anonymous
  • anonymous
any body can derive the formula for a surface area and volume of cone through integration(calculus)??????????
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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amistre64
  • amistre64
its been done already :)
anonymous
  • anonymous
hahaha
amistre64
  • amistre64
the typical radius is what.... is what, the slope of the linear between radius and height?

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More answers

amistre64
  • amistre64
you are adding up all the circumferences of the circles from 0 to h
yuki
  • yuki
volume is easy
yuki
  • yuki
the base of a cone is shaped as a circle
amistre64
  • amistre64
[S] 2pi [f(x)] dx ...right?
anonymous
  • anonymous
i tried integrating it bt i guess well have 2 use the volume of a frustum
yuki
  • yuki
so the base area is \[\pi r^2\]
yuki
  • yuki
if you use the disc method to integrate it into a cone, first we need to find the volume of each disc
amistre64
  • amistre64
do we include the area of the base into surface area? or is it an open base?
anonymous
  • anonymous
the disc method dint wrk out
yuki
  • yuki
if dx is the width of each disc, then the volume would be \[\pi r^2dx\]
anonymous
  • anonymous
i tried this
amistre64
  • amistre64
r = [f(x)] right?
yuki
  • 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
  • amistre64
f(x) = h = linear equation from point (0,r) to (r,h)
amistre64
  • amistre64
or... (0,x) to (x,y) :)
amistre64
  • amistre64
ack...(x,0.. my mistake)
anonymous
  • anonymous
dV= πr2^2dx
amistre64
  • amistre64
him; deriving to get volume?
amistre64
  • amistre64
integrate
yuki
  • yuki
if a cone has radius r and height h, the radius of the disc will be \[({-r \over h}x +r)\]
anonymous
  • anonymous
r=xtant
amistre64
  • amistre64
do we have a height and a radius?
yuki
  • yuki
so the volume of the disc is \[\pi ({-r \over h}x+r)^2dx\]
yuki
  • yuki
now we will find the limits of integration
yuki
  • 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
  • yuki
when you integrate it you will get \[{1 \over 3}\pi r^2\]
anonymous
  • anonymous
dV=πr^2dh r=Rh/H dh=Hdr/R dV=πr^2Hdr/R V= πR^3 x H/3R
yuki
  • yuki
did it help at all ?
anonymous
  • anonymous
yeah i think this takes care f d volume
yuki
  • yuki
now let's work on the surface area
anonymous
  • anonymous
dS=2πrdl l=rL/R dl= Ldr/R S= 2Lπrdr/R πR^2 L/R = πRL
anonymous
  • anonymous
DONE!!!!!!!!!!
yuki
  • 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
  • anonymous
thanks a lot fr the solution
yuki
  • 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
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
  • yuki
now for this one, you have to be careful with the limit of integration
yuki
  • 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
  • yuki
so if we could find the slanted height we are good to go
yuki
  • 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
  • yuki
so if I call L my slanted height, \[L = \sqrt{h^2 + r^2}\]
anonymous
  • 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
  • yuki
so the integration we have to do is \[\int\limits_{0}^{L} 2\pi[({-r \over h})x+r]dx\]
yuki
  • yuki
which ends up being \[\pi rL\]
yuki
  • yuki
or \[\pi r \sqrt{h^2+r^2}\]
yuki
  • yuki
I hope that helps :)
yuki
  • yuki
Give me about 5min to get the picture of why the radius is \[{-r \over h}x+r\]
anonymous
  • anonymous
dV=πr^2dh r=Rh/H dh=Hdr/R dV=πr^2Hdr/R V= πR^3 x H/3R
anonymous
  • anonymous
thats the vol
yuki
  • yuki
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yuki
  • yuki
It would be nice if you could tell me if it helped or not ... :)
anonymous
  • anonymous
ya it helped thanks

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