## anonymous 5 years ago any body can derive the formula for a surface area and volume of cone through integration(calculus)??????????

1. amistre64

2. anonymous

hahaha

3. amistre64

the typical radius is what.... is what, the slope of the linear between radius and height?

4. amistre64

you are adding up all the circumferences of the circles from 0 to h

5. anonymous

volume is easy

6. anonymous

the base of a cone is shaped as a circle

7. amistre64

[S] 2pi [f(x)] dx ...right?

8. anonymous

i tried integrating it bt i guess well have 2 use the volume of a frustum

9. anonymous

so the base area is $\pi r^2$

10. anonymous

if you use the disc method to integrate it into a cone, first we need to find the volume of each disc

11. amistre64

do we include the area of the base into surface area? or is it an open base?

12. anonymous

the disc method dint wrk out

13. anonymous

if dx is the width of each disc, then the volume would be $\pi r^2dx$

14. anonymous

i tried this

15. amistre64

r = [f(x)] right?

16. anonymous

now since the radius of a cone changes as the position of the disc changes, we need to find r in terms of x

17. amistre64

f(x) = h = linear equation from point (0,r) to (r,h)

18. amistre64

or... (0,x) to (x,y) :)

19. amistre64

ack...(x,0.. my mistake)

20. anonymous

dV= πr2^2dx

21. amistre64

him; deriving to get volume?

22. amistre64

integrate

23. anonymous

if a cone has radius r and height h, the radius of the disc will be $({-r \over h}x +r)$

24. anonymous

r=xtant

25. amistre64

do we have a height and a radius?

26. anonymous

so the volume of the disc is $\pi ({-r \over h}x+r)^2dx$

27. anonymous

now we will find the limits of integration

28. anonymous

this one is easy, the height is h, so the limit of integration is from x=0 to x=h thus the integral we are looking for to find the volume is $\int\limits_{0}^{h} \pi ({-r \over h}x+r)^2dx$

29. anonymous

when you integrate it you will get ${1 \over 3}\pi r^2$

30. anonymous

dV=πr^2dh r=Rh/H dh=Hdr/R dV=πr^2Hdr/R V= πR^3 x H/3R

31. anonymous

did it help at all ?

32. anonymous

yeah i think this takes care f d volume

33. anonymous

now let's work on the surface area

34. anonymous

dS=2πrdl l=rL/R dl= Ldr/R S= 2Lπrdr/R πR^2 L/R = πRL

35. anonymous

DONE!!!!!!!!!!

36. anonymous

the same idea is applied to the problem, but now instead of a volume, we are looking for an area. so instead of a disc, we will consider the lateral area of the disc, which can be found by $2\pi r *dx$

37. anonymous

thanks a lot fr the solution

38. anonymous

again, the radius of a cone changes, so we need to find the radius in terms of x. But we already found it, which is again ${-r \over h}x+r$

39. anonymous

again, the radius of a cone changes, so we need to find the radius in terms of x. But we already found it, which is again ${-r \over h}x+r$

40. anonymous

so the lateral area of the discs can be found by $2\pi[ {({-r \over h} )x+r}]dx$

41. anonymous

now for this one, you have to be careful with the limit of integration

42. anonymous

the length of the slanted height is going to be my limit of integration because we are finding the lateral area of each discs so that all dx's add up to the slanted height.

43. anonymous

so if we could find the slanted height we are good to go

44. anonymous

now you can see that the slanted height makes a right triangle with the height of the cone and the radius of the base

45. anonymous

so if I call L my slanted height, $L = \sqrt{h^2 + r^2}$

46. anonymous

S.A = L*2pi*r, where L is the slant length V = A*h, in this case height is dx, r = x so we have; $\pi\int_0^x x^2 dx = (1/3)\pi x^3$

47. anonymous

so the integration we have to do is $\int\limits_{0}^{L} 2\pi[({-r \over h})x+r]dx$

48. anonymous

which ends up being $\pi rL$

49. anonymous

or $\pi r \sqrt{h^2+r^2}$

50. anonymous

I hope that helps :)

51. anonymous

Give me about 5min to get the picture of why the radius is ${-r \over h}x+r$

52. anonymous

dV=πr^2dh r=Rh/H dh=Hdr/R dV=πr^2Hdr/R V= πR^3 x H/3R

53. anonymous

thats the vol

54. anonymous

55. anonymous

It would be nice if you could tell me if it helped or not ... :)

56. anonymous

ya it helped thanks