Question

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

Answer 1

where do x^4 and x^3 match up? 0 and 1 right?

Answer 2

try again:

2pi {S} x(x^4) - x(x^3) dx ; [0,1]

Answer 3

x^5 ints to x^6/6 and x^4 ints up to x^5/5

Answer 4

2pi/6 - 2pi/5 = ?

Answer 5

What's the {S}?

Answer 6

its the symbol I use to type in the integration elongated 's'

Answer 7

Ah, okay ^^
but doesn't 2pi/6 - 2pi/5 = -pi/15?

Answer 8

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

Answer 9

take the absolute value |-pi/15|

Answer 10

But when graphed, between 0 and 1, y = x^4 is the one with the larger volume, no? So why would it be backwards?

Answer 11

x^4 is flatter than x^3; so its actually below it

try .5^4 and .5^3 for trial

Answer 12

.5^4 is actually a smaller number than .5^3 i think

Answer 13

.625 and .125 right?

Answer 14

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

Answer 15

.0625 and .125 maybe

Answer 16

0.0625 and 0.125 the ^4 is less than the ^3

Answer 17

since we are working with numbers that are fractions; we get x^4 under x^3

Answer 18

anything greater than 1 reverses it

Answer 19

That's the graph I'm working with. The yellow highlighted area is what I'm revolving.

Answer 20

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

Answer 21

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

Answer 22

x^4 is under x^3 between 0 and 1; so it simply means you subtracted the lesser from the greater

Answer 23

and i proved that by:

(1/2)^4 < (1/2)^3

1/16 < 1/8

Answer 24

and when you subtract a lesser from a greater; it gives you a negative...

Answer 25

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?

Answer 26

(3/4)^4 < (3/4)^3

81/256 < 27/64

Answer 27

x^3 is bigger then x^4 between 0 and 1

Answer 28

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!

Answer 29

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|

Answer 30

we simply get a - value rather than a positive value whih indicates that we simply reversed the numbers

Answer 31

Ah! Okay, I get it now! ^^ Thanks so much!