Find the equation of a surface obtained by rotating of a circle $^2 + ^2 − 4 + 3 = 0$ around $$-axis.

Question

1. Find the equation of a surface obtained by rotating of a circle $^2 + ^2 − 4 + 3 = 0$ around $$-axis.

2. Find the intersection of surfaces $^2 + ^2 = 2$ and $^2 + ^2 + ^2 = 8$. What kind of curve is it?

What I supposed to do? In 1) I need to find some transformation, which I don't know. In second I just tried to substitute equation variables one into another? but still not sure.

Answer 1

1. $x²+y²-4x+3=0$ => $x²-4x+4+y²=0=(x-2)²+y²=1$; it is equation of circle which centre in O(2;0; 0) and radius R=1. After rotation around y-axis we will have sphere (it's obvious, we will have a GLP that equidistant from O, what is the sphere by the definition); radius of sphere R=1; centre of sphere O. So we can write the equation of this sphere: $(x-2)²+y²+z²=1²$.
2. We can find set of points of intersection just by solving the system of equations (SOE): $x²+y²=2z$; $x²+y²+z²=8=2z+z²=8$ => $z²+2z-8=(z+4)(z-2)=0$; so $z=-4$ or $z=2$. What about $z=-4$ we can see that $x²+y²=2z=-8$ impossible, so it doesn't satisfy SOE; If $z=2$ then $x²+y²=2z=4$ => $x²+y²=4$; it's the only set of points of intersection and it's the equation of circle with centre O(0; 0; 4), radius R=2 and located in the plane $z=4$.

Answer 2

$$ \begin{cases} x^2+y^2=2 z\\ x^2+y^2+z^2=8\\ \end{cases} $$ is the equation of the circle with center $(0,0,2)$ and radius $2$ in the plane $z=2$, intersection of the paraboloid $2z=x^2+y^2$ and the sphere $x^2+y^2+z^2=8$.

Indeed subtracting the two equation we get $$z^2-2z-8=0\to z_1=2;\;z_2=-4$$ $z_2=-4$ must be discarded because $z\ge0$. So we have

$$ \begin{cases} x^2+y^2=4\\ z=2\\ \end{cases} $$ The curve is shown in the picture below

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Answer 3

1. The point $(a,b)$ traces a circle of radius $a$ in the plane $y=b$ after rotating around the $y$-axis, i.e. $$x^2+z^2=a^2,\>\>\>\>\>y=b$$ Then, substitute $a$ and $b$ into $(x-2)^2+y^2=1$ to obtain the 3D surface $$(\sqrt{x^2+z^2}-2)^2+y^2 =1$$ which is a ring torus.

2. Substitute $z=\frac12(x^2+y^2) $ into $x^2+y^2+z^2=8$ to get

$$(x^2+y^2)^2 +4(x^2+y^2)-32=0$$ which leads to $x^2 +y^2 = 4 $ and $z=2$, i.e. a circle of radius $2$ in the plane $z=2$.