Here's the question you clicked on:
How do you solve it?
how do i integrate: \[π \int\limits_{0}^{6} (y ^{2}/12 +1)^{2} dy\]
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
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.
I've been taught to integrate like this when the limits are on the y axis: \[\int\limits_{a}^{b} π x ^{2} dy\]
It generates a cylinder with radius 6 and height 4
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.
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?
Find the volume generated by revolving the following area and then subtract it from the vol of the cylinder.
|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.
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
The final answer is suppose to be 90π.
so it is supposed to be the same problem
...which means the integral is wrong
No, there are two diff problems here as far as i can tell, the integral is a completely diff question
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
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.
\[2\pi\int_{0}^{6}y(\frac{y^2}{12}+1)dy\] is shell method, which is how I would do it
\[\pi \int\limits_{0}^{6}(\frac{y^4}{144}+\frac{y^2}{6}+1) dy=\pi(\frac{y^5}{5 \cdot 144}+\frac{y^3}{3 \cdot 6}+y)|_0^6\] \[=\pi(\frac{6^5}{5 \cdot 144}+\frac{6^3}{3 \cdot 6}+6)=\pi(\frac{6^4 \cdot 6}{5 \cdot 144}+\frac{6^2}{3}+6)\] \[=\pi(\frac{6^4 }{5 \cdot 24}+\frac{ 6 \cdot 6}{3}+6)=\pi(\frac{6^4}{5 \cdot 24} +6(2)+6)\] \[=\pi(\frac{6 \cdot 6^3}{5 \cdot 24}+12+6)=\pi(\frac{6^3}{5 \cdot 4}+18)\] \[=\pi(\frac{6 \cdot 6^2}{5 \cdot 4}+18)=\pi(\frac{3 \cdot 6^2}{5 \cdot 2}+18)\] \[=\pi(\frac{3 \cdot 36}{10}+18)=\pi(\frac{12 \cdot 9}{10}+18)=\pi(\frac{108}{10}+18)\] 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
shell method:|dw:1329325465495:dw|Look at the radius of each cylinder( y) and the height of each cylinder (x(y)) and we get\[2\pi\int_{0}^{6}rhdr=2\pi\int_{0}^{6}y(\frac{y^2}{12}+1)=90\pi\]
oh wait so it was this question?
I am integrating along y because I used shell method, disk method would be a pain here because we would need two integrals
@myininaya did it right. I think you used the same method as me, but you expanded x^2 in the beginning? @bahrom7893 yes
I'm still quite confused about the shell method.
myin did the integral right, but it is not a solid about the x-axis
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
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<x<1 rings of inner radius y(x)=sqrt(12x-1) and outer radius y=6 between 1<x<6
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).
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.