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Ok which function is on top?
y=sqrt2, bounded on left by y -axis and below by x=y^2
just confirming it is y=\[y=\sqrt(2)\] x= x^2 right functions?
-- correction x=y^2
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
The intersection point will have y=sqrt(2)
when x = 2
what is x intersection is 2
good so far?
So the region we are rotating has these points(0,0)-- origin (0,sqrt(2)top left corner, (2,sqrt(2) intersection
so far good, i have a drawing set up
So we are going to use shell method which is just add up bunch off cylinder
Volume of cylinder is pi*radius^2*h
So if you look your drawing(may be draw horizonal rectangle) you will see that the y value of function represent radius
and the function x=y^2 represent h
yea from the integral of 0 to sqrt2 right?
you got the limit of the integration right
so just integrate that? with the height as y^2?
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:)