Use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region about the y-axis for
y=x y=0 and x=2

- anonymous

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- Zale101

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

- anonymous

Yes

- Zale101

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## More answers

- Zale101

Since we are revolving the shell around the y-axis, we will be using the example i circled.
|dw:1434059042077:dw|

- Zale101

Makes sense?

- anonymous

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

- anonymous

But yeah I understand.

- Zale101

But before anything else. You will analyze the rectangle and see what function along the x-axis is higher and what functions where the x-axis is lower. |dw:1434059232453:dw|

- anonymous

ok I think I get it. The one thing im really confused on is the integral. do you use the numbers on the x-axis for the integral?

- anonymous

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

- anonymous

i used that and then plugged the integral in so i got \[18\pi \] as my answer is that right. I think we may have learned it a different way because we never used that Big X little x thing we use a set equation that combines circumference and height

- Zale101

|dw:1434059661157:dw|

- Zale101

I forgot the mention, the intervals are based on where the spot is enclosed on the y-axis because we have dy and everything should be based on Y.

- Zale101

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

- anonymous

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

- anonymous

i told you 2 but its 3 sorry about that

- Zale101

okay

- Zale101

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

- anonymous

I don't know because the formula we learned dose not have that anywhere in there so i don't understand that.

- anonymous

I honestly thought i had it right and just wanted to double check my answers and make sure but now im super confused

- Zale101

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

- anonymous

Yeah!

- Zale101

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

- Zale101

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

- anonymous

ok i still don't get the X-x thing i thought it was just the equation which is in this case just x but then u times that by x from 2pi move the 2 pi out front because its a constant take the anti derivative of the x^2 so then you end up with \[2\pi \int\limits_{0}^{3} 1/3 x^{3}\]

- Zale101

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

- anonymous

thanks for helping btw i know its a pain

- Zale101

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

- Zale101

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

- anonymous

ok yeah that makes sense.

- anonymous

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

- Zale101

Yes.

- Zale101

Use the fundamental theorem of calculus to find the definite integral.
\[\Large \int\limits_{a}^{b}f(y)dy=F(b)-F(a)\]

- anonymous

so 9pi or am i still not getting it?

- anonymous

omg! ok thank you so much!!

- Zale101

Np!

- courtneygraley009

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