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Derive-a-formula-of-volume-of-right-circular-cone-when-the-formula-of-volume-of-cyllinder-is-given-

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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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