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First, draw the graphs and identify the enclosed area. Have you done that already?

Yes

|dw:1434058993462:dw|

Makes sense?

yeah! except your axis kind of confuse me I thought the y and x were switched

But yeah I understand.

\[\int\limits_{0}^{3} 2\pi (x ^{2})\] so like this ?

|dw:1434059661157:dw|

First step: Let's identify the intervals. What do you think the intervals are?

i gave you the wrong one by accident the problem says y=0 to x=3 so i think those are the intervals

i told you 2 but its 3 sorry about that

okay

So the intervals were right. But what did you get for the big x and the small x?

\(\large a(x)=2\pi(radius)(height)\)

Yeah!

|dw:1434060538490:dw|
\(\large a(x)=2pi(y)(X-x)\)

\[\Large \int\limits_{0}^{3}2 \pi y(X-x)dy\]

|dw:1434060977099:dw|
The radius is y, and the height is (X-x)

thanks for helping btw i know its a pain

|dw:1434061121877:dw|
The big X is x=3
and the small x is y=x

Makes sense? Hope I didn't confused you...

ok yeah that makes sense.

ok! so then the next step is the anti derivative of that?

Yes.

so 9pi or am i still not getting it?

omg! ok thank you so much!!

Np!