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

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amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0its been done already :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0the typical radius is what.... is what, the slope of the linear between radius and height?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0you are adding up all the circumferences of the circles from 0 to h

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the base of a cone is shaped as a circle

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0[S] 2pi [f(x)] dx ...right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i tried integrating it bt i guess well have 2 use the volume of a frustum

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so the base area is \[\pi r^2\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if you use the disc method to integrate it into a cone, first we need to find the volume of each disc

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0do we include the area of the base into surface area? or is it an open base?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the disc method dint wrk out

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if dx is the width of each disc, then the volume would be \[\pi r^2dx\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now since the radius of a cone changes as the position of the disc changes, we need to find r in terms of x

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0f(x) = h = linear equation from point (0,r) to (r,h)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0or... (0,x) to (x,y) :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0ack...(x,0.. my mistake)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0him; deriving to get volume?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if a cone has radius r and height h, the radius of the disc will be \[({r \over h}x +r)\]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0do we have a height and a radius?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so the volume of the disc is \[\pi ({r \over h}x+r)^2dx\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now we will find the limits of integration

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0this 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 \[\int\limits_{0}^{h} \pi ({r \over h}x+r)^2dx\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0when you integrate it you will get \[{1 \over 3}\pi r^2\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0dV=πr^2dh r=Rh/H dh=Hdr/R dV=πr^2Hdr/R V= πR^3 x H/3R

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah i think this takes care f d volume

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now let's work on the surface area

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0dS=2πrdl l=rL/R dl= Ldr/R S= 2Lπrdr/R πR^2 L/R = πRL

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the 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
 5 years ago
Best ResponseYou've already chosen the best response.0thanks a lot fr the solution

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0again, 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 \[{r \over h}x+r\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0again, 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 \[{r \over h}x+r\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so the lateral area of the discs can be found by \[2\pi[ {({r \over h} )x+r}]dx\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now for this one, you have to be careful with the limit of integration

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so if we could find the slanted height we are good to go

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now you can see that the slanted height makes a right triangle with the height of the cone and the radius of the base

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so if I call L my slanted height, \[L = \sqrt{h^2 + r^2}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0S.A = L*2pi*r, where L is the slant length V = A*h, in this case height is dx, r = x so we have; \[\pi\int_0^x x^2 dx = (1/3)\pi x^3 \]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so the integration we have to do is \[\int\limits_{0}^{L} 2\pi[({r \over h})x+r]dx\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0which ends up being \[\pi rL\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0or \[\pi r \sqrt{h^2+r^2}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Give me about 5min to get the picture of why the radius is \[{r \over h}x+r\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0dV=πr^2dh r=Rh/H dh=Hdr/R dV=πr^2Hdr/R V= πR^3 x H/3R

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It would be nice if you could tell me if it helped or not ... :)
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