Sketching a 3-D surface using plane traces. $x = 16-y^2-z^2$

Question

If I have something like this: $x = 16-y^2-z^2$ how would I go about making three different plane traces so that a rough 3-D surface could be constructed/sketched using them?

I have a slight understanding that you need to set each $x,y,z$ to any number $k$ and then see what is left over for each, but I'm having trouble figuring out what the equations represent.

For example, $$Let \;\;\; x=k \;\;\; now \;\;\; k = 16-y^2-z^2$$ What is that exactly? A circle? Maybe it has been too long and I don't recognize these things anymore, but I went and looked through my book and can't find any examples that relate well to the one above.

Thanks for any help!

EDIT: Thinking about this, should I maybe set each variable to zero and then see what is left over?

Answer 1

The graph of $$x = 16-y^2-z^2$$ is a paraboloid with the $x$-axis as its axis of symmetry.

For any $ x\le 16 $ we get a circle of radius $\sqrt {16-x}.$

For example for $x=0$ we get $ y^2 +z^2 =16$ which is a circle of radius $4$ in $yz$ plane.

Answer 2

You're only one step away. We'll take the example you gave of setting $x=k$ for some constant $k$. The last step is to manipulate the resulting equation into a form you are familiar with. With a bit of rearranging, we can find your intuition was correct: $$y^2+z^2=16-k$$ Definitely a circle! This tells us that the $yz$ trace is a circle of radius $\sqrt{16-k}$ about the origin. Notice that when $k>16$, the radius becomes negative, so there is no solution set and the circle disappears. I'll let you work out the other two traces for yourself, but they can be found in a similar manner.