## YUNoWorkOS 2 years ago Find the volume of the solid of revolution generated by revolving the equation $$y=\sqrt(x)$$ about the line x=4 (The region is bounded by the x axis and x=4)

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1. inkyvoyd

I understand if I rewrite the equation as y=$$\sqrt{x-4}$$ then integrate with appropriate bounds it will work, but there is some sort of alternate interpretation I am not understanding.

2. inkyvoyd

*rewrite the equation as y=$$\sqrt{x+4}$$

3. amistre64

|dw:1365190324765:dw| $\int 2\pi ~r~h$

4. amistre64

r = c-x, h = sqrt(x), and dx from 0 to c

5. YUNoWorkOS

omg thank you - that applies for shell method as well correct?

6. amistre64

that IS the shell method :)

7. YUNoWorkOS

oh shoot - is there any way to do thi with the disk method?

8. YUNoWorkOS

this problem is listed in the disk method section of my textbook, but i'm not sure how it's done...

9. amistre64

$\int~2\pi r$ $r=g(y)\text{ ; which is the inverse of f(x)}$and dy from 0 to f(c)

10. amistre64

inverse of y = sqrt(x); x = y^2

11. amistre64

inky suggests that we move the graph to something simpler to play with; shift if left by "c" units given us y = sqrt(x+c) r = x = sqrt(y) - c

12. amistre64

r = x = y^2 - c :)

13. amistre64

the only issue that may present itself with that setup tho is that you get a negative summation

14. amistre64

why did i do 2pi r? thats circumference :/ .... area is $\int pi r^2$