Volumes Question (1 Viewer)

kaz1 · Oct 5, 2009

jet · Oct 5, 2009

Pretty sure that's right.
$\text{First we find the volume of the bottom half of the cone. Each circular disk has radius } r\left (h \right ) = \frac{75\sqrt{2}}{\sqrt{h^2 + 50}} \text{ and height ( or thickness) } \delta h \\ \text{Therefore the disk has volume } \delta V = \pi \left (r\left (h \right ) \right )^2 \delta h \\ = \left (\frac{11250\pi}{h^2 + 50} \right ) \delta h \\ \text{To find the volume of the bottom half, we sum the slices between }h = 0 \text{ and } h = 20 \\ \therefore V_{\text{bottom}} = \lim_{\delta h \to 0} \sum^{20}_{0} \left (\frac{11250\pi}{h^2 + 50} \right ) \delta h \\ = 11250\pi \int^{20}_0 \frac{1}{h^2 + 50} dh \\ = 11250\pi \left [\frac{1}{\sqrt{50}}arctan \left (\frac{h}{\sqrt{50}} \right ) \right ]^{20}_{0} \\ = \frac{11250\pi}{\sqrt{50}}arctan \left (\frac{20}{\sqrt{50}} \right ) \therefore \text{ the entire volume is twice this assuming symmetry} \\ \text{Thus, } V_{\text{whole}} = \frac{22500\pi}{\sqrt{50}}arctan \left (\frac{20}{\sqrt{50}}\right ) \\ = 12305$

Volumes Question (1 Viewer)

kaz1

et tu

jet

Banned

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