Question

How do you graph \[F(x,y) = \sqrt {{x^2} + {y^2} - 1} + \ln (4 - {x^2} - {y^2})\] ?

Answer 1

Hint: In the above given equation \[F(x,y) = \sqrt {{x^2} + {y^2} - 1} + \ln (4 - {x^2} - {y^2})\] we will observe two independent variables \[x\] and \[y\] . Now we will replace these variables \[x\] and \[y\] as \[r\] so \[F(x,y) = f(r)\] m also denoted as \[z\] .In order to plot the graph we need to convert the equation in polar form. Later on we will sketch a two dimensional graph for \[z = \sqrt {{r^2} - 1} + \ln (4 - {r^2})\] and finally we will rotate it about \[z\] axis.

Complete step-by-step answer:
As we know that in the above problem for which we need to sketch the graph consists of the given function which we have is
  \[F(x,y) = \sqrt {{x^2} + {y^2} - 1} + \ln (4 - {x^2} - {y^2})\]
Here we will observe that \[F\] is the given function of the independent variables \[x\] and \[y\] .
Now we will let \[\sum {} \] be the surface of the given function \[F\]
Here it the \[F(x,y) = f(r)\] in which
  \[r = \sqrt {{x^2} + {y^2}} \]
Precisely we will observe that
  \[f(r) = \sqrt {{r^2} - 1} + \ln (4 - {r^2})\]
Or \[z = \sqrt {{r^2} - 1} + \ln (4 - {r^2})\]

Now if we will revolve the graph around the \[z\] axis we will get the graph of the given function as a surface of revolution. Hence the graph of the given function is sketched


Note: By substituting the values and equating it keeping the equation balanced we can draw the graph for the given equation.
Notes: The graph which is obtained in the above equation is the 3-Dimensional graph as it shows involvement of three variables \[x,y{\text{ }}and{\text{ }}z\] . It is difficult to obtain these kinds of 3-Dimensional graphs through simple derivatives and coordinate concepts .To draw this graph knowledge of advanced mathematics concepts like solids of revolution is required.