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Question Number 3714 by Rasheed Soomro last updated on 19/Dec/15

$$\mathcal{D}\mathrm{erive}\mathrm{a}\mathrm{formula}\mathrm{of}\mathrm{volume}\mathrm{of}\mathbf{right}\mathbf{circular}\mathbf{cone}$$$$\mathrm{when}\mathrm{the}\mathrm{formula}\mathrm{of}\mathrm{volume}\mathrm{of}\mathbf{cyllinder}\mathrm{is}\mathrm{given}.$$

Answered by Filup last updated on 19/Dec/15

$${\mathrm{I}}^{\prime }\mathrm{m}\mathrm{not}\mathrm{sure}\mathrm{how}\mathrm{to}\mathrm{answer}\mathrm{the}\mathrm{question}$$$$\mathrm{given}.\mathrm{But}\mathrm{I}\mathrm{can}\mathrm{show}\mathrm{it}\mathrm{this}\mathrm{way}:$$$$$$$$\mathrm{Line}y=mx$$$$height=r$$$$baselength=h$$$$\left(itsasidewaysconeagainstthexaxis\right)$$$$\therefore m=\frac{r}{h}$$$$y=\frac{r}{h}x$$$$$$$$V=\pi {\int }_a^b{y}^{2}dx$$$$\therefore V=\pi {\int }_0^h{\left(\frac{r}{h}x\right)}^{2}dx$$$$V=\pi r^2\frac{1}{h^2}{\int }_0^h{x}^{2}dx$$$$V=\pi r^2\frac{1}{h^2}(\frac{1}{3}{h}^3-0)$$$$\therefore V=\frac{1}{3}\pi r^{2}h$$$$$$

Commented by RasheedSindhi last updated on 19/Dec/15

$$Integralisusedforarea.$$$$Howisitusedforvolume?$$$$Whichlineisy=mx?$$

Commented by Filup last updated on 19/Dec/15

$$fory=f\left(x\right),\mathrm{volume}\mathrm{is}\mathrm{found}\mathrm{by}$$$$\mathrm{rotating}\mathrm{around}\mathrm{either}\mathrm{the}\mathrm{x}\mathrm{or}\mathrm{y}\mathrm{axis}.$$$$$$$$\mathrm{Rotate}\mathrm{around}x\mathrm{axis}:$$$$V=\pi {\int }_a^b{y}^{2}dx$$$$$$$$yaxis:$$$$V=\pi {\int }_a^b{x}^{2}dy$$$$$$$$a\mathrm{and}b\mathrm{are}\mathrm{values}\mathrm{on}\mathrm{the}\mathrm{axis}\mathrm{which}$$$$\mathrm{you}\mathrm{are}\mathrm{rotating}\mathrm{around}.Similarly:$$$$V=\pi {\int }_{x_1}^{x_2}{y}^{2}dx$$$$V=\pi {\int }_{f\left(x_1\right)}^{f\left(x_2\right)}{x}^{2}dy$$$$$$$$$$$$\mathrm{For}\mathrm{example}:$$$$y=x^2$$$$\mathrm{Rotating}\mathrm{the}\mathrm{curve}\mathrm{around}\mathrm{the}\mathrm{y}\mathrm{axis},$$$$\mathrm{you}\mathrm{have}\mathrm{a}\mathrm{shape}\mathrm{that}\mathrm{resembles}\mathrm{a}\mathrm{bowl}.$$$$\mathrm{Say}\mathrm{the}\mathrm{bowl}\mathrm{is}4\mathrm{units}\mathrm{high}.$$$$y=x^2(rotatearoundy\therefore haveto$$$$transposeforx)$$$$V=\pi {\int }_a^b{x}^{2}dy$$$$y=x^2$$$$x^2=y$$$$V=\pi {\int }_0^4{x}^{2}dy$$$$V=\pi {\int }_0^{4}ydy$$$$V=\frac{1}{2}\pi 16$$$$V=8\pi$$

Commented by Filup last updated on 19/Dec/15

$$\mathrm{Go}\mathrm{onto}\mathrm{the}\mathrm{KhanAcademy}\mathrm{website}.$$$$\mathrm{They}\mathrm{have}\mathrm{amazing}\mathrm{tutorials}\mathrm{and}\mathrm{lessons}$$$$\mathrm{on}\mathrm{calculating}\mathrm{the}\mathrm{volume}\mathrm{of}\mathrm{curves}$$$$\mathrm{rotated}\mathrm{around}\mathrm{an}\mathrm{axis}.$$$$$$$$\mathrm{Just}\mathrm{search}\mathrm{it}\mathrm{on}\mathrm{google}:)$$

Commented by RasheedSindhi last updated on 19/Dec/15

$$\mathcal{T}{h}^{\alpha }nk\mathcal{S}!$$

Commented by Rasheed Soomro last updated on 22/Dec/15

$$Anotherapproachmaybeasgiven$$$$below:$$$$Consideracyllenderandaconeboth$$$$erectupontheircircularbases.$$$$Nowaswegoabove,theradiusremains$$$$constantincyllender\mathit{b}\mathit{u}\mathit{t}radiusgoes$$$$todecreasemoreandmoreandultimately$$$$becomeszeroatitsvertexincaseofcone.$$$$Thuswecansay\mathit{r}=\mathit{c}incyllenderand$$$$\mathit{r}\to 0incone.$$$$Sothevolumeofconeisalimitofcyllender$$$$volumeasr\to 0???$$

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