volume of cylindrical shells (calculus):
y^2=x, x=2y rotate about y axis

- anonymous

volume of cylindrical shells (calculus):
y^2=x, x=2y rotate about y axis

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- anonymous

bro whats the base equation again, something like pi int .... of the derivative squared?

- anonymous

idk if that matters

- anonymous

i have my integral calculus last semester.

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## More answers

- anonymous

y=+or- sqrtx , y=x/2 to make it simpler

- anonymous

come on guys, this isnt that hard.

- anonymous

wait for a while, i'm solving it.

- anonymous

what is really the given? it is lacking.

- anonymous

thats it

- anonymous

really?

- anonymous

"find the volume of the solid obtained by rotating the regions bounded by the curves about the specified line.

- anonymous

you have to find the point where the two curves meet, and then solve the integral from 0 to that point

- anonymous

they meet at (0,0)

- anonymous

got it.

- anonymous

thats one yes, and then they cross again

- anonymous

the answer is

- anonymous

8pi over 3

- anonymous

where do the two curves cross?

- anonymous

the limits are 0 and 2.

- anonymous

how

- anonymous

just substitute x=y^2 to x=2y
that will become, y^2-2y=0
y(y-2)=0
y=0
y=2

- anonymous

okay, let me double check now

- anonymous

i had a mistake, the answer must be 10pi/3

- anonymous

almost have it

- anonymous

okay,
i have the solution but i have a hard time encoding it.

- anonymous

whats (2^1.5)/(2/3) -1

- anonymous

i dont have my calculator :/

- anonymous

3.242640687

- anonymous

okay thats what i got... where did you get pi from?

- anonymous

im looking in the back of the book for the answer to see whos right.

- anonymous

2pi integral of (x^1/2-x/2)dx
2pi (x^3/2)/(3/2)-x^2/4, substitute x=2
2pi(2^3/2)(2/3)-2^2/4
2pi(4(2/3)-1)
2pi(8/3 - 1)
2 pi (5/3)
(10pi)/3

- anonymous

okay i did it waaaay wrong haha

- anonymous

back of the book says 64pi/15

- anonymous

is it correct now?

- anonymous

is it the answer on the book?

- anonymous

64pi/15, so its close

- anonymous

okay im too tired for this, thanks for the help

- anonymous

you are welcome, actually i can explain it, i think i just miss some terms in my solution. by the way, do you have skype account?

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