7:51PM
0 0
profile picture
ROOS.
Posts: 1992 - Karma: 39
it is not right, i think youve done two different things. there is no upper function so it should just be the pi times the integral of (root x) -2 all squared because you have to account for the fact that you are rotating around y=2 instead of y=0, im not sure but i think thats right
Mar 13 2011 7:57PM
0 0
profile picture
ROOS.
Posts: 1992 - Karma: 39
i know that is not right, you use pi for this formula because you are just moving pi, aka the constant, out of the integral which you can do then you square f(x) inside the intergral
Mar 13 2011 7:59PM
0 0
profile picture
moarplease
Posts: 293 - Karma: 15
you have to derive two new equations based on the information you are given.  put them in terms of x.  so... y=sqrt(x) becomes x=y^2.   then when y=0 the radius is 2. 2nd equation becomes (2-y).  Since its volume you just solve this integral.   V=2pi int((2-y)(y^2))dy from 0 to 2.  answer should be: 8pi/3.
Mar 13 2011 8:06PM
0 0
profile picture
shaggy.
Posts: 433 - Karma: 299
this thread just convinced me NOT to take calc next year
Mar 13 2011 8:09PM
0 0
profile picture
moarplease
Posts: 293 - Karma: 15
its not that hard. you can do it! dont short change yourself.
Mar 13 2011 8:14PM
0 0
profile picture
Bar_Down
Posts: 5896 - Karma: 395
holy fuck i tripped out hard, i think i read the thread really fast and didnt understand what you were saying ahaha

but basically, all i know is that, to find the volume of the solid resulting from the revolution of a function around an axis, you do pi times the integral of the function squared
Mar 13 2011 8:14PM
0 0
profile picture
Bar_Down
Posts: 5896 - Karma: 395
and you can just take it from there and apply it to whatever youre doing
Mar 13 2011 8:16PM
0 0
profile picture
t3hb055
Posts: 813 - Karma: 15
Damn not looking forward to this. AP calc blows. We just passed integrals and the log functions or whatever
Mar 13 2011 8:18PM
0 0
profile picture
An10a_Eng
Posts: 2505 - Karma: 26
look up the shell method for volumes of revolution. dont remember it off the top of my head but its pretty easy
Mar 13 2011 8:33PM
0 0
profile picture
shocker611
Posts: 2599 - Karma: 423
i dont know what calc 3 you took but mine was hell
Mar 13 2011 8:38PM
0 0
profile picture
An10a_Eng
Posts: 2505 - Karma: 26
most of the math in calc 3 is easier than in calc 2 but its pretty tough trying to wrap your mind around the theory behind stokes theorem and stuff like that
Mar 13 2011 8:40PM
0 0
profile picture
shocker611
Posts: 2599 - Karma: 423
i would disagree, i thought calc 2 was easy as hell with calc 3 being the opposite.. guess it just depends on professors.
Mar 13 2011 9:44PM
0 0
profile picture
Gnarshred
Posts: 5670 - Karma: 18
Who did you have for Calc 2/3?
Mar 13 2011 9:45PM
0 0
profile picture
Sklar
Posts: 8864 - Karma: 31,612
so am i right or is this right?
Mar 13 2011 9:47PM
0 0
profile picture
Gnarshred
Posts: 5670 - Karma: 18
Ah, actually you are wrong, You found the volume if it is revolved about the origin. check it:
Click for larger image

As you can see, the volume is the same as if you were to rotate -sqrt[x]+2 about the origin. Then a simple method of disks shows us that the volume is 8pi/3

Check me, but I think thats right...
Mar 13 2011 10:24PM
0 0
profile picture
Sklar
Posts: 8864 - Karma: 31,612
hmm, okay. looks like my teacher taught it to me wrong... again

thanks
Mar 13 2011 10:31PM
0 0
All times are Eastern (-5)