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How do you solve it?

how do i integrate: \[π \int\limits_{0}^{6} (y ^{2}/12 +1)^{2} dy\]

and for that integral, just multiply it out.

And then integrate term by term that will make your life easy.

actually finding the vol of the solid generated by the rectangle is easy.

It generates a cylinder with radius 6 and height 4

What rectangle are you referring to?

Imagine a line passing through (4,6) and (4,0)

that will form a rectangle right? with the line y = 6, x = 4, and x and y axes

yes, i see. But what do you after that?

you can do this with disk or shell method. Do you know both?|dw:1329324049040:dw|

So find the volume of the solid generated by that rectangle - it will be a cylinder

I haven't heard of the 'disk' or 'shell' method?

What do i do after working out the vol. of solid rectangle?

|dw:1328460177913:dw|

How would you integrate this? http://d.pr/2Eo2

Just expand and then integrate term by term, it's not hard.

without expanding? Like chain rule?

I'm not sure if you can. Does the question force you to integrate without expanding?

no but that is one technique i have to learn.

your formula is for a revolution about y, so it's not right

no that's a diff question

yeah bahrom is right i'm pretty sure the only way to integrate that is by expanding it

that is the only way i see anyways

I mean it's not u-sub, and I don't recognize any inverse formulas.. inverse tan won't work

and integration by parts is gonna give u a horrible mess

...but this integral is not supposed to represent the area in the above problem? separate Q you say?

it is not so nasty to expand this
it is the easiest method to expand

Yea turing, it's separate

I think expanding is the only way.

The final answer is suppose to be 90π.

so it is supposed to be the same problem

...which means the integral is wrong

\[2\pi\int_{0}^{6}y(\frac{y^2}{12}+1)dy\] is shell method, which is how I would do it

oh wait so it was this question?

I'm still quite confused about the shell method.

myin did the integral right, but it is not a solid about the x-axis

Yes it is. ?

no, hers are disks about the y-axis

|dw:1329325925783:dw|like this but centered on the y-axis

area of each disk is pi*r^2, but if r is x(y) we are going around the y-axis

Oh I see.

This is rotation 360 degree about y-axis?

if you want shell method for your integral it can be done, but you will need to split the integral in two, or subtract the second area as bahrom did
disks of radius y(x)=6 between 0

Ok, let me try.

that'll work :D