Question

Refer to the figure and find the volume V generated by rotating the given region about the specified line.
R3 about AB

Answer 1

Here's the figure

Answer 2

So I know you have to use the formula A(x) = pi(outer radius)^2 - pi(inner radius)^2 and put things in terms of x but I'm having a little bit of trouble going from there.

Answer 3

r3 about ab ... hmm

Answer 4

disc method (or washer as youve addressed) would use an inverse

whereas a shell method would use the given stuff

Answer 5

can you define the line from O to B ?

Answer 6

Yeah. O = (0,0) so I just found the slope to find the equation for that line which would be y = 2x. So, if I'm correct, I would have x = (y/2)^4 for my outer radius and x = (y/2) for the inner radius. Or would I have to subtract them from one?

Answer 7

if we define the radius in terms of y ...

R = (y/2)^4
r = y/2

seem appropriate

as the y moves from 0 to 2

Answer 8

if we did this in parts:\[\pi R^2-\pi r^2\]
would be correct, or you can do it all together by factoring out the pi:\[\int_{0}^{2}\pi R^2-\int_{0}^{2}\pi r^2~\color{red}{\to}~\pi\int_{0}^{2}R^2-r^2 \]

Answer 9

might have to adjust the radius for the axis tho

Answer 10

1 - f(y) to adjust for the offset rotation

Answer 11

Okay. so R = 1 - (y/2)^4 and r = 1 - (y/2) ?

Answer 12

one way to look at this is to consider moving the whole thing such that the center of rotation is about the y axis instead

y = 2* 4rt(1-x) and y = 2(1-x)

Answer 13

y = 2* 4rt(1-x)
y/2 = 4rt(1-x)
(y/2)^4 = 1-x
x = 1 - (y/2)^4


y = 2(1-x)
y/2 = 1-x
x = 1 - y/2

Answer 14

to see if the adjustment is valid; when x=0, r=1 ... when x=1, r=0

Answer 15

lol, somethings prolly still goofy. But yeah, these things can be cumbersome to address when the axis of rotation is not "normal"

Answer 16

Okay, I think I see what you're saying. So you took your x values and subtracted them from one to pull the function to x = 0 and then just found your values from there. Makes sense.

Answer 17

yep

Answer 18

So, to get the volume, we'd set it up V=\[\pi \int\limits_{0}^{2}[(1-(y/2)^{4})^{2} - (1 - (y/2))^{2}]dx\]
Right ?

Answer 19

** dy

Answer 20

looks good to me ...

when y=0, x = 1
when y=2, x = 0

and our radius is defined by the value of x

Answer 21

Okay so once you square the terms you'll have\[\pi \int\limits_{0}^{2}(1-2(y/2)^{4}+(y/2)^{6})-(1-y+(y/2)^{2})dy\]
Right ?

Answer 22

not (y/2)^6, (y/2)^8, right ? Exponents raised to another exponent multiply

Answer 23

should expand out to say:
\[\frac{x^8}{256}-\frac{x^4}{8}-\frac{x^2}{4}+x\]

Answer 24

if we did a shell run:\[\int 2\pi(RH-rh)\]
\[2\pi\int_{0}^{1}(1-x)(2x^{1/4})-(1-x)(2x)~dx\]

Answer 25

We haven't gone over shell yet, that's why I was doing washer. :b So shell just lets you put a volume integration thats over why in terms of x ?

Answer 26

** y
lol don't know why I typed out why XD

Answer 27

shell adds up the area of "sheets" that are 2pi(x) wide, and f(x) tall