anonymous
  • anonymous
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 y-axis. 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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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amistre64
  • amistre64
where do x^4 and x^3 match up? 0 and 1 right?
amistre64
  • amistre64
try again: 2pi {S} x(x^4) - x(x^3) dx ; [0,1]
amistre64
  • amistre64
x^5 ints to x^6/6 and x^4 ints up to x^5/5

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amistre64
  • amistre64
2pi/6 - 2pi/5 = ?
anonymous
  • anonymous
What's the {S}?
amistre64
  • amistre64
its the symbol I use to type in the integration elongated 's'
anonymous
  • anonymous
Ah, okay ^^ but doesn't 2pi/6 - 2pi/5 = -pi/15?
amistre64
  • amistre64
10pi - 12pi ---------- = -2pi/30 = -pi/15 yes; which simply means that 30 you got the numbers backwards watch tho: 5-4 = 1 4-5 = -1 so just disregard the sign
amistre64
  • amistre64
take the absolute value |-pi/15|
anonymous
  • anonymous
But when graphed, between 0 and 1, y = x^4 is the one with the larger volume, no? So why would it be backwards?
amistre64
  • amistre64
x^4 is flatter than x^3; so its actually below it try .5^4 and .5^3 for trial
amistre64
  • amistre64
.5^4 is actually a smaller number than .5^3 i think
amistre64
  • amistre64
.625 and .125 right?
anonymous
  • anonymous
but 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
  • amistre64
.0625 and .125 maybe
amistre64
  • amistre64
0.0625 and 0.125 the ^4 is less than the ^3
amistre64
  • amistre64
since we are working with numbers that are fractions; we get x^4 under x^3
amistre64
  • amistre64
anything greater than 1 reverses it
anonymous
  • anonymous
That's the graph I'm working with. The yellow highlighted area is what I'm revolving.
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anonymous
  • anonymous
So it should be The area enclosed between y=x^4 and the line y = 1, rotated about the y-axis, MINUS The area enclosed between y=x^3 and the line y = 1 rotated about the y-axis
anonymous
  • anonymous
I 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
  • amistre64
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amistre64
  • amistre64
x^4 is under x^3 between 0 and 1; so it simply means you subtracted the lesser from the greater
amistre64
  • amistre64
and i proved that by: (1/2)^4 < (1/2)^3 1/16 < 1/8
amistre64
  • amistre64
and when you subtract a lesser from a greater; it gives you a negative...
anonymous
  • anonymous
Ah, I see what happened now. This is what confused me: if you imagine the solids formed when the curves are rotated about the y-axis, the one formed by y=x^4 is bigger, isn't it?
amistre64
  • amistre64
(3/4)^4 < (3/4)^3 81/256 < 27/64
amistre64
  • amistre64
x^3 is bigger then x^4 between 0 and 1
anonymous
  • anonymous
I see that the y-values 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
  • amistre64
spose 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
  • amistre64
we simply get a - value rather than a positive value whih indicates that we simply reversed the numbers
anonymous
  • anonymous
Ah! Okay, I get it now! ^^ Thanks so much!

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