Use the shell method to find the volume of the solid generated by revolving the region bounded by the line y=x+2 and the parabola y=x^2 about the following line. y=4

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Use the shell method to find the volume of the solid generated by revolving the region bounded by the line y=x+2 and the parabola y=x^2 about the following line. y=4

Mathematics
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draw it out, please
|dw:1436218798400:dw|
The intersection points:\(x^2 = x+2\) gives ust x =-1 and x =2 then the integral is defined as \[\int_{-1}^2( area~outer -area~inner)dx\\\pi* \int_{-1}^2 ((4-x^2)^2 -(4-(x+2))^2dx\]

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The answer is correct, but that is the washer method.
oh, shell method!! ooooooooook
I think we have to break it into 2 parts, the first one from -1 to 0 and the second one from 0 to 2 (w.r.t.x)
:)
Clearly I don't know or I wouldn't be asking, but with the shell method aren't we taking it with respect to y. With that said, I've broken it down from 4-1 1-0 and still am not getting the correct answer
|dw:1436221417680:dw|
Got it
|dw:1436222834363:dw| the shape is like this i have some good drawing skills don't yo think hahah
|dw:1436222936259:dw|
Thats why I don't bother with attempting to draw it online, I'm terrible. When I was taking the second integral between 1-0 I was doing y(2sqrty) I was leaving off the 4- for the radius of that one stupid simple mistakes
Thank you though
you integrated with respect to x yes?
dy
hmm why? the height of the cylinder goes in x direction
correct this is the shell method
the shell method is the cylinder method can you show your work, how you integrated with dy
\[\int\limits_{1}^{4}2\pi(4-y)(\sqrt{y}-(y-2)dy +\int\limits_{0}^{1}2\pi(4-y)(2\sqrt{y})dy\]
let me draw that nicely with a program! this drawing confuses me lol
You can use the washer method with respect to X to get the same answer \[\int\limits_{-1}^{2}\pi((4-x)^2-(4-(x+2)))^2dx\]
why do you have 0 to 1?
because you actually have 3 functions, but its "hidden" If you look at the graph at the point -1,1 you can no longer take the height using y-2. It changes to \[-\sqrt{y}. So \int\limits_{0}^{1} is taken with positive \sqrt{y} - -\sqrt{y}\]
sorry can't focus, i will check this later :)
its correct the correct answer is 108pi/5
|dw:1436225174941:dw|
\[y=x+2 \\ x=y-2 \\ \text{ and } y=x^2 \\ \text{ gives } x=\sqrt{y} \text{ or } x=-\sqrt{y}\]
|dw:1436225286527:dw|
|dw:1436225325924:dw|
we have two difference things to look at because the left function switches at y=1
we've gone through I'd like to close
like from y=0 to y=1 we have right function is x=sqrt(y) and left function x=-sqrt(y) and from y=1 to y=4 we have right function x=sqrt(y) and left function x=y-2
you can close it just trying to help anyone who is still trying to understand it
Clicking on close and nothing is happening. Continue on if you want
Since this question is closed, I wanted to document for those interested the details of @Loser66's washer solution. Move the graphs down by 4 to make symmetric about the x-axis: |dw:1436229534071:dw| Setup and evaluate your integral: $$ \large{ \int_{-1}^2\int_{x-2}^{x^2-4}2\pi y~dy~dx\\ =\left.\begin{matrix}\pi\int_{-1}^2y^2 \\ \end{matrix}\right|^{x^2-4}_{x-2}~dx\\ =\pi\int_{-1}^2\left((x^2-4)^2-(x-2)^2\right)\\ =\pi\int_{-1}^2\left(x^4-9 x^2+4 x+12\right)~dx\\ =\pi \left(\frac{x^5}{5}-3 x^3+2 x^2+12 x\right)|^2_{-1}\\ } $$ Evaluation using wolfram gives (control+click ->): \(\href{http:///www.wolframalpha.com/input/?i=pi+*+integral++%28%28x%5E2-4%29%5E2+-+%28x-2%29%5E2%29+dx%2Cx%3D-1+to+x%3D2}{\cfrac{108\pi}{5}}\).
And... here is the solution using the shell method:\( \href{http:///www.wolframalpha.com/input/?i=2*pi*+%28integral+y*%28sqrt%284-y%29+-+%282-y%29+%29+dy%2Cy%3D0+to+y%3D3%29+%2B+%284*pi*integral+y*sqrt%284-y%29+++dy%2C+y%3D3+to+y%3D4%29}{\cfrac{108\pi}{5}\text{ (control+click me)}}. \) $$ ~~\!\bf\color{blue}{o}~\color{blue}{o}\\ ~~~\!\bullet \\ ~\smile \\ ~~\!\!/||\text{\\} $$

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