I've bee given 2 equations in spherical form. One of a sphere $(s_1)$ and the other of a cylinder $(s_2)$.

$$S_1 : \rho = 4\cos \phi \text{ and } S_2 : \rho \sin \phi = 1$$

I need to find the volume inside the sphere and inside the cylinder using both spherical and cylindrical coordinates.

So far I was able to convert surfaces to Cartesian form

$$S_1 : x^2 + y^2 + (z-2)^2 = 4 \text{ and } S_2: x^2 + y^2 = 1$$

But now when I try calculating the volume, I get values that dont make sense.

Im expecting a volume of a bit less than $4\pi $ $units^3$ $(4 \cdot \pi \cdot 1^2)$, but I cant seem to come near that. Im getting values around $5$ and $6$ (depending on if I calculated it in cylindrical or cartesian)

Please help. Thanks

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  • $\begingroup$ @AhmedS.Attaalla I added a drawing to show you my reasoning $\endgroup$ – Liam F-A Jul 20 '17 at 23:53
  • $\begingroup$ @AhmedS.Attaalla Using cartesian I get 12.015 but in cylindrical I get 11.901 $\endgroup$ – Liam F-A Jul 20 '17 at 23:55
  • $\begingroup$ The actual volume is, I believe, $\left(\frac{32}{3}-4\sqrt{3}\right)\pi \doteq 3.738\pi \doteq 11.745$. $\endgroup$ – Brian Tung Jul 21 '17 at 0:00
  • $\begingroup$ @BrianTung. Thanks. Would you be able to give me a hint to how you got that ? $\endgroup$ – Liam F-A Jul 21 '17 at 0:02
  • 1
    $\begingroup$ I actually subtracted the donut-shaped exclusion from the sphere as a whole. The sphere's volume is, of course, $\frac{32}{3}\pi$. As for the volume of the donut, there's a trick: The volume of that donut is a function only of the length of the interior "tunnel"; it is the volume of a sphere of equal diameter. The length of the tunnel is $2\sqrt{3}$, so the donut has the same volume as a sphere of radius $\sqrt{3}$, which is $4\sqrt{3}$. The trick can be proven rigorously, but this comment is too small to contain the proof. :-P $\endgroup$ – Brian Tung Jul 21 '17 at 0:04

The equation of the sphere is $r^2+(z-2)^2=4$, so $z=2 \pm \sqrt{4-r^2}$. The solid of interest is bounded below and above by the sphere and by the cylinder $r=1$ on the side.


$$V=\int_{0}^{2\pi} \int_{0}^{1} \int_{2-\sqrt{4-r^2}}^{2+\sqrt{4-r^2}} r dz dr d\theta$$

$$=\int_{0}^{2\pi} \int_{0}^{1} 2r\sqrt{4-r^2} dr d \theta$$

$$=-2\pi \int_{4}^{3} u^{\frac{1}{2}} du$$

$$=\frac{4}{3}\pi (8-3\sqrt{3})$$

  • $\begingroup$ How is this possible? The cylinder of radius $1$ and height $4$ has volume $4\pi$, and yet the volume of interest is a proper subset of that cylinder, isn't it? $\endgroup$ – Brian Tung Jul 20 '17 at 23:59
  • $\begingroup$ I added instead of subtracted, fixed now @Brian Tung $\endgroup$ – Ahmed S. Attaalla Jul 21 '17 at 0:32
  • $\begingroup$ @AhmedS.Attaalla Ok thats very good thank you! But why dont I get the same answer when I calculate it in cartesian ? $\endgroup$ – Liam F-A Jul 21 '17 at 0:36
  • $\begingroup$ @AhmedS.Attaalla Actually, my mistake. Thats perfect ! Thank you $\endgroup$ – Liam F-A Jul 21 '17 at 0:45
  • $\begingroup$ You're welcome. ${}{}{}{}$ $\endgroup$ – Ahmed S. Attaalla Jul 21 '17 at 0:51

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