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amistre64
 5 years ago
Find the volume of the solid created by the bounded region of (x=yy^2) and (x=y^2 3) when it is spun around the line (x=4).
amistre64
 5 years ago
Find the volume of the solid created by the bounded region of (x=yy^2) and (x=y^2 3) when it is spun around the line (x=4).

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ah this question still o.O

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah :) but I found out what I was doing wrong.... other than trying to drink and derive :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So we at least know you can add 4 to all terms to change the bounds and have it revolved around x = 0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0And it's a torus, right? So you need to integrate the area of the torus around the line in a full circle.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0A cross section of the torus.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Can't you just integrate it?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0its pi (S) [4+F(y)]^2  [4+G(y)]^2  1, 3/2 I kept trying: pi (S) [(4+F(y))  (4+G(y))]^2  1, 3/2

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0math: yes, but couldnt figure out how to last night :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[x = y  y^2 + 4\] \[x = y^2 + 1\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Okay so the bounds are y = 3, y = 1/2

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0Krb: goog good. the bounds of integration are 1 and 3/2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah those are the bounds ^

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i need to figure out how to use the equations to find the solids of revolution..if this is x=4, then the line we get is a vertical one right? so its been revolved upwards?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0you got the equation right; now square them individually before you subtract them..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So \[\int\limits_{(1)}^{(3/2)}(y  y^2 + 4)  (y^2 + 1)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0That's what i would think

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Krb you're always on this site. I like how when I look at paul's notes, you were the first one on open study.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0think; the rotation is around the "y axis" so to speak

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah..thats what i mean, so its like integrating the difference of squares of the equations right? and these equations signify ?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0Krb: getting closer. we need to integrate [F(y)]^2  [G(x)]^2 in the bounds

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes, normally when integrating with respect to x you are finding the are underneath the curve that drops to the x axis. So in this case you are finding the area "to the side of the curve" that drops to the y axis

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm confused as to why it would be squared

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i guess it should be squared...shouldn't it?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0Krb: if we do it seperately for each on, there is no real "area" so to speak until we combine them. Does that make sense?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0pi* (S) F(y)^2 dy  (S) G(y)^2 dy The equations ARE our Radiusessess

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if y1 & y2 are the curves then , \[\int\limits_{}^{}[y1^{2}y2^{2}]\] gives the volume of the solid? where y1=upper curve & y2 the lower

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0SO we are looking at it from a top down view, and then you need pi * r^2 for both "circles"

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0on efor inner and the other for outer 'circle' i guess

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0think: thats the gyst of it yeah

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[[\pi \int\limits_{1}^{(3/2)}(y  y^2 + 4)^2]  [\pi \int\limits_{1}^{(3/2)}(y^2 + 1)^2]\]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0Krb: your a genius :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0we should start a mensa club right here lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the open study math mensa

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0:) :) so krb ur'e pursuing your masters too?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0C1  C2 is what we were missing last night. 2pi(R)  2pi(r) does not equal 2pi(Rr)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0it does, but it doesnt when you integrate lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wer those constants that u'll missed out yesterday?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ends up being 85.9029

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0think: I'm honestly not sure

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i dunno why it gets posted twice always, when i click only once!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I think I probably would like to get a Masters
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