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Suppose we have the paraboloid $z=x^2+y^2$ and the plane $z=y$. Their intersection produces a curve $C$, and certain surfaces bounded by it, for example the disc $S$ which directly fills the area of $C$ and the paraboloid $S'$ given by $z=x^2+y^2$ which extends from $C$ downwards and is bounded by $C$.

My question is on how to parameterise these objects. My initial instinct is to substitute one equation into the other giving the circle $(y-\frac{1}{2})^2+x^2=\frac{1}{4}$. Now I could go ahead and parameterise this circle using polars or even do integrals directly using $x$ and $y$. But this gives the projection of the intersection of the plane and paraboloid in the xy-plane. But I have not directly parameterised $C$. Since the $z=y$ plane makes a $45$ degree angle with the xy-plane, I think that we can introduce a factor of $\sqrt2$ into our parameterisation, but I would like to know if there is a way to directly parameterise $C$ (and also the surfaces $S$ and $S'$)?

Note my question is similar to this one except instead of being level, my disc is slanted (and I would also like to find out how to parameterise the 'bowl shaped' surface).

Many thanks for any help.

2 Answers 2

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The surfaces intersect when $y = x^2 + y^2$ so the curve of intersection is

$$ C = \left \{ (x, y, z) \, | \, \left( y - \frac{1}{2} \right)^2 + x^2 = \frac{1}{4}, z = y \right \}. $$

Assume for a minute that you would have the equation $x^2 - \left( y - \frac{1}{2} \right)^2 + = \frac{1}{4}$ in $\mathbb{R}^2$. This is the equation of a circle with radius $\frac{1}{2}$ centered at $\left( 0, \frac{1}{2} \right)$. To parametrize this circle, you can use polar coordinates centered at $\left( 0, \frac{1}{2} \right)$. Namely, let $$x = \frac{1}{2} \cos \theta, \,\, y = \frac{1}{2} + \frac{1}{2} \sin \theta = \frac{1}{2}( 1 + \sin \theta)$$ and then the circle is described by $$\left \{ \frac{1}{2}(\cos \theta, 1 + \sin \theta) \, | \, \theta \in [0,2\pi] \right \}.$$

Returning to your original problem, you can do the same and then use the equation $z = y$ to get the "slanted circle"

$$ C = \left \{ \frac{1}{2} \left( \cos \theta, 1 + \sin \theta, 1 + \sin \theta \right) \, | \, \theta \in [0, 2\pi] \right \}. $$

Note that this curve is actually an ellipse on the $z = y$ plane and not a circle (precisely because you intersect the paraboloid with a slanted plane). Only the projection of this ellipse to the $xy$ plane results in a circle.

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    Thank you for you answer. All of it makes sense (in fact I ended up with similar reasoning on paper). Although I do have a quick question: $ C = \left \{ \frac{1}{2} \left( \cos \theta, 1 + \sin \theta, 1 + \sin \theta \right) \, | \, \theta \in [0, 2\pi] \right \} $ obviously satisfies the $z=y$ condition but how come $z=x^2+y^2$ doesn't hold here? Is it because we're only interested in what is happening in the *plane* $z=y$ and we don't care about the bowl-shaped region below? – 2017-02-05
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    It actually satisfies both conditions (otherwise it wouldn't be the intersection). Namely, $x^2 + y^2 = \frac{1}{4}(\cos^2 \theta + (1 + \sin \theta)^2) = \frac{1}{4} ( 2 + 2 \sin \theta ) = \frac{1}{2}(1 + \sin \theta) = z$. – 2017-02-05
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    Ah forgive me yes that was obvious! And finally, to get the full disc, I guess we would introduce a further 'radius' parameter? I originally thought we'd vary this from $0$ to $1/2$ since this is the radius of the circle in the $xy$ plane but since this is an ellipse, I doubt it's that simple. I guess really we'd have something like $0 \leq r \leq \text{trig function}$? – 2017-02-05
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    Ah, I guess we would use the standard parameterisation for an ellipse, i.e. $x=a \cos \theta$ and $y=b \sin \theta$ where $a, b$ are the radius on the x and y axes respectively. – 2017-02-05
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    You won't need a trig function, you can do a parametrization with $0 \leq r \leq \frac{1}{2}$. Namely, $S = \{ r \left( \cos \theta, 1 + \sin \theta, 1 + \sin \theta \right) \, | \, 0 \leq r \leq \frac{1}{2}, 0 \leq \theta \leq 2\pi \}$. But it still doesn't mean that $S$ is a disc, it will be an ellipse. – 2017-02-05
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    Indeed - I suppose I need to get used to the fact that the varying $z$ coordinate induces the ellipse. Many thanks for all your help! – 2017-02-05
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    @mathphys: Sure. If you draw the situation (by hand, or using a software), you'll see that it makes sense. You have a circle lying on the $xy$ plane and above it, on the $z = y$ there is an ellipse that projects to this circle. – 2017-02-05
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Let,

$$y-\frac{1}{2}=\frac{1}{2}\sin (\theta)$$

$$y=\frac{1}{2}(\sin (\theta)+1)$$

$$x=\frac{1}{2}\cos (\theta)$$

That way $(y-\frac{1}{2})^2+x^2=\frac{1}{4}$.

With $\theta \in [0,2\pi]$. Because we also intersect $z=x^2+y^2$. Then,

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

$$=(\frac{1}{2}(\sin (\theta)+1))^2+(\frac{1}{2}\cos (\theta))^2$$