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Question Number 3714 by Rasheed Soomro last updated on 19/Dec/15
Derive a formula of volume of right circular cone  when the formula of volume of cyllinder  is given.
$$\mathcal{D}\mathrm{erive}\:\mathrm{a}\:\mathrm{formula}\:\mathrm{of}\:\mathrm{volume}\:\mathrm{of}\:\boldsymbol{\mathrm{right}}\:\boldsymbol{\mathrm{circular}}\:\boldsymbol{\mathrm{cone}} \\ $$$$\mathrm{when}\:\mathrm{the}\:\mathrm{formula}\:\mathrm{of}\:\mathrm{volume}\:\mathrm{of}\:\boldsymbol{\mathrm{cyllinder}}\:\:\mathrm{is}\:\mathrm{given}. \\ $$
Answered by Filup last updated on 19/Dec/15
I′m not sure how to answer the question  given. But I can show it this way:    Line y=mx  height = r  base length = h  (its a sideways cone against the x axis)  ∴m=(r/h)  y=(r/h)x    V=π∫_a ^( b) y^2 dx  ∴V=π∫_0 ^( h) ((r/h)x)^2 dx  V=πr^2 (1/h^2 )∫_0 ^( h) x^2 dx  V=πr^2 (1/h^2 )((1/3)h^3 −0)  ∴V=(1/3)πr^2 h
$$\mathrm{I}'\mathrm{m}\:\mathrm{not}\:\mathrm{sure}\:\mathrm{how}\:\mathrm{to}\:\mathrm{answer}\:\mathrm{the}\:\mathrm{question} \\ $$$$\mathrm{given}.\:\mathrm{But}\:\mathrm{I}\:\mathrm{can}\:\mathrm{show}\:\mathrm{it}\:\mathrm{this}\:\mathrm{way}: \\ $$$$ \\ $$$$\mathrm{Line}\:{y}={mx} \\ $$$${height}\:=\:{r} \\ $$$${base}\:{length}\:=\:{h} \\ $$$$\left({its}\:{a}\:{sideways}\:{cone}\:{against}\:{the}\:{x}\:{axis}\right) \\ $$$$\therefore{m}=\frac{{r}}{{h}} \\ $$$${y}=\frac{{r}}{{h}}{x} \\ $$$$ \\ $$$${V}=\pi\int_{{a}} ^{\:{b}} {y}^{\mathrm{2}} {dx} \\ $$$$\therefore{V}=\pi\int_{\mathrm{0}} ^{\:{h}} \left(\frac{{r}}{{h}}{x}\right)^{\mathrm{2}} {dx} \\ $$$${V}=\pi{r}^{\mathrm{2}} \frac{\mathrm{1}}{{h}^{\mathrm{2}} }\int_{\mathrm{0}} ^{\:{h}} {x}^{\mathrm{2}} {dx} \\ $$$${V}=\pi{r}^{\mathrm{2}} \frac{\mathrm{1}}{{h}^{\mathrm{2}} }\left(\frac{\mathrm{1}}{\mathrm{3}}{h}^{\mathrm{3}} −\mathrm{0}\right) \\ $$$$\therefore{V}=\frac{\mathrm{1}}{\mathrm{3}}\pi{r}^{\mathrm{2}} {h} \\ $$$$ \\ $$
Commented by RasheedSindhi last updated on 19/Dec/15
Integral is used for area.  How is it used for volume?  Which line is y=mx ?
$${Integral}\:{is}\:{used}\:{for}\:{area}. \\ $$$${How}\:{is}\:{it}\:{used}\:{for}\:{volume}? \\ $$$${Which}\:{line}\:{is}\:{y}={mx}\:? \\ $$
Commented by Filup last updated on 19/Dec/15
for y=f(x), volume is found by  rotating around either the x or y axis.    Rotate around x axis:  V=π∫_a ^( b) y^2 dx    y axis:  V=π∫_a ^( b) x^2 dy    a and b are values on the axis which  you are rotating around. Similarly:  V=π∫_x_1  ^( x_2 ) y^2 dx  V=π∫_(f(x_1 )) ^( f(x_2 )) x^2 dy      For example:  y=x^2   Rotating the curve around the y axis,  you have a shape that resembles a bowl.  Say the bowl is 4 units high.  y=x^2       (rotate around y  ∴have to                      transpose for x)  V=π∫_a ^( b) x^2 dy  y=x^2   x^2 =y  V=π∫_0 ^( 4) x^2 dy  V=π∫_0 ^( 4) ydy  V=(1/2)π16  V=8π
$${for}\:{y}={f}\left({x}\right),\:\mathrm{volume}\:\mathrm{is}\:\mathrm{found}\:\mathrm{by} \\ $$$$\mathrm{rotating}\:\mathrm{around}\:\mathrm{either}\:\mathrm{the}\:\mathrm{x}\:\mathrm{or}\:\mathrm{y}\:\mathrm{axis}. \\ $$$$ \\ $$$$\mathrm{Rotate}\:\mathrm{around}\:{x}\:\mathrm{axis}: \\ $$$${V}=\pi\int_{{a}} ^{\:{b}} {y}^{\mathrm{2}} {dx} \\ $$$$ \\ $$$${y}\:{axis}: \\ $$$${V}=\pi\int_{{a}} ^{\:{b}} {x}^{\mathrm{2}} {dy} \\ $$$$ \\ $$$${a}\:\mathrm{and}\:{b}\:\mathrm{are}\:\mathrm{values}\:\mathrm{on}\:\mathrm{the}\:\mathrm{axis}\:\mathrm{which} \\ $$$$\mathrm{you}\:\mathrm{are}\:\mathrm{rotating}\:\mathrm{around}.\:{Similarly}: \\ $$$${V}=\pi\int_{{x}_{\mathrm{1}} } ^{\:{x}_{\mathrm{2}} } {y}^{\mathrm{2}} {dx} \\ $$$${V}=\pi\int_{{f}\left({x}_{\mathrm{1}} \right)} ^{\:{f}\left({x}_{\mathrm{2}} \right)} {x}^{\mathrm{2}} {dy} \\ $$$$ \\ $$$$ \\ $$$$\mathrm{For}\:\mathrm{example}: \\ $$$${y}={x}^{\mathrm{2}} \\ $$$$\mathrm{Rotating}\:\mathrm{the}\:\mathrm{curve}\:\mathrm{around}\:\mathrm{the}\:\mathrm{y}\:\mathrm{axis}, \\ $$$$\mathrm{you}\:\mathrm{have}\:\mathrm{a}\:\mathrm{shape}\:\mathrm{that}\:\mathrm{resembles}\:\mathrm{a}\:\mathrm{bowl}. \\ $$$$\mathrm{Say}\:\mathrm{the}\:\mathrm{bowl}\:\mathrm{is}\:\mathrm{4}\:\mathrm{units}\:\mathrm{high}. \\ $$$${y}={x}^{\mathrm{2}} \:\:\:\:\:\:\left({rotate}\:{around}\:{y}\:\:\therefore{have}\:{to}\right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{transpose}\:{for}\:{x}\right) \\ $$$${V}=\pi\int_{{a}} ^{\:{b}} {x}^{\mathrm{2}} {dy} \\ $$$${y}={x}^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} ={y} \\ $$$${V}=\pi\int_{\mathrm{0}} ^{\:\mathrm{4}} {x}^{\mathrm{2}} {dy} \\ $$$${V}=\pi\int_{\mathrm{0}} ^{\:\mathrm{4}} {ydy} \\ $$$${V}=\frac{\mathrm{1}}{\mathrm{2}}\pi\mathrm{16} \\ $$$${V}=\mathrm{8}\pi \\ $$
Commented by Filup last updated on 19/Dec/15
Go onto the KhanAcademy website.  They have amazing tutorials and lessons  on calculating the volume of curves  rotated around an axis.    Just search it on google :)
$$\mathrm{Go}\:\mathrm{onto}\:\mathrm{the}\:\mathrm{KhanAcademy}\:\mathrm{website}. \\ $$$$\mathrm{They}\:\mathrm{have}\:\mathrm{amazing}\:\mathrm{tutorials}\:\mathrm{and}\:\mathrm{lessons} \\ $$$$\mathrm{on}\:\mathrm{calculating}\:\mathrm{the}\:\mathrm{volume}\:\mathrm{of}\:\mathrm{curves} \\ $$$$\mathrm{rotated}\:\mathrm{around}\:\mathrm{an}\:\mathrm{axis}. \\ $$$$ \\ $$$$\left.\mathrm{Just}\:\mathrm{search}\:\mathrm{it}\:\mathrm{on}\:\mathrm{google}\::\right) \\ $$
Commented by RasheedSindhi last updated on 19/Dec/15
Th^α nkS!
$$\mathcal{T}{h}^{\alpha} {nk}\mathcal{S}! \\ $$
Commented by Rasheed Soomro last updated on 22/Dec/15
An other approach may be as given  below:  Consider a cyllender and a cone both  erect upon their circular bases.  Now as we go above,the radius remains  constant in cyllender but radius goes  to decrease more and more and ultimately  becomes zero at its vertex in case of cone.  Thus we can say r=c in cyllender and  r→0 in cone.  So the volume of cone is a limit of cyllender  volume as r→0  ???
$${An}\:{other}\:{approach}\:{may}\:{be}\:{as}\:{given} \\ $$$${below}: \\ $$$${Consider}\:{a}\:{cyllender}\:{and}\:{a}\:{cone}\:{both} \\ $$$${erect}\:{upon}\:{their}\:{circular}\:{bases}. \\ $$$${Now}\:{as}\:{we}\:{go}\:{above},{the}\:{radius}\:{remains} \\ $$$${constant}\:{in}\:{cyllender}\:\boldsymbol{{but}}\:{radius}\:{goes} \\ $$$${to}\:{decrease}\:{more}\:{and}\:{more}\:{and}\:{ultimately} \\ $$$${becomes}\:{zero}\:{at}\:{its}\:{vertex}\:{in}\:{case}\:{of}\:{cone}. \\ $$$${Thus}\:{we}\:{can}\:{say}\:\boldsymbol{{r}}=\boldsymbol{{c}}\:{in}\:{cyllender}\:{and} \\ $$$$\boldsymbol{{r}}\rightarrow\mathrm{0}\:{in}\:{cone}. \\ $$$${So}\:{the}\:{volume}\:{of}\:{cone}\:{is}\:{a}\:{limit}\:{of}\:{cyllender} \\ $$$${volume}\:{as}\:{r}\rightarrow\mathrm{0}\:\:??? \\ $$

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