anonymous
  • anonymous
Refer to the figure and find the volume V generated by rotating the given region about the specified line. R3 about AB
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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anonymous
  • anonymous
Here's the figure
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anonymous
  • anonymous
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.
amistre64
  • amistre64
r3 about ab ... hmm

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amistre64
  • amistre64
disc method (or washer as youve addressed) would use an inverse whereas a shell method would use the given stuff
amistre64
  • amistre64
can you define the line from O to B ?
anonymous
  • anonymous
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?
amistre64
  • amistre64
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
amistre64
  • amistre64
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:\[\int_{0}^{2}\pi R^2-\int_{0}^{2}\pi r^2~\color{red}{\to}~\pi\int_{0}^{2}R^2-r^2 \]
amistre64
  • amistre64
might have to adjust the radius for the axis tho
amistre64
  • amistre64
1 - f(y) to adjust for the offset rotation
anonymous
  • anonymous
Okay. so R = 1 - (y/2)^4 and r = 1 - (y/2) ?
amistre64
  • amistre64
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)
amistre64
  • amistre64
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
amistre64
  • amistre64
to see if the adjustment is valid; when x=0, r=1 ... when x=1, r=0
amistre64
  • amistre64
lol, somethings prolly still goofy. But yeah, these things can be cumbersome to address when the axis of rotation is not "normal"
anonymous
  • anonymous
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.
amistre64
  • amistre64
yep
anonymous
  • anonymous
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 ?
anonymous
  • anonymous
** dy
amistre64
  • amistre64
looks good to me ... when y=0, x = 1 when y=2, x = 0 and our radius is defined by the value of x
anonymous
  • anonymous
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 ?
anonymous
  • anonymous
not (y/2)^6, (y/2)^8, right ? Exponents raised to another exponent multiply
amistre64
  • amistre64
should expand out to say: \[\frac{x^8}{256}-\frac{x^4}{8}-\frac{x^2}{4}+x\]
amistre64
  • amistre64
if we did a shell run:\[\int 2\pi(RH-rh)\] \[2\pi\int_{0}^{1}(1-x)(2x^{1/4})-(1-x)(2x)~dx\]
anonymous
  • anonymous
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 ?
anonymous
  • anonymous
** y lol don't know why I typed out why XD
amistre64
  • amistre64
shell adds up the area of "sheets" that are 2pi(x) wide, and f(x) tall
amistre64
  • amistre64
|dw:1377546365897:dw|
anonymous
  • anonymous
Alrighty. So, after expanding your formula you would get\[2\pi \int\limits_{0}^{1}(2x ^{1/4}-2x ^{1/3}) - (2x - 2x ^{2})dx\]
anonymous
  • anonymous
And you can pull the 2's out in front of each, right ? so it'd be 2(x^(1/4) - x^(1/3) - 2(x - x^2)
amistre64
  • amistre64
\[2\pi\int_{0}^{1}(1-x)(2x^{1/4})-(1-x)(2x)~dx\] \[2\pi\int_{0}^{1}2x^{1/4}-2x^{5/4}-(1-x)(2x)~dx\] \[2\pi\int_{0}^{1}2x^{1/4}-2x^{5/4}-(2x-2x^2)~dx\] \[2\pi\int_{0}^{1}2x^{1/4}-2x^{5/4}+2x^2-2x~dx\] \[4\pi\int_{0}^{1}x^{1/4}-x^{5/4}+x^2-x~dx\]
anonymous
  • anonymous
Oh okay, you treated the first x as x^(4/4) and added cause of the multiplication. Cool.
amistre64
  • amistre64
at 0, the integral is 0, so this amounts to: \[4\pi(\frac45-\frac49+\frac13-\frac12)\] with any luck :)
amistre64
  • amistre64
34pi/45 either way we go ... which is what is expected :)
anonymous
  • anonymous
Okay, great, that totally makes sense. Thank youuuu. :3
amistre64
  • amistre64
youre welcome

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