Math Help - Sketching graph in R3

1. Sketching graph in R3

Let $f(x,y) = x^{1/3}y^{1/3}$, and let C be the curve of intersection of $z = f(x,y)$ with the plane $y = x$.

Sketch the graph of the curve C.

I know how to graph y = x in R3, but I am not sure how to graph z = f(x,y). Would you use the same steps as you would when sketching a graph in R2? (e.g. first, second derivatives, asymptotes etc).

2. Originally Posted by SyNtHeSiS
Let $f(x,y) = x^{1/3}y^{1/3}$, and let C be the curve of intersection of $z = f(x,y)$ with the plane $y = x$. Sketch the graph of the curve C.
The intersection $S$ is:

$S=\{(x,x,\sqrt[3]{x^2}):\;x\in \mathbb{R}\}\subset \mathbb{R}^3$

You can determine the position of $z=\sqrt[3]{x^2}$ drawing the curve, as usual.

Fernando Revilla

3. Thanks. I tried checking the graph on wolfram alpha and I dont get why when x < 0 the graph (real part) is negative. I mean if you let x = -2 in z = x^{2/3} you get a positive value.