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Mathematics 8 Online
OpenStudy (anonymous):

determine the volume generated by revoling the region bounded by y=2x^2, y=0, x=0, and x=5 around the line x=6

OpenStudy (anonymous):

let's plot the functions

OpenStudy (anonymous):

I did it correctly for the one that revoled the same area around the y axis but the new line is throwing me for a loop

OpenStudy (anonymous):

Are you familar with volumes of revolution at all?

OpenStudy (mathmate):

It looks like x=0 is extraneous.

OpenStudy (anonymous):

yes I have identified the region in need and just cannot figure out how to change the equation all my class was told to move the revoultion line to either the x or y axis

OpenStudy (mathmate):

Do you use the washer method or the shell method?

OpenStudy (anonymous):

ap cal?

OpenStudy (anonymous):

I cannot remeber the name but we set it up like this \[V=\int\limits_{a}^{b}(f(x)^2-g(x)^2)\]

OpenStudy (anonymous):

I forgot the time by pi on the right side sorry

OpenStudy (mathmate):

So can you first find the limits of integration (dx, I suppose), and you need pi there somewhere.

OpenStudy (anonymous):

okay, first think of axis or rotation minus the function in parentheses squared and don't forget the pi out front. and the dy in the back....you will need to change this to x=terms of y.

OpenStudy (anonymous):

okay but its in setting up the new graph that I stumble I cannot for some reason seem to get the values to match the way they need to

OpenStudy (anonymous):

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OpenStudy (anonymous):

that plus the reflection matchs the picture I drew for Part One Part Two is to caculate it out using the method my math teacher taugh

OpenStudy (anonymous):

*taught

OpenStudy (anonymous):

|dw:1324441032953:dw|

OpenStudy (anonymous):

This should get you started.

OpenStudy (anonymous):

|dw:1324441341625:dw|

OpenStudy (anonymous):

I know my stuff about the graph and how it looks its the new rotation line thats tnrowing me off all other questions have been about the x or y axis

OpenStudy (anonymous):

I understand, First think of axis of rotation minus the function.

OpenStudy (anonymous):

the largest radius comes first.

OpenStudy (anonymous):

so its the line of rotation minus sqrt of y/2 its furthest away and therefore the larger radius.

OpenStudy (anonymous):

then is smaller radius, the line of rotation minus the line x=5.

OpenStudy (anonymous):

both are squared. since this is revolved about a vertical line you will use dy

OpenStudy (anonymous):

if this is a calculator problem, you could now enter it.....if not there is more work to do.

OpenStudy (anonymous):

its not but i got help with the part I needed thanks

OpenStudy (anonymous):

k

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