Graph natural log of x^2+y^2

Problem

z=ln(x^2+y^2)

Solution

1. Identify the function as a surface in three dimensions, where z=ƒ(x,y)=ln(x^2+y^2)

2. Determine the domain by noting that the natural logarithm is only defined for positive arguments, so x^2+y^2>0 which means the function is defined everywhere except at the origin (0,0)

3. Analyze symmetry by observing that the expression x^2+y^2 represents the square of the distance from the origin in the x*yplane, indicating that the surface has rotational symmetry around the zaxis.

4. Convert to cylindrical coordinates using r^2=x^2+y^2 to simplify the expression to z=ln(r^2) or z=2*ln(r)

5. Examine vertical cross-sections (traces) by looking at the graph of z=2*ln(r) for r>0 which shows that as r→0 z→−∞ creating a deep "well" or spike pointing downward at the origin.

6. Examine horizontal cross-sections (level curves) by setting z=k which gives ln(x^2+y^2)=k or x^2+y^2=e^k These are circles centered at the origin with radius √(,e^k)

7. Describe the surface as a rotationally symmetric, funnel-like shape that opens upward as x and y increase and plunges toward negative infinity as they approach the origin.

Final Answer

z=ln(x^2+y^2)

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