## YUNoWorkOS Group Title 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) one year ago one year ago

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1. inkyvoyd Group Title

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 Group Title

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

3. amistre64 Group Title

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

4. amistre64 Group Title

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

5. YUNoWorkOS Group Title

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

6. amistre64 Group Title

that IS the shell method :)

7. YUNoWorkOS Group Title

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

8. YUNoWorkOS Group Title

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

9. amistre64 Group Title

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

10. amistre64 Group Title

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

11. amistre64 Group Title

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 Group Title

r = x = y^2 - c :)

13. amistre64 Group Title

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

14. amistre64 Group Title

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