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anonymous
 one year ago
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
anonymous
 one year ago
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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Loser66
 one year ago
Best ResponseYou've already chosen the best response.3The intersection points:\(x^2 = x+2\) gives ust x =1 and x =2 then the integral is defined as \[\int_{1}^2( area~outer area~inner)dx\\\pi* \int_{1}^2 ((4x^2)^2 (4(x+2))^2dx\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The answer is correct, but that is the washer method.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.3oh, shell method!! ooooooooook

Loser66
 one year ago
Best ResponseYou've already chosen the best response.3I 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)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Clearly 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 41 10 and still am not getting the correct answer

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.1dw:1436222834363:dw the shape is like this i have some good drawing skills don't yo think hahah

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.1dw:1436222936259:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thats why I don't bother with attempting to draw it online, I'm terrible. When I was taking the second integral between 10 I was doing y(2sqrty) I was leaving off the 4 for the radius of that one stupid simple mistakes

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.1you integrated with respect to x yes?

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.1hmm why? the height of the cylinder goes in x direction

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0correct this is the shell method

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.1the shell method is the cylinder method can you show your work, how you integrated with dy

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{1}^{4}2\pi(4y)(\sqrt{y}(y2)dy +\int\limits_{0}^{1}2\pi(4y)(2\sqrt{y})dy\]

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.1let me draw that nicely with a program! this drawing confuses me lol

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You can use the washer method with respect to X to get the same answer \[\int\limits_{1}^{2}\pi((4x)^2(4(x+2)))^2dx\]

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.1why do you have 0 to 1?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0because 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 y2. It changes to \[\sqrt{y}. So \int\limits_{0}^{1} is taken with positive \sqrt{y}  \sqrt{y}\]

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.1sorry can't focus, i will check this later :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0its correct the correct answer is 108pi/5

freckles
 one year ago
Best ResponseYou've already chosen the best response.0dw:1436225174941:dw

freckles
 one year ago
Best ResponseYou've already chosen the best response.0\[y=x+2 \\ x=y2 \\ \text{ and } y=x^2 \\ \text{ gives } x=\sqrt{y} \text{ or } x=\sqrt{y}\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.0dw:1436225286527:dw

freckles
 one year ago
Best ResponseYou've already chosen the best response.0dw:1436225325924:dw

freckles
 one year ago
Best ResponseYou've already chosen the best response.0we have two difference things to look at because the left function switches at y=1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0we've gone through I'd like to close

freckles
 one year ago
Best ResponseYou've already chosen the best response.0like 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=y2

freckles
 one year ago
Best ResponseYou've already chosen the best response.0you can close it just trying to help anyone who is still trying to understand it

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Clicking on close and nothing is happening. Continue on if you want

ybarrap
 one year ago
Best ResponseYou've already chosen the best response.1Since 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 xaxis: dw:1436229534071:dw Setup and evaluate your integral: $$ \large{ \int_{1}^2\int_{x2}^{x^24}2\pi y~dy~dx\\ =\left.\begin{matrix}\pi\int_{1}^2y^2 \\ \end{matrix}\right^{x^24}_{x2}~dx\\ =\pi\int_{1}^2\left((x^24)^2(x2)^2\right)\\ =\pi\int_{1}^2\left(x^49 x^2+4 x+12\right)~dx\\ =\pi \left(\frac{x^5}{5}3 x^3+2 x^2+12 x\right)^2_{1}\\ } $$ Evaluation using wolfram gives (control+click >): \(\href{http:///www.wolframalpha.com/input/?i=pi+*+integral++%28%28x%5E24%29%5E2++%28x2%29%5E2%29+dx%2Cx%3D1+to+x%3D2}{\cfrac{108\pi}{5}}\).

ybarrap
 one year ago
Best ResponseYou've already chosen the best response.1And... here is the solution using the shell method:\( \href{http:///www.wolframalpha.com/input/?i=2*pi*+%28integral+y*%28sqrt%284y%29++%282y%29+%29+dy%2Cy%3D0+to+y%3D3%29+%2B+%284*pi*integral+y*sqrt%284y%29+++dy%2C+y%3D3+to+y%3D4%29}{\cfrac{108\pi}{5}\text{ (control+click me)}}. \) $$ ~~\!\bf\color{blue}{o}~\color{blue}{o}\\ ~~~\!\bullet \\ ~\smile \\ ~~\!\!/\text{\\} $$
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