# Cross section views of $f(x,y)=(x+y)^4$

In the ZY-plane, y=0 so the section is $S_1 : \{y=0\} \bigcap \{z=x^4\}$ and in the ZX-plane, x=0 so the section is $S_2 : \{x=0\} \bigcap \{z=y^4\}$. I sketched parabolas for both section views but I'm not sure if that's right after looking at the 3D plot of the function on Wolfram: http://www.wolframalpha.com/input/?i=plot+%28x%2By%29%5E4

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I think you mean XZ for the first set and YZ for the second. And they are not parabolas; plot $y=x^4$ and you'll see that it's a bit steeper than $y=x^2$, which is a parabola. –  Mario Carneiro Dec 12 '12 at 6:21
Thanks, I did take that into account when sketching the views, so they have a steeper slope than x^2 - I was just conveying the rough shape by calling them parabolas. But are the sketches still accurate? –  user51462 Dec 12 '12 at 6:32
The sets you wrote are correct, so I imagine so, if you drew them as you say. Of course, there is also the XY-plane intersection plot, which is $\{x-y=z=0\}$. –  Mario Carneiro Dec 12 '12 at 6:40
For the XY-plane I sketched the lines $y=-x-1$ and $y=-x+1$ and then the level curves, with the space between each getting gradually smaller. So the level curves would look like diagonal lines on an XY-plane? –  user51462 Dec 12 '12 at 7:03
Yes. The intersection with $z=0$ is $x+y=0$ (correction to earlier comment), and the other level curves would be $x+y=c$ for other values of $c=z^{1/4}$. –  Mario Carneiro Dec 12 '12 at 8:44