<p>Can anybody explain this for myself and a friend? I think that the first part of the integral is either x or y if its a revolution around the y or x axis, respectively. If it's a revolution around another given line then you have to subtract to get the right region...i think?! Anybody able to explain this in words to me?</p>
<p>yeah I'm in the same class and I second this...there's gotta be some shell experts out there.</p>
<p>squak, squak.</p>
<p>only I'm better than the OP b/c i'm not a junior member.</p>
<p>but yeah: shell method help?</p>
<p>but im better then ziggy because i came out of nowhere after i got a D and he an A on the last report card and now have a better grade then him.</p>
<p>anyway, HELP!!!!</p>
<p>No bird/man hybrid is better than me.</p>
<p>Anyway: BUMP</p>
<p>correct 10 char</p>
<p>alright, im not familiar with the term "shell method" but i do know what you are talking about. </p>
<p>when you revolve the area around an axis, you integrate along that axis and multiply the integral by pi. for revolving around a line, you add or subtract depending on whether the area being revolved is above or below the line. for example from a previous BC exam:</p>
<p>if you were to have two equations:
f(x)= 0.25 +sin(pix)
g(x)= 4^-x
and you were trying to find the volume of the solid generated when an area is revolved around the x axis, when the area is from x=0.1792 to x=1 (x coordinates of intersection of the two graphs), then you would integrate along the x axis and subtract the squares of the functions, </p>
<p>S=integral</p>
<p>pi*S from 0.172 to 1 of (f(x)^2)-(g(x)^2)dx</p>
<p>However, if you were going to determin the volume of the solid generated when the area is revolved about the horizontal line y=1, you would subtract -1 to each function inside the square because you are increasing the length of the radius of the solid</p>
<p>=pi*S from 0.172 to 1 of ((f(x)--1)^2)-((g(x)--1)^2)dx</p>
<p>does this make sense?</p>
<p>that makes sense, our book titles that something else...but on this test you HAVE to use shell, you take some integral off the radius x height of the region and mulitple by 2pi</p>
<p>i understand...but i believe you just explained the "disk" or "washer" method...the shell method involves multiplying by 2 pi</p>
<p>ah! i know what you are talking about. let me get my notes. ill explain with a problem.</p>
<p>thanks so much!</p>
<p>actually i just found this, and it explains it pretty well. <a href="http://archives.math.utk.edu/visual.calculus/5/volumes.6/index.html%5B/url%5D">http://archives.math.utk.edu/visual.calculus/5/volumes.6/index.html</a>
go to the first flash/java ex.
take a look and if you still have questions then maybe i can explain it differently.</p>
<p>i didnt understand what you meant in the beginning, but what I explained above isn't the shell method. the website has a good illustration, though. the shell method basically takes a "slice" and uses it to represent k number of slices of the area you are revolving. the reason you multiply the integral by 2pi is because you're accounting for the circumference of the revolution you're performing around the axis.</p>
<p>thanks..that is really helpful.</p>
<p>now that i see the flash app i sort of understand where PR is coming from (a little easier to understand than the text).</p>
<p>that site, in addition to my PR, really helped.</p>
<p>i think i did really well on today's quiz (for once).</p>
<p>thanks</p>
<p>Shell method isn't that much different from the disc method.</p>
<p>yeah, our teacher just explained like it was way different and didnt explain it well. hmm, one more lecture and we're done for the year :)</p>
<p>Is this for AB?</p>
<p>yes, considering there is a FR question on area (although you could just as easily use the disk ("washer") method, so i guess it isnt really required for AB). In a lot of cases it is easier to use than the other method, although you do not technically need to know it. I'm not sure if it's in the official AB curriculum or not.</p>