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Could anyone help me find the volume of the solid paraboloid of revolution $$y=10-x^2-z^2, y\geq0$$

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  • $\begingroup$ If you are ok, you can set as solved. Thanks! $\endgroup$ – user Dec 13 '17 at 7:53
  • $\begingroup$ If you are ok, you can set as solved. Thanks! $\endgroup$ – user Dec 15 '17 at 21:36
  • $\begingroup$ I do not how to set as solved, could you help me ? $\endgroup$ – Roon Ann Ho Jan 5 '18 at 10:26
  • $\begingroup$ Take a look here math.meta.stackexchange.com/questions/3286/… $\endgroup$ – user Jan 5 '18 at 10:30
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It would help to visualize this differently. This is a body of revolution about the $y$-axis. Therefore we can think of this as the function $y(x)=10-x^2$ to be rotated about the $y$-axis. By Pappus's $2^{nd}$ Centroid theorem the volume is given the by the area multiplied by the path length of it centroid upon rotation, or

$$V=2\pi RA$$

For rotation about the $y$-axis$ the centroid is given by

$$R=\frac{1}{A}\int x\cdot y(x)~dx$$

thus

$$ \begin{align} V &=2\pi \int_0^{\sqrt{10}} x(10-x^2)~dx\\ &=2\pi ~\left(5x^2-\frac{x^4}{4}\right)\biggr|_0^{\sqrt{10}}\\ &=50\pi \end{align} $$

This result has been verified numerically.

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HINT

Let's proceed by cylindrical coordinates:

$$x = r \cos{\theta}$$ $$z = r \sin{\theta}$$ $$y = y$$

That is

$$\int_{0}^{2\pi}\int_0^{10}\int_0^{\sqrt{10-y}}r \, dr \, dy\, d\theta $$

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  • $\begingroup$ Could you give me more details about the calculation? $\endgroup$ – Roon Ann Ho Dec 12 '17 at 18:20

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