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adnanchowdhury
 3 years ago
Best ResponseYou've already chosen the best response.0How do you solve it?

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

bahrom7893
 3 years ago
Best ResponseYou've already chosen the best response.2Rotate 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

bahrom7893
 3 years ago
Best ResponseYou've already chosen the best response.2and for that integral, just multiply it out.

bahrom7893
 3 years ago
Best ResponseYou've already chosen the best response.2And then integrate term by term that will make your life easy.

bahrom7893
 3 years ago
Best ResponseYou've already chosen the best response.2actually finding the vol of the solid generated by the rectangle is easy.

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

bahrom7893
 3 years ago
Best ResponseYou've already chosen the best response.2It generates a cylinder with radius 6 and height 4

bahrom7893
 3 years ago
Best ResponseYou've already chosen the best response.2Yea 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.

adnanchowdhury
 3 years ago
Best ResponseYou've already chosen the best response.0What rectangle are you referring to?

bahrom7893
 3 years ago
Best ResponseYou've already chosen the best response.2Imagine a line passing through (4,6) and (4,0)

bahrom7893
 3 years ago
Best ResponseYou've already chosen the best response.2that will form a rectangle right? with the line y = 6, x = 4, and x and y axes

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

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

bahrom7893
 3 years ago
Best ResponseYou've already chosen the best response.2So find the volume of the solid generated by that rectangle  it will be a cylinder

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

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

bahrom7893
 3 years ago
Best ResponseYou've already chosen the best response.2Find the volume generated by revolving the following area and then subtract it from the vol of the cylinder.

bahrom7893
 3 years ago
Best ResponseYou've already chosen the best response.2dw:1328460177913:dw

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

bahrom7893
 3 years ago
Best ResponseYou've already chosen the best response.2Just expand and then integrate term by term, it's not hard.

adnanchowdhury
 3 years ago
Best ResponseYou've already chosen the best response.0without expanding? Like chain rule?

bahrom7893
 3 years ago
Best ResponseYou've already chosen the best response.2I'm not sure if you can. Does the question force you to integrate without expanding?

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

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

bahrom7893
 3 years ago
Best ResponseYou've already chosen the best response.2no that's a diff question

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

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.0that is the only way i see anyways

bahrom7893
 3 years ago
Best ResponseYou've already chosen the best response.2I mean it's not usub, and I don't recognize any inverse formulas.. inverse tan won't work

bahrom7893
 3 years ago
Best ResponseYou've already chosen the best response.2and integration by parts is gonna give u a horrible mess

TuringTest
 3 years ago
Best 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?

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

bahrom7893
 3 years ago
Best ResponseYou've already chosen the best response.2Yea turing, it's separate

bahrom7893
 3 years ago
Best ResponseYou've already chosen the best response.2I think expanding is the only way.

adnanchowdhury
 3 years ago
Best ResponseYou've already chosen the best response.0Basicallly, I ended up with this integral, by rearranging the given equation to make x the subject. Then substituting into this: http://d.pr/PR8z

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

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

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

bahrom7893
 3 years ago
Best ResponseYou've already chosen the best response.2No, there are two diff problems here as far as i can tell, the integral is a completely diff question

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.2No ,they just said they got the integral by subbing in for y, but they are integrating on the wrong axis with the wrong method

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

TuringTest
 3 years ago
Best 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

myininaya
 3 years ago
Best 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

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

bahrom7893
 3 years ago
Best ResponseYou've already chosen the best response.2oh wait so it was this question?

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

adnanchowdhury
 3 years ago
Best 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

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

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

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

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

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

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

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.2if 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

adnanchowdhury
 3 years ago
Best ResponseYou've already chosen the best response.0Yes! 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).

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.2Shell 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.
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