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I'm trying to solve an integral from my homework, however, to do that, I need to find the circle cut of the next sphere and plane:


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

$x + y + z = 1 $

I found the cutting points with the x,y,z axis, but didn't manage to find the radius or the center of the circle cut.

I would appreciate any help, Thanks.

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Step 1: From the plane, $\;z=1-(x+y)\;$ ;

Step 2: Substitute the above in sphere :

$$x^2+y^2+\left(1-(x+y)\right)^2=1\implies x^2+y^2-x-y+xy=0\;\;\color{red}{(*)}$$

Step 3: Knowing that we always get a circle as intersection if it is not empty or only one point, we try to parametrize Step 2 as a circle, for example by means of completing the square:

$$\color{red}{(*)}\;\;\left(x+\frac{y-1}2\right)^2+\frac34y^2-\frac y2-\frac14=0\implies\left(x+\frac{y-1}2\right)^2+\frac34\left(y-\frac13\right)^2=\frac13\;\;\color{blue}{(**)}$$$${}$$

Step 4: "Force" a circle parametrization here, by means of:

$$x':=\left(x+\frac{y-1}2\right)\;,\;\;y':=\frac{\sqrt3}2\left(y-\frac13\right)\implies\color{blue}{(**)}\;\;x'^2+y'^2=\left(\frac1{\sqrt3}\right)^2$$

or with polar parametrization:

$$\begin{cases} x+\cfrac{y-1}2=\cfrac1{\sqrt3}\cos\theta\implies x+\cfrac12y=\cfrac1{\sqrt3}\cos\theta+\cfrac12\\{}\\ \cfrac{\sqrt3}2\left(y-\cfrac13\right)=\cfrac1{\sqrt3}\sin\theta\implies\cfrac12y=\cfrac13\sin\theta+\cfrac16\end{cases}$$

and substracting we get

$$\begin{cases}x=\cfrac13\left(\sqrt3\,\cos\theta-\sin\theta\right)+\cfrac13\\{}\\ y=\cfrac23\sin\theta+\cfrac13\end{cases}$$

and finally the last coordinate is

$$z=1-x-y=1-\cfrac13\left(\sqrt3\,\cos\theta-\sin\theta\right)-\cfrac13-\cfrac23\sin\theta-\cfrac13=$$

$$=\frac13-\frac13\left(\sqrt3\,\cos\theta+\sin\theta\right)$$

and our circle is

$$r(\theta)=\left(\,\cfrac13\left(\sqrt3\,\cos\theta-\sin\theta\right)+\cfrac13\,,\,\,\cfrac23\sin\theta+\cfrac13\,,\,\,\frac13-\frac13\left(\sqrt3\,\cos\theta+\sin\theta\right)\,\right)$$

Please do check the above is a circle with center $\;\left(\cfrac13,\,\cfrac13,\,\cfrac13\right)\;$ (this should have been expected. Why?), and radius $\;\sqrt\frac23\;$

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  • $\begingroup$ Hey, I got the same radius and center, but I used a different way. However, I now try to calculate first linear integral of (x^2) on this intersection. Do you think I should use this parametrization? It seems like there's even easier way. Thanks $\endgroup$ – kfiros Jun 6 '17 at 20:56
  • $\begingroup$ @kfiros Well, it doesn't look that hard, either... $\endgroup$ – DonAntonio Jun 6 '17 at 21:01
  • $\begingroup$ There is a hint: Look on the integral of (x^2 +y^2 +z^2).. (same curve). I guess maybe it's 1/3 of (x^2) integral, but why would it help me? $\endgroup$ – kfiros Jun 6 '17 at 21:02
  • $\begingroup$ I meant 3 times the integral of (x^2) $\endgroup$ – kfiros Jun 6 '17 at 21:12

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