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anonymous
 5 years ago
Hey guys, really need help with a Volumes question.
The volume enclosed between the curves y = x^4 and y = x^3 is rotated about the yaxis. Find the volume swept out using cylindrical shells
I'm getting a negative answer for some reason...
I'm doing Volume of y=x^4 revolved minus the volume of y=x^3 revolved.
delta(V) = 2*pi*r*h*delta(r).
For y=x^4, r = x, h = x^4, so V = Integral from 0 to 1 of 2*pi*x^5
For y=x^4, r = x, h = x^3, V = Integral from 0 to 1 of 2*pi*x^4
Volume of region enclosed = Integral from 0 to 1 of [2*pi*x^5  2*pi*x^4]
I'm getting a negative number
anonymous
 5 years ago
Hey guys, really need help with a Volumes question. The volume enclosed between the curves y = x^4 and y = x^3 is rotated about the yaxis. Find the volume swept out using cylindrical shells I'm getting a negative answer for some reason... I'm doing Volume of y=x^4 revolved minus the volume of y=x^3 revolved. delta(V) = 2*pi*r*h*delta(r). For y=x^4, r = x, h = x^4, so V = Integral from 0 to 1 of 2*pi*x^5 For y=x^4, r = x, h = x^3, V = Integral from 0 to 1 of 2*pi*x^4 Volume of region enclosed = Integral from 0 to 1 of [2*pi*x^5  2*pi*x^4] I'm getting a negative number

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amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1where do x^4 and x^3 match up? 0 and 1 right?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1try again: 2pi {S} x(x^4)  x(x^3) dx ; [0,1]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1x^5 ints to x^6/6 and x^4 ints up to x^5/5

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1its the symbol I use to type in the integration elongated 's'

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ah, okay ^^ but doesn't 2pi/6  2pi/5 = pi/15?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.110pi  12pi  = 2pi/30 = pi/15 yes; which simply means that 30 you got the numbers backwards watch tho: 54 = 1 45 = 1 so just disregard the sign

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1take the absolute value pi/15

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0But when graphed, between 0 and 1, y = x^4 is the one with the larger volume, no? So why would it be backwards?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1x^4 is flatter than x^3; so its actually below it try .5^4 and .5^3 for trial

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1.5^4 is actually a smaller number than .5^3 i think

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but 5>1. between 0 and 1, y=x^4 is below y=x^3, which means y=x^4 has a larger volume. So I shouldn't be getting a negative number when done in that order

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.10.0625 and 0.125 the ^4 is less than the ^3

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1since we are working with numbers that are fractions; we get x^4 under x^3

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1anything greater than 1 reverses it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0That's the graph I'm working with. The yellow highlighted area is what I'm revolving.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So it should be The area enclosed between y=x^4 and the line y = 1, rotated about the yaxis, MINUS The area enclosed between y=x^3 and the line y = 1 rotated about the yaxis

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I know taking the absolute value would be the logical thing to do, but I really don't see why the answer is negative. I'm subtracting a smaller area from a larger area. I should be getting a positive answer without having to absolute value it

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1x^4 is under x^3 between 0 and 1; so it simply means you subtracted the lesser from the greater

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1and i proved that by: (1/2)^4 < (1/2)^3 1/16 < 1/8

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1and when you subtract a lesser from a greater; it gives you a negative...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ah, I see what happened now. This is what confused me: if you imagine the solids formed when the curves are rotated about the yaxis, the one formed by y=x^4 is bigger, isn't it?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1(3/4)^4 < (3/4)^3 81/256 < 27/64

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1x^3 is bigger then x^4 between 0 and 1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I see that the yvalues are bigger, but I really don't see how the volume is bigger. Oh well... I don't see it, but it makes sense. Thank you!

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1spose this was the volume of a box with a smaller box inside... 5 cubic feet  4 cubic feet = 1 cubic feet left over but if we do the opposite 4 cubic feet  5 cubic feet = 1 cubic feet its the same absvalue

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1we simply get a  value rather than a positive value whih indicates that we simply reversed the numbers

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ah! Okay, I get it now! ^^ Thanks so much!
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