Suppose you have a sphere of radius R centered in the origin: $$x^2 + y^2 + z^2 = R^2$$ and through this sphere, completely centered in the origin passes a rectangular parallepiped with dimensions: $$2l*2l*3R$$ (I don't know how to provide a visual representation for that, but imagine two lateral sides being squares with sides $2l$ and the other two lateral sides being rectangles. The length of those rectangles is bigger that $2R$ so it completely passes through the sphere.)

We also suppose that $$R<\sqrt{2}l$$ meaning the square does not englobe the whole sphere. I want to calculate the length of the trace left by the rectangular parallepiped on the surface of the sphere.

I tried: $$L_{c} = \int_{C} \mathbf{ds} $$ In polar coordinates and considering the symmetry of the problem, for $x,y,z>0$ I found: $$\mathbf{s}(\theta) = lcos(\theta)\mathbf{e}_{r} + \sqrt{R^{2} - l^{2}(cos(\theta))^{2}}\mathbf{e}_{z}, 0\le\theta\le\pi/4$$ which corresponds to the trace left by just one side of the part of the square that's in the x, y, z > 0 region. $$\mathbf{s}'(\theta) = -lsin(\theta)\mathbf{e}_{r} + lcos(\theta)\mathbf{e}_{\theta} + \frac{l^{2}sin(2\theta)} {2\sqrt{R^{2} - l^{2}cos^{2}(\theta)}}\mathbf{e}_{z}$$ $$\left|\mathbf{s}(\theta)\right| = l\sqrt{\frac{1 + l^2sin^{2}(2\theta)}{4(R^{2} - l^{2}cos^{2}(\theta))}}$$ And so finally: $$\frac{1}{8}L_{c} = 2\int_{0}^{\pi/4}\left|\mathbf{s}(\theta)\right|d\theta$$

However the resulting integral seems too difficult to compute and hving an idea of the result I should get, I think my description of the curve ( the trace left on the sphere ) was wrong. Could someone point out where I went wrong or provide an alternative parametric description of the trace?

  • $\begingroup$ Do I understand this to be the intersection of 4 planes and a sphere? $\endgroup$ – Narlin May 13 '18 at 18:28
  • $\begingroup$ yes, this is exactly what it is, and I have considered half of the plane parallel to x axis ( positive axis ) which has length $l$ for my work. And I want to calculate the length if this intersection. @Narlin $\endgroup$ – Desperados May 13 '18 at 18:44
  • $\begingroup$ What do you mean by the "length of the trace left by the ... parallelepiped on the ... sphere"? There are arcs, pieces of surface, what have you. $\endgroup$ – Christian Blatter May 13 '18 at 19:49
  • $\begingroup$ The length of the arc of the intersection of the parallelepiped with the sphere...see $[IM]$ or $[JK]$ in the answer below by Narlin. @ChristianBlatter $\endgroup$ – Desperados May 13 '18 at 20:08

Hint. Work out the curve segments individually. sphere punctured by square rod

| cite | improve this answer | |
  • $\begingroup$ There is an error in the figure. It should read "The line between G and H is (x,y,z)=H+t*(G-H)" without the absolute value bars. $\endgroup$ – Narlin May 14 '18 at 1:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.