# Math Help - multivariable functions

1. ## multivariable functions

COncerning finding the domain of functions in two variables, I'm unsure as to how to sketch the graph of an equation with two square roots. i have sketched the graph of the domain of each of the terms simple enough for this function:

f(x,y)= (sqrt(4-x^2-y^2)) - (sqrt(1-x^2))

i end up with a circle for the first term and a parabola for the second. when i sketch the graph of the domain of the entire function, should i get the intersection? or just put one sketch on top of the other? please help. thank you

2. Let me see. To be exact, the domain of the first term is all $(x,y)$ such that

$4-x^{2}-y^{2}\ge 0$, which implies

$4\ge x^{2}+y^{2}.$ So that's all the points $(x,y)$ inside or on the boundary of the circle of radius 2 centered at the origin.

The domain of the second term is all $(x,y)$ such that

$1-x^{2}\ge 0,$ which implies $x^{2}-1\le 0,$ or $(x-1)(x+1)\le 0.$

There are three regions involved here: $(-\infty,-1), (-1,1), (1,\infty).$ You also have the boundary points $\pm 1,$ which are clearly in the domain, since equality is allowed. If you plug in numbers, you can see that it is only the $x$ values in $[-1,1]$ that are allowed. $y$ doesn't matter in this second piece.

So, finally, to answer your question: you must take the intersection of these two regions, because if you go outside either of them, one part of your function will not be defined (if you're assuming you have real functions, which I am assuming).

Does this make sense?

3. Thanks for your reply. And i believe that does make sense. thanks very much. although im quite confused about the boundaries of the circle. would it not be the sqaure root of 2? instead of 2. although the radius would be two. anyway, i think i got the important bit which is to just take the intersection then.
Highly appreciate your help. More questions to come.

4. No, if you've got the equation $x^{2}+y^{2}=4,$ that is the equation of a circle of radius 2. To see this, suppose $x=0.$ What $y$ values satisfy the equation? Well, the equation is now $0+y^{2}=4,$ implying that $y^{2}=4$, or $y=\pm 2.$ So two points on the circle are $(0,2)$ and $0,-2).$ The distance away from the origin for both points is 2. Hence the radius is 2.

You're very welcome for whatever help I could provide. Have a good one!