Question

find the volume of the solid using shell method, region of y=sqrt2, x=y^2 , x=0. about the x-axis

Answer 1

Ok which function is on top?

Answer 2

y=sqrt2, bounded on left by y -axis and below by x=y^2

Answer 3

just confirming it is
y=\[y=\sqrt(2)\]
x= x^2

right functions?

Answer 4

-- correction
x=y^2

Answer 5

yes

Answer 6

Ok so x=y^2 is a side way parabola facing in the positive x direction(right)
y=sqrt(2) is just a horizonal striaght line
We need to find intersection of of lines

Answer 7

y=sqrt2 right?

Answer 8

The intersection point will have y=sqrt(2)

Answer 9

when x = 2

Answer 10

what is x intersection is 2

Answer 11

so (2,sqrt(2)

Answer 12

good so far?

Answer 13

yes

Answer 14

So the region we are rotating has these points(0,0)-- origin (0,sqrt(2)top left corner, (2,sqrt(2) intersection

Answer 15

any questions?

Answer 16

so far good, i have a drawing set up

Answer 17

So we are going to use shell method which is just add up bunch off cylinder

Answer 18

Volume of cylinder is pi*radius^2*h

Answer 19

got ya

Answer 20

So if you look your drawing(may be draw horizonal rectangle) you will see that the y value of function represent radius

Answer 21

and the function x=y^2 represent h

Answer 22

yea from the integral of 0 to sqrt2 right?

Answer 23

Yes

Answer 24

\[2\pi\int\limits(y*y^2 dy(\]

Answer 25

you got the limit of the integration right

Answer 26

so just integrate that? with the height as y^2?

Answer 27

Yes

Answer 28

oooh thank you, i kept thinking of subtracting sqrt2 from the height wasn't really paying attention that this was respect to y. Thank you for all your help:)