anonymous
  • anonymous
Integration question: http://d.pr/RTc6
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
How do you solve it?
anonymous
  • anonymous
how do i integrate: \[π \int\limits_{0}^{6} (y ^{2}/12 +1)^{2} dy\]
bahrom7893
  • 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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bahrom7893
  • bahrom7893
and for that integral, just multiply it out.
bahrom7893
  • bahrom7893
And then integrate term by term that will make your life easy.
bahrom7893
  • bahrom7893
actually finding the vol of the solid generated by the rectangle is easy.
anonymous
  • 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
  • bahrom7893
It generates a cylinder with radius 6 and height 4
bahrom7893
  • 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
  • anonymous
What rectangle are you referring to?
bahrom7893
  • bahrom7893
Imagine a line passing through (4,6) and (4,0)
bahrom7893
  • bahrom7893
that will form a rectangle right? with the line y = 6, x = 4, and x and y axes
anonymous
  • anonymous
yes, i see. But what do you after that?
TuringTest
  • TuringTest
you can do this with disk or shell method. Do you know both?|dw:1329324049040:dw|
bahrom7893
  • bahrom7893
So find the volume of the solid generated by that rectangle - it will be a cylinder
anonymous
  • anonymous
I haven't heard of the 'disk' or 'shell' method?
anonymous
  • anonymous
What do i do after working out the vol. of solid rectangle?
bahrom7893
  • bahrom7893
Find the volume generated by revolving the following area and then subtract it from the vol of the cylinder.
bahrom7893
  • bahrom7893
|dw:1328460177913:dw|
anonymous
  • anonymous
How would you integrate this? http://d.pr/2Eo2
bahrom7893
  • bahrom7893
Just expand and then integrate term by term, it's not hard.
anonymous
  • anonymous
without expanding? Like chain rule?
bahrom7893
  • bahrom7893
I'm not sure if you can. Does the question force you to integrate without expanding?
anonymous
  • anonymous
no but that is one technique i have to learn.
TuringTest
  • TuringTest
your formula is for a revolution about y, so it's not right
bahrom7893
  • bahrom7893
no that's a diff question
myininaya
  • myininaya
yeah bahrom is right i'm pretty sure the only way to integrate that is by expanding it
myininaya
  • myininaya
that is the only way i see anyways
bahrom7893
  • bahrom7893
I mean it's not u-sub, and I don't recognize any inverse formulas.. inverse tan won't work
bahrom7893
  • bahrom7893
and integration by parts is gonna give u a horrible mess
TuringTest
  • TuringTest
...but this integral is not supposed to represent the area in the above problem? separate Q you say?
myininaya
  • myininaya
it is not so nasty to expand this it is the easiest method to expand
bahrom7893
  • bahrom7893
Yea turing, it's separate
bahrom7893
  • bahrom7893
I think expanding is the only way.
anonymous
  • 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
  • anonymous
The final answer is suppose to be 90π.
TuringTest
  • TuringTest
so it is supposed to be the same problem
TuringTest
  • TuringTest
...which means the integral is wrong
bahrom7893
  • bahrom7893
No, there are two diff problems here as far as i can tell, the integral is a completely diff question
TuringTest
  • 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
  • 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
  • TuringTest
\[2\pi\int_{0}^{6}y(\frac{y^2}{12}+1)dy\] is shell method, which is how I would do it
myininaya
  • 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
  • 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
  • bahrom7893
oh wait so it was this question?
TuringTest
  • 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
  • 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
  • anonymous
I'm still quite confused about the shell method.
TuringTest
  • TuringTest
myin did the integral right, but it is not a solid about the x-axis
anonymous
  • anonymous
Yes it is. ?
TuringTest
  • TuringTest
no, hers are disks about the y-axis
TuringTest
  • TuringTest
|dw:1329325925783:dw|like this but centered on the y-axis
TuringTest
  • TuringTest
area of each disk is pi*r^2, but if r is x(y) we are going around the y-axis
anonymous
  • anonymous
Oh I see.
anonymous
  • anonymous
This is rotation 360 degree about y-axis?
TuringTest
  • 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
  • anonymous
Ok, let me try.
anonymous
  • 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
  • TuringTest
that'll work :D
TuringTest
  • 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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