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Guys I am really having trouble constraining the region between these three surfaces. I am imagining a sort of "Dome", or a "muffin head" sort of shape. Is this correct ? Anyway, I need to be able to write the following volume integral in rectangular, cylindrical and spherical coordinates:

Consider the region that is between $x^2 + y^2 + z^2 = 1$, $x^2 + y^2 + z^2 = 9$, and finally above the upper nappe of the cone $z^2 = 3(x^2 + y ^2)$

upon further consideration, does the smaller sphere even matter ? wouldn't it just represent a hole in the larger sphere, an area im not even worried about finding the volume of ?

Anyway, thanks for looking!

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  • $\begingroup$ Yes, the smaller sphere is just a hollow point within the larger sphere, so it will not be taken into consideration when calculating the volume. Only the volume between them will. $\endgroup$ – rurouniwallace May 10 '13 at 4:21
  • $\begingroup$ that makes sense... so perhaps i can just find the volume between the larger sphere and the cone... which is like a muffin top? $\endgroup$ – Neo May 10 '13 at 4:24
  • $\begingroup$ Yes, it will look like a muffin top, but you are not finding the volume between the larger sphere and the cone. You need to find the volume between the larger sphere and the smaller sphere above the cone. $\endgroup$ – rurouniwallace May 10 '13 at 4:28
  • $\begingroup$ Right so i need to take the integral over the top of the cone, the integral of the larger sphere's radius - the top of the cone, and a Pi/2 integral from the top of the cone to the z axis right ? $\endgroup$ – Neo May 10 '13 at 4:32
  • $\begingroup$ Not quite...it's pretty much impossible to explain in words xD Check my answer, and let me know if anything is unclear. $\endgroup$ – rurouniwallace May 10 '13 at 4:49
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First find where the cone intersects the inner sphere:

$$x^2+y^2=1-z^2$$

$$z^2=3-3z^2$$

$$z^2=\frac{3}{4}$$

This means that the radius of the boundary between the cone and inner sphere is $\frac{1}{2}$, and the radius of the boundary between the cone and the radius of the boundary for the outer sphere can be found to be $\frac{\sqrt{33}}{2}$, using the same process. Converting the sphere equations into cylindrical coordinates and using as bounds $0 \le \theta < 2\pi$ and $\frac{1}{2}\le r < \frac{\sqrt{33}}{2}$, the volume integral is as follows:

$$V=\int_0^{2\pi}\int_{\frac{1}{2}}^{\frac{\sqrt{33}}{2}}(\sqrt{9-r^2}-\sqrt{1-r^2})rdr d\theta$$

A similar procedure can be followed to find the integral in rectangular and spherical coordinates.

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  • $\begingroup$ thanks for this... i was pretty close in my mind. i should be able to do the rest! +1 for you =D $\endgroup$ – Neo May 10 '13 at 4:49
  • $\begingroup$ @Neo Glad to help! $\endgroup$ – rurouniwallace May 10 '13 at 4:50
  • $\begingroup$ zetta, are u sure this is volume and not area ? $\endgroup$ – Neo May 10 '13 at 5:05
  • $\begingroup$ @Neo crap, you're right, I forgot to add the r term. I've corrected it, see my latest edit. $\endgroup$ – rurouniwallace May 10 '13 at 5:07
  • $\begingroup$ what about the z term ? thats what im confused about! $\endgroup$ – Neo May 10 '13 at 5:09

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