## CoconutJJ one year ago The bounded area by the graph y=sqrt(x), y=x-4 and the x axis is rotated along the x axis to form a solid. Use Integration By Shells to find the Volume of the Solid.

1. misssunshinexxoxo

This might help utilize the "Shell" method; https://www.khanacademy.org/math/integral-calculus/solid_revolution_topic/solid_of_revolution/v/solid-of-revolution-part-5

2. CoconutJJ

|dw:1435377786774:dw| This is the bounded area

3. CoconutJJ

Wait so how does the integral change when you rotate it around the x axis instead of y?

4. misssunshinexxoxo
5. misssunshinexxoxo

With "washer method"

6. misssunshinexxoxo

To rotate about the y-axis, swap the roles of x and y throughout the above discussion. To rotate about a different axis, change coordinates so that the axis of rotation is one of the coordinate axes, and then do the computation.

7. misssunshinexxoxo

Follow rules of "shell" and "washer" method to guide throughout finding volume

8. anonymous

been a while, and I was never great at these so take this with a grain of salt. I believe you have to change the formulas around and integrate in terms of y. |dw:1435378239112:dw|

9. anonymous

and then the limits will be the y coordinates of the intersection points

10. CoconutJJ

so then the thickness of the cylinder is dy?

11. CoconutJJ

|dw:1435378522268:dw|

12. UsukiDoll

@peachpi it's been a while for me too.

13. misssunshinexxoxo

Peachpi seems absolutely right; this is a tough area

14. anonymous

this is a solid. so there's no hole, right?

15. anonymous

the area lies on the axis it's being rotated on

16. UsukiDoll

but eventually we have to integrate this thing... who is going to be a and who is going to be b on the integral symbol ?

17. CoconutJJ

well they intersect when x=6.562

18. anonymous

yeah, but the integral is in y. a is 0. b you solve y² = y + 4 to get

19. CoconutJJ

How is the integral with respect to y ?

20. UsukiDoll

maybe it has something to do with the x axis being rotated so we can't use that? I have no idea x.x

21. UsukiDoll

oh I see where the a and b for the integral is |dw:1435378968903:dw|

22. UsukiDoll

those intersection points are the values for a and b

23. CoconutJJ

|dw:1435378996664:dw|

24. UsukiDoll

well a = 0 is correct.. but for the b we need to find the value of that intersection point but as @peachpi mentioned we have to solve " y² = y + 4 to get"

25. anonymous

I think it is simpler to do it horizontally than vertically, which would require two integrals.

26. UsukiDoll

hmmm y^2-y-4 = 0 and then factor ? OH NO! I DO NOT WANT TO DO 2 INTEGRALS

27. UsukiDoll

(-1)^2-4(1)(-4) = 1-4(-4) = 1+ 16 = 17 and it's not a perfect square.. dahhhh

28. anonymous

What is the point of intersection?

29. CoconutJJ

|dw:1435379165728:dw|

30. UsukiDoll

you graph the two equations first. and then they should have common points.

31. CoconutJJ

|dw:1435379200509:dw|

32. UsukiDoll

with 0 being one of them

33. anonymous

they're not nice numbers. about (6.56, 2.56)

34. anonymous

|dw:1435379227291:dw|

35. CoconutJJ

So if we use the shell method from 0 to 4, we are basically using the disc method?

36. UsukiDoll

double integrals.. double the anti derivatives :S

37. anonymous

|dw:1435379315446:dw|

38. anonymous

$\int dV = 2\pi\int rh~dy = 2\pi\int \limits_0^{y_1} y(f(y)-g(y))~dy$

39. CoconutJJ

THANK YOU! I think I can do it from here now :)

40. anonymous

The volume of a cylinder is: $V = \pi r^2h$When you differentiate, it is: $\frac{dV}{dr} = 2\pi r h \implies \int dV = 2\pi\int rh~dr$That's how I think about it.

41. CoconutJJ

Thankks