Question

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

Answer 1

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

Answer 2

idk if that matters

Answer 3

i have my integral calculus last semester.

Answer 4

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

Answer 5

come on guys, this isnt that hard.

Answer 6

wait for a while, i'm solving it.

Answer 7

what is really the given? it is lacking.

Answer 8

thats it

Answer 9

really?

Answer 10

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

Answer 11

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

Answer 12

they meet at (0,0)

Answer 13

got it.

Answer 14

thats one yes, and then they cross again

Answer 15

the answer is

Answer 16

8pi over 3

Answer 17

where do the two curves cross?

Answer 18

the limits are 0 and 2.

Answer 19

how

Answer 20

just substitute x=y^2 to x=2y

that will become, y^2-2y=0
y(y-2)=0

y=0
y=2

Answer 21

okay, let me double check now

Answer 22

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

Answer 23

almost have it

Answer 24

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

Answer 25

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

Answer 26

i dont have my calculator :/

Answer 27

3.242640687

Answer 28

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

Answer 29

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

Answer 30

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

Answer 31

okay i did it waaaay wrong haha

Answer 32

back of the book says 64pi/15

Answer 33

is it correct now?

Answer 34

is it the answer on the book?

Answer 35

64pi/15, so its close

Answer 36

okay im too tired for this, thanks for the help

Answer 37

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?