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|dw:1435377786774:dw|
This is the bounded area

Wait so how does the integral change when you rotate it around the x axis instead of y?

https://www.wyzant.com/resources/lessons/math/calculus/integration/finding_volume

With "washer method"

Follow rules of "shell" and "washer" method to guide throughout finding volume

and then the limits will be the y coordinates of the intersection points

so then the thickness of the cylinder is dy?

|dw:1435378522268:dw|

Peachpi seems absolutely right; this is a tough area

this is a solid. so there's no hole, right?

the area lies on the axis it's being rotated on

well they intersect when x=6.562

yeah, but the integral is in y.
a is 0.
b you solve yÂ² = y + 4 to get

How is the integral with respect to y ?

maybe it has something to do with the x axis being rotated so we can't use that? I have no idea x.x

oh I see where the a and b for the integral is |dw:1435378968903:dw|

those intersection points are the values for a and b

|dw:1435378996664:dw|

I think it is simpler to do it horizontally than vertically, which would require two integrals.

hmmm y^2-y-4 = 0 and then factor ?
OH NO! I DO NOT WANT TO DO 2 INTEGRALS

(-1)^2-4(1)(-4) = 1-4(-4) = 1+ 16 = 17 and it's not a perfect square.. dahhhh

What is the point of intersection?

|dw:1435379165728:dw|

you graph the two equations first. and then they should have common points.

|dw:1435379200509:dw|

with 0 being one of them

they're not nice numbers. about (6.56, 2.56)

|dw:1435379227291:dw|

So if we use the shell method from 0 to 4, we are basically using the disc method?

double integrals.. double the anti derivatives :S

|dw:1435379315446:dw|

\[
\int dV = 2\pi\int rh~dy = 2\pi\int \limits_0^{y_1} y(f(y)-g(y))~dy
\]

THANK YOU! I think I can do it from here now :)

Thankks