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

- anonymous

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- amistre64

its been done already :)

- anonymous

hahaha

- amistre64

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

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## More answers

- amistre64

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

- yuki

volume is easy

- yuki

the base of a cone is shaped as a circle

- amistre64

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

- anonymous

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

- yuki

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

- yuki

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

- amistre64

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

- anonymous

the disc method dint wrk out

- yuki

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

- anonymous

i tried this

- amistre64

r = [f(x)] right?

- yuki

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

- amistre64

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

- amistre64

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

- amistre64

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

- anonymous

dV= πr2^2dx

- amistre64

him; deriving to get volume?

- amistre64

integrate

- yuki

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

- anonymous

r=xtant

- amistre64

do we have a height and a radius?

- yuki

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

- yuki

now we will find the limits of integration

- yuki

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

- yuki

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

- anonymous

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

- yuki

did it help at all ?

- anonymous

yeah i think this takes care f d volume

- yuki

now let's work on the surface area

- anonymous

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

- anonymous

DONE!!!!!!!!!!

- yuki

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 \[2\pi r *dx\]

- anonymous

thanks a lot fr the solution

- yuki

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

- yuki

- yuki

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

- yuki

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

- yuki

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.

- yuki

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

- yuki

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

- yuki

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

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

- yuki

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

- yuki

which ends up being
\[\pi rL\]

- yuki

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

- yuki

I hope that helps :)

- yuki

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

- anonymous

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

- anonymous

thats the vol

- yuki

##### 1 Attachment

- yuki

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

- anonymous

ya it helped thanks

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