Question 783392
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Assume center of the base is at the origin and axis of the cone is coincident with the *[tex \Large y]-axis.


Height at any point is given by:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ \left(4r\ -\ x\right)\left(\frac{9h}{4r}\right)\ =\ 9h\ -\ \frac{9h}{4r}x]


Then:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ V\ =\ \int_0^{4r}\,dV\ =\ \int_0^{4r}\,2\pi{xy}\,dx\ =\ 2\pi\int_0^{4r}\,x\left(9h\ -\ \frac{9h}{4r}x\right)\,dx]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2\pi\left\[\frac{9h}{2}x^2\ -\ \frac{9h}{3\cdot{4h}}x^3\right\]^{4r}_0\ =\ 2\pi\left(\left(\frac{9h\cdot{16 r^2}}{2}\ -\ \frac{6h\cdot{16 r^2}}{2}\right)-0\right)\ =\ 48\pi{r^2}h]


Check:  Volume of a cone:  *[tex \LARGE \ \frac{1}{3}\pi{R^2}H]


Substitute *[tex \LARGE 4r] and *[tex \LARGE 9h]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{1}{3}\pi{(4r)^2\cdot{9h}\ =\ \frac{\pi\cdot 16r^2\cdot{9h}}{3}\ =\ 48\pi{r^2}h]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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