Question

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

Answer 1

its been done already :)

Answer 2

hahaha

Answer 3

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

Answer 4

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

Answer 5

volume is easy

Answer 6

the base of a cone is shaped as a circle

Answer 7

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

Answer 8

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

Answer 9

so the base area is \[\pi r^2\]

Answer 10

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

Answer 11

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

Answer 12

the disc method dint wrk out

Answer 13

if dx is the width of each disc, then the volume would be

\[\pi r^2dx\]

Answer 14

i tried this

Answer 15

r = [f(x)] right?

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

dV= πr2^2dx

Answer 21

him; deriving to get volume?

Answer 22

integrate

Answer 23

if a cone has radius r and height h, the radius of the disc will be

\[({-r \over h}x +r)\]

Answer 24

r=xtant

Answer 25

do we have a height and a radius?

Answer 26

so the volume of the disc is

\[\pi ({-r \over h}x+r)^2dx\]

Answer 27

now we will find the limits of integration

Answer 28

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

\[\int\limits_{0}^{h} \pi ({-r \over h}x+r)^2dx\]

Answer 29

when you integrate it you will get

\[{1 \over 3}\pi r^2\]

Answer 30

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

Answer 31

did it help at all ?

Answer 32

yeah i think this takes care f d volume

Answer 33

now let's work on the surface area

Answer 34

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

Answer 35

DONE!!!!!!!!!!

Answer 36

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 \[2\pi r *dx\]

Answer 37

thanks a lot fr the solution

Answer 38

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

\[{-r \over h}x+r\]

Answer 39

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

\[{-r \over h}x+r\]

Answer 40

so the lateral area of the discs can be found by

\[2\pi[ {({-r \over h} )x+r}]dx\]

Answer 41

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

Answer 42

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.

Answer 43

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

Answer 44

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

Answer 45

so if I call L my slanted height,

\[L = \sqrt{h^2 + r^2}\]

Answer 46

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;
\[\pi\int_0^x x^2 dx = (1/3)\pi x^3 \]

Answer 47

so the integration we have to do is

\[\int\limits_{0}^{L} 2\pi[({-r \over h})x+r]dx\]

Answer 48

which ends up being

\[\pi rL\]

Answer 49

or
\[\pi r \sqrt{h^2+r^2}\]

Answer 50

I hope that helps :)

Answer 51

Give me about 5min to get the picture of why the radius is

\[{-r \over h}x+r\]

Answer 52

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

Answer 53

thats the vol

Answer 54

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

Answer 55

ya it helped thanks