Integration question: http://d.pr/RTc6

- anonymous

Integration question: http://d.pr/RTc6

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- anonymous

How do you solve it?

- anonymous

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

- bahrom7893

Rotate the rectangle first, then rotate the parabola itself, find individual volumes then subtract the volume of the solid generated by the parabola from the solid generated by the rectangle

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## More answers

- bahrom7893

and for that integral, just multiply it out.

- bahrom7893

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

- bahrom7893

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

- anonymous

I've been taught to integrate like this when the limits are on the y axis: \[\int\limits_{a}^{b} π x ^{2} dy\]

- bahrom7893

It generates a cylinder with radius 6 and height 4

- bahrom7893

Yea but you don't really have to do that for the solid generated by the rectangle. If you rotate it around it generates a cylinder.

- anonymous

What rectangle are you referring to?

- bahrom7893

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

- bahrom7893

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

- anonymous

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

- TuringTest

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

- bahrom7893

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

- anonymous

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

- anonymous

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

- bahrom7893

Find the volume generated by revolving the following area and then subtract it from the vol of the cylinder.

- bahrom7893

|dw:1328460177913:dw|

- anonymous

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

- bahrom7893

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

- anonymous

without expanding? Like chain rule?

- bahrom7893

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

- anonymous

no but that is one technique i have to learn.

- TuringTest

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

- bahrom7893

no that's a diff question

- myininaya

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

- myininaya

that is the only way i see anyways

- bahrom7893

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

- bahrom7893

and integration by parts is gonna give u a horrible mess

- TuringTest

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

- myininaya

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

- bahrom7893

Yea turing, it's separate

- bahrom7893

I think expanding is the only way.

- anonymous

Basicallly, I ended up with this integral, by rearranging the given equation to make x the subject. Then substituting into this: http://d.pr/PR8z

- anonymous

The final answer is suppose to be 90π.

- TuringTest

so it is supposed to be the same problem

- TuringTest

...which means the integral is wrong

- bahrom7893

No, there are two diff problems here as far as i can tell, the integral is a completely diff question

- TuringTest

No ,they just said they got the integral by subbing in for y, but they are integrating on the wrong axis with the wrong method

- anonymous

That is the formula i am given in the book for integrating on the y-axis. So, i can't be integrating on the wrong axis.

- TuringTest

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

- myininaya

\[\pi \int\limits_{0}^{6}(\frac{y^4}{144}+\frac{y^2}{6}+1) dy=\pi(\frac{y^5}{5 \cdot 144}+\frac{y^3}{3 \cdot 6}+y)|_0^6\]
\[=\pi(\frac{6^5}{5 \cdot 144}+\frac{6^3}{3 \cdot 6}+6)=\pi(\frac{6^4 \cdot 6}{5 \cdot 144}+\frac{6^2}{3}+6)\]
\[=\pi(\frac{6^4 }{5 \cdot 24}+\frac{ 6 \cdot 6}{3}+6)=\pi(\frac{6^4}{5 \cdot 24} +6(2)+6)\]
\[=\pi(\frac{6 \cdot 6^3}{5 \cdot 24}+12+6)=\pi(\frac{6^3}{5 \cdot 4}+18)\]
\[=\pi(\frac{6 \cdot 6^2}{5 \cdot 4}+18)=\pi(\frac{3 \cdot 6^2}{5 \cdot 2}+18)\]
\[=\pi(\frac{3 \cdot 36}{10}+18)=\pi(\frac{12 \cdot 9}{10}+18)=\pi(\frac{108}{10}+18)\]
so this is the same problem or different? lol
i don't know anymore
by the way getting the answer is hard without a calculator as you can see

- TuringTest

shell method:|dw:1329325465495:dw|Look at the radius of each cylinder( y) and the height of each cylinder (x(y)) and we get\[2\pi\int_{0}^{6}rhdr=2\pi\int_{0}^{6}y(\frac{y^2}{12}+1)=90\pi\]

- bahrom7893

oh wait so it was this question?

- TuringTest

I am integrating along y because I used shell method, disk method would be a pain here because we would need two integrals

- anonymous

@myininaya did it right. I think you used the same method as me, but you expanded x^2 in the beginning?
@bahrom7893 yes

- anonymous

I'm still quite confused about the shell method.

- TuringTest

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

- anonymous

Yes it is. ?

- TuringTest

no, hers are disks about the y-axis

- TuringTest

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

- TuringTest

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

- anonymous

Oh I see.

- anonymous

This is rotation 360 degree about y-axis?

- TuringTest

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

- anonymous

Ok, let me try.

- anonymous

Yes! I got it. I did what bahrom7893 told me to do. Find volume of cylinder made by rectange on this graph. Then subtact it by the volume between where the line cross x-axis and where the rectangle ends (at x =4).

- TuringTest

that'll work :D

- TuringTest

Shell method\[2\pi\int_{0}^{6}y(\frac{y^2}{12}+1)dy\]Disk method (my way):\[\pi\int_{0}^{1}6^2dx+\pi\int_{1}^{6}6^2-(\sqrt{12(x-1)})^2dx\]Disk method bahrom's way\[\pi\int_{0}^{6}6^2dx-\pi\int_{1}^{6}(\sqrt{12(x-1)})^2dx\]all should give the same answer.
Try to think about the difference in the concept of each formula if you can, that will broaden your understanding.

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