Determine the volume of the solid obtained by rotating the portion of the region bounded by and that lies in the first quadrant about the y-axis.

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Determine the volume of the solid obtained by rotating the portion of the region bounded by and that lies in the first quadrant about the y-axis.

Mathematics
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your trying to tempt me arent ya :)
Nope. Not in the least. ;)
is this saying the volume of the first quadrant spun around the y axis?

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Yup. Which is to say, bound by both axis with the limits being each intersection
i appear to be missing a function to determine proper bound lol; or have gone blind :)
Oh you're right they didn't post... My bad! It's y=(x^1/3) and y =x/4
x^3 = x/4 would be the bounds then right...
cbrt(x)=x/4 typoed it lol
im going with 0 to 4
gotta redo that lol
0 to 8 ...
Yeah try again lol
There you go
my stupidity got in the way
I forgive you. 64 is awfully close to one. ;)
spin around the y axis....
we can do this from 0 to 2 on the y right
Less bored?
lol nerds!
less.... :)
Yes, if you mean in terms of y.
one radius is x=4y the other is .....
y^3?
I wouldn't do it that way.
\[\pi \int\limits_{0}^{2} [4y]^2 - [y^3]^2 dy\]
Though yes. You can. The top one being?
why is it gonna break?
Haha no,but yeah that's right. Let's pretend y just put inthe calculator and got it right because you can push buttons well. :)
i get 128/3 - 128/7
512/21 = 24.381 ??
if I do shelss; i wonder if i get the same result :)
That was fun :) it's 512 pi/21you forgot pi lol
i think i dropped a pi lol
ack!!.... yeah, just realized it :)
76.59 closer?
yes!! shell and disk are the same yay!!
\[2\pi \int\limits_{0}^{8} x(x^{4/3}) -x(\frac{x}{4}) dx\] = 76.59

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