anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Loser66
  • Loser66
draw it out, please
Loser66
  • Loser66
|dw:1436218798400:dw|
Loser66
  • Loser66
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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anonymous
  • anonymous
The answer is correct, but that is the washer method.
Loser66
  • Loser66
oh, shell method!! ooooooooook
Loser66
  • Loser66
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)
dan815
  • dan815
:)
anonymous
  • anonymous
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
Loser66
  • Loser66
|dw:1436221417680:dw|
anonymous
  • anonymous
Got it
xapproachesinfinity
  • xapproachesinfinity
|dw:1436222834363:dw| the shape is like this i have some good drawing skills don't yo think hahah
xapproachesinfinity
  • xapproachesinfinity
|dw:1436222936259:dw|
anonymous
  • anonymous
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
anonymous
  • anonymous
Thank you though
xapproachesinfinity
  • xapproachesinfinity
you integrated with respect to x yes?
anonymous
  • anonymous
dy
xapproachesinfinity
  • xapproachesinfinity
hmm why? the height of the cylinder goes in x direction
anonymous
  • anonymous
correct this is the shell method
xapproachesinfinity
  • xapproachesinfinity
the shell method is the cylinder method can you show your work, how you integrated with dy
anonymous
  • anonymous
\[\int\limits_{1}^{4}2\pi(4-y)(\sqrt{y}-(y-2)dy +\int\limits_{0}^{1}2\pi(4-y)(2\sqrt{y})dy\]
xapproachesinfinity
  • xapproachesinfinity
let me draw that nicely with a program! this drawing confuses me lol
anonymous
  • anonymous
You can use the washer method with respect to X to get the same answer \[\int\limits_{-1}^{2}\pi((4-x)^2-(4-(x+2)))^2dx\]
xapproachesinfinity
  • xapproachesinfinity
why do you have 0 to 1?
anonymous
  • anonymous
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 \[-\sqrt{y}. So \int\limits_{0}^{1} is taken with positive \sqrt{y} - -\sqrt{y}\]
xapproachesinfinity
  • xapproachesinfinity
sorry can't focus, i will check this later :)
anonymous
  • anonymous
its correct the correct answer is 108pi/5
freckles
  • freckles
|dw:1436225174941:dw|
freckles
  • freckles
\[y=x+2 \\ x=y-2 \\ \text{ and } y=x^2 \\ \text{ gives } x=\sqrt{y} \text{ or } x=-\sqrt{y}\]
freckles
  • freckles
|dw:1436225286527:dw|
freckles
  • freckles
|dw:1436225325924:dw|
freckles
  • freckles
we have two difference things to look at because the left function switches at y=1
anonymous
  • anonymous
we've gone through I'd like to close
freckles
  • freckles
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
freckles
  • freckles
you can close it just trying to help anyone who is still trying to understand it
anonymous
  • anonymous
Clicking on close and nothing is happening. Continue on if you want
ybarrap
  • ybarrap
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 ->): \(\href{http:///www.wolframalpha.com/input/?i=pi+*+integral++%28%28x%5E2-4%29%5E2+-+%28x-2%29%5E2%29+dx%2Cx%3D-1+to+x%3D2}{\cfrac{108\pi}{5}}\).
ybarrap
  • ybarrap
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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