My question is in the attachment.Sorry for any inconvenience.

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- Mar 22nd 2008, 11:46 AMprescott2006Double integral:volume
My question is in the attachment.Sorry for any inconvenience.

- Mar 22nd 2008, 03:23 PMmr fantastic
There's no symmetry when calculating this volume.

The graph of $\displaystyle 1 - \cos^3 \theta$ is not symmetric over $\displaystyle 0 \leq \theta \leq \pi$ ......

$\displaystyle \int_{0}^{\pi/2} 1 - \cos^3 \theta \, d \theta = \frac{\pi}{2} - \frac{2}{3} \neq \int_{\pi/2}^{\pi} 1 - \cos^3 \theta \, d \theta = \frac{\pi}{2} + \frac{2}{3}$ .....