## El_Arrow one year ago Need help please!

1. El_Arrow

find the volume of the solid formed by revolving the region bounded by the graph of f(x) = x^2 and g(x) = 12-x to the right of x=1 about the y-axis

2. El_Arrow

first thing i did was set f(x) and g(x) equal to one another and i got -4 and 3

3. El_Arrow

|dw:1440982705148:dw|

4. freckles

$x^2=12-x \\ x^2+x-12=0 \\ (x+4)(x-3)=0 \\ x=-4 \text{ or } x=3$ (-4)^2=16 (3)^2=9 |dw:1440982735696:dw|

5. freckles

|dw:1440982829162:dw| we are looking at this region (the shaded region) right?

6. El_Arrow

yes

7. El_Arrow

so it would be the integral from 1 to 3 @freckles ?

8. freckles

yea

9. freckles

so it looks like you were trying to use shells method

10. El_Arrow

yes thats correct

11. freckles

|dw:1440983126638:dw|

12. El_Arrow

so something like this |dw:1440983268893:dw|

13. freckles

yes. |dw:1440983363623:dw|

14. El_Arrow

but why do you have the square root over the x why not x^2

15. freckles

lol

16. freckles

I forgot what function we were using

17. freckles

sqrt(x) is suppose to be x^2

18. El_Arrow

oh okay lol

19. freckles

i'm sorry i do idiot things sometimes

20. El_Arrow

its okay

21. freckles

|dw:1440983615720:dw|

22. El_Arrow

the answer is 116pi/3 but i got 16pi?

23. freckles

hmm so the radius should have been just x

24. El_Arrow

i think so let me see if it works

25. freckles

|dw:1440983787477:dw|

26. freckles

|dw:1440983907300:dw| so you were right to begin with you just needed 1 to 3 instead of 0 to 1

27. El_Arrow

yeah i got right now thanks alot

28. freckles

it does make sense I think if I had looked at that whole thing from 0 to 3 and subtract out the part that is 0 to 1 we would have seen it better $\int\limits_0^3 2\pi(x)(12-x-x^2) dx-\int\limits_0^1 2\pi(x)(12-x-x^2)dx \\ =\int\limits_1^3 2 \pi (x) (12-x-x^2) dx$