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adnanchowdhuryBest ResponseYou've already chosen the best response.0
How do you solve it?
 2 years ago

adnanchowdhuryBest ResponseYou've already chosen the best response.0
how do i integrate: \[π \int\limits_{0}^{6} (y ^{2}/12 +1)^{2} dy\]
 2 years ago

bahrom7893Best ResponseYou've already chosen the best response.2
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
 2 years ago

bahrom7893Best ResponseYou've already chosen the best response.2
and for that integral, just multiply it out.
 2 years ago

bahrom7893Best ResponseYou've already chosen the best response.2
And then integrate term by term that will make your life easy.
 2 years ago

bahrom7893Best ResponseYou've already chosen the best response.2
actually finding the vol of the solid generated by the rectangle is easy.
 2 years ago

adnanchowdhuryBest ResponseYou've already chosen the best response.0
I've been taught to integrate like this when the limits are on the y axis: \[\int\limits_{a}^{b} π x ^{2} dy\]
 2 years ago

bahrom7893Best ResponseYou've already chosen the best response.2
It generates a cylinder with radius 6 and height 4
 2 years ago

bahrom7893Best ResponseYou've already chosen the best response.2
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.
 2 years ago

adnanchowdhuryBest ResponseYou've already chosen the best response.0
What rectangle are you referring to?
 2 years ago

bahrom7893Best ResponseYou've already chosen the best response.2
Imagine a line passing through (4,6) and (4,0)
 2 years ago

bahrom7893Best ResponseYou've already chosen the best response.2
that will form a rectangle right? with the line y = 6, x = 4, and x and y axes
 2 years ago

adnanchowdhuryBest ResponseYou've already chosen the best response.0
yes, i see. But what do you after that?
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.2
you can do this with disk or shell method. Do you know both?dw:1329324049040:dw
 2 years ago

bahrom7893Best ResponseYou've already chosen the best response.2
So find the volume of the solid generated by that rectangle  it will be a cylinder
 2 years ago

adnanchowdhuryBest ResponseYou've already chosen the best response.0
I haven't heard of the 'disk' or 'shell' method?
 2 years ago

adnanchowdhuryBest ResponseYou've already chosen the best response.0
What do i do after working out the vol. of solid rectangle?
 2 years ago

bahrom7893Best ResponseYou've already chosen the best response.2
Find the volume generated by revolving the following area and then subtract it from the vol of the cylinder.
 2 years ago

bahrom7893Best ResponseYou've already chosen the best response.2
dw:1328460177913:dw
 2 years ago

adnanchowdhuryBest ResponseYou've already chosen the best response.0
How would you integrate this? http://d.pr/2Eo2
 2 years ago

bahrom7893Best ResponseYou've already chosen the best response.2
Just expand and then integrate term by term, it's not hard.
 2 years ago

adnanchowdhuryBest ResponseYou've already chosen the best response.0
without expanding? Like chain rule?
 2 years ago

bahrom7893Best ResponseYou've already chosen the best response.2
I'm not sure if you can. Does the question force you to integrate without expanding?
 2 years ago

adnanchowdhuryBest ResponseYou've already chosen the best response.0
no but that is one technique i have to learn.
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.2
your formula is for a revolution about y, so it's not right
 2 years ago

bahrom7893Best ResponseYou've already chosen the best response.2
no that's a diff question
 2 years ago

myininayaBest ResponseYou've already chosen the best response.0
yeah bahrom is right i'm pretty sure the only way to integrate that is by expanding it
 2 years ago

myininayaBest ResponseYou've already chosen the best response.0
that is the only way i see anyways
 2 years ago

bahrom7893Best ResponseYou've already chosen the best response.2
I mean it's not usub, and I don't recognize any inverse formulas.. inverse tan won't work
 2 years ago

bahrom7893Best ResponseYou've already chosen the best response.2
and integration by parts is gonna give u a horrible mess
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.2
...but this integral is not supposed to represent the area in the above problem? separate Q you say?
 2 years ago

myininayaBest ResponseYou've already chosen the best response.0
it is not so nasty to expand this it is the easiest method to expand
 2 years ago

bahrom7893Best ResponseYou've already chosen the best response.2
Yea turing, it's separate
 2 years ago

bahrom7893Best ResponseYou've already chosen the best response.2
I think expanding is the only way.
 2 years ago

adnanchowdhuryBest ResponseYou've already chosen the best response.0
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
 2 years ago

adnanchowdhuryBest ResponseYou've already chosen the best response.0
The final answer is suppose to be 90π.
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.2
so it is supposed to be the same problem
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.2
...which means the integral is wrong
 2 years ago

bahrom7893Best ResponseYou've already chosen the best response.2
No, there are two diff problems here as far as i can tell, the integral is a completely diff question
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.2
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
 2 years ago

adnanchowdhuryBest ResponseYou've already chosen the best response.0
That is the formula i am given in the book for integrating on the yaxis. So, i can't be integrating on the wrong axis.
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.2
\[2\pi\int_{0}^{6}y(\frac{y^2}{12}+1)dy\] is shell method, which is how I would do it
 2 years ago

myininayaBest ResponseYou've already chosen the best response.0
\[\pi \int\limits_{0}^{6}(\frac{y^4}{144}+\frac{y^2}{6}+1) dy=\pi(\frac{y^5}{5 \cdot 144}+\frac{y^3}{3 \cdot 6}+y)_0^6\] \[=\pi(\frac{6^5}{5 \cdot 144}+\frac{6^3}{3 \cdot 6}+6)=\pi(\frac{6^4 \cdot 6}{5 \cdot 144}+\frac{6^2}{3}+6)\] \[=\pi(\frac{6^4 }{5 \cdot 24}+\frac{ 6 \cdot 6}{3}+6)=\pi(\frac{6^4}{5 \cdot 24} +6(2)+6)\] \[=\pi(\frac{6 \cdot 6^3}{5 \cdot 24}+12+6)=\pi(\frac{6^3}{5 \cdot 4}+18)\] \[=\pi(\frac{6 \cdot 6^2}{5 \cdot 4}+18)=\pi(\frac{3 \cdot 6^2}{5 \cdot 2}+18)\] \[=\pi(\frac{3 \cdot 36}{10}+18)=\pi(\frac{12 \cdot 9}{10}+18)=\pi(\frac{108}{10}+18)\] 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
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.2
shell method:dw:1329325465495:dwLook at the radius of each cylinder( y) and the height of each cylinder (x(y)) and we get\[2\pi\int_{0}^{6}rhdr=2\pi\int_{0}^{6}y(\frac{y^2}{12}+1)=90\pi\]
 2 years ago

bahrom7893Best ResponseYou've already chosen the best response.2
oh wait so it was this question?
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.2
I am integrating along y because I used shell method, disk method would be a pain here because we would need two integrals
 2 years ago

adnanchowdhuryBest ResponseYou've already chosen the best response.0
@myininaya did it right. I think you used the same method as me, but you expanded x^2 in the beginning? @bahrom7893 yes
 2 years ago

adnanchowdhuryBest ResponseYou've already chosen the best response.0
I'm still quite confused about the shell method.
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.2
myin did the integral right, but it is not a solid about the xaxis
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.2
no, hers are disks about the yaxis
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.2
dw:1329325925783:dwlike this but centered on the yaxis
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.2
area of each disk is pi*r^2, but if r is x(y) we are going around the yaxis
 2 years ago

adnanchowdhuryBest ResponseYou've already chosen the best response.0
This is rotation 360 degree about yaxis?
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.2
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<x<1 rings of inner radius y(x)=sqrt(12x1) and outer radius y=6 between 1<x<6
 2 years ago

adnanchowdhuryBest ResponseYou've already chosen the best response.0
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 xaxis and where the rectangle ends (at x =4).
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.2
Shell method\[2\pi\int_{0}^{6}y(\frac{y^2}{12}+1)dy\]Disk method (my way):\[\pi\int_{0}^{1}6^2dx+\pi\int_{1}^{6}6^2(\sqrt{12(x1)})^2dx\]Disk method bahrom's way\[\pi\int_{0}^{6}6^2dx\pi\int_{1}^{6}(\sqrt{12(x1)})^2dx\]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.
 2 years ago
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