## 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

1. Loser66

2. Loser66

|dw:1436218798400:dw|

3. Loser66

The intersection points:$$x^2 = x+2$$ gives ust x =-1 and x =2 then the integral is defined as $\int_{-1}^2( area~outer -area~inner)dx\\\pi* \int_{-1}^2 ((4-x^2)^2 -(4-(x+2))^2dx$

4. anonymous

The answer is correct, but that is the washer method.

5. Loser66

oh, shell method!! ooooooooook

6. 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)

7. dan815

:)

8. 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

9. Loser66

|dw:1436221417680:dw|

10. anonymous

Got it

11. xapproachesinfinity

|dw:1436222834363:dw| the shape is like this i have some good drawing skills don't yo think hahah

12. xapproachesinfinity

|dw:1436222936259:dw|

13. 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

14. anonymous

Thank you though

15. xapproachesinfinity

you integrated with respect to x yes?

16. anonymous

dy

17. xapproachesinfinity

hmm why? the height of the cylinder goes in x direction

18. anonymous

correct this is the shell method

19. xapproachesinfinity

the shell method is the cylinder method can you show your work, how you integrated with dy

20. anonymous

$\int\limits_{1}^{4}2\pi(4-y)(\sqrt{y}-(y-2)dy +\int\limits_{0}^{1}2\pi(4-y)(2\sqrt{y})dy$

21. xapproachesinfinity

let me draw that nicely with a program! this drawing confuses me lol

22. anonymous

You can use the washer method with respect to X to get the same answer $\int\limits_{-1}^{2}\pi((4-x)^2-(4-(x+2)))^2dx$

23. xapproachesinfinity

why do you have 0 to 1?

24. 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 $-\sqrt{y}. So \int\limits_{0}^{1} is taken with positive \sqrt{y} - -\sqrt{y}$

25. xapproachesinfinity

sorry can't focus, i will check this later :)

26. anonymous

its correct the correct answer is 108pi/5

27. freckles

|dw:1436225174941:dw|

28. freckles

$y=x+2 \\ x=y-2 \\ \text{ and } y=x^2 \\ \text{ gives } x=\sqrt{y} \text{ or } x=-\sqrt{y}$

29. freckles

|dw:1436225286527:dw|

30. freckles

|dw:1436225325924:dw|

31. freckles

we have two difference things to look at because the left function switches at y=1

32. anonymous

we've gone through I'd like to close

33. 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

34. freckles

you can close it just trying to help anyone who is still trying to understand it

35. anonymous

Clicking on close and nothing is happening. Continue on if you want

36. 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 ->): $$\href{http:///www.wolframalpha.com/input/?i=pi+*+integral++%28%28x%5E2-4%29%5E2+-+%28x-2%29%5E2%29+dx%2Cx%3D-1+to+x%3D2}{\cfrac{108\pi}{5}}$$.

37. 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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