Visualizing 3d graph of two variables

Beamsbox
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I have a question in my book that states:
"T or F The natural domain of f(x,y,z) = sqrt(1-x^2-y^2) is a disk of radius 1 centered at the origin in the xy-plane."

This is F as the graph is an infinite solid cylinder. But I can't visualize it. If I let f(x,y,z) be z, and square both sides, I get:

x^2 + y^2 + z^2 = 1 which is the graph of a sphere centered at the origin, right?

Not sure where the 'solid cylinder' comes in...
 
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I think it is true, since the shape is a sphere radius one centered at the origin (the infinite cylinder equation would be x^{2}+y^{2}=1).

Think of projecting the sphere onto the xy-plane. What would it's shadow look like?
 
Beamsbox said:
I have a question in my book that states:
"T or F The natural domain of f(x,y,z) = sqrt(1-x^2-y^2) is a disk of radius 1 centered at the origin in the xy-plane."

This is F as the graph is an infinite solid cylinder. But I can't visualize it. If I let f(x,y,z) be z, and square both sides, I get:

x^2 + y^2 + z^2 = 1 which is the graph of a sphere centered at the origin, right?

Not sure where the 'solid cylinder' comes in...

You have f(x,y,z)=\sqrt{1-x^2-y^2}

That's a function of three variables right? But the value of the function is only in terms of x and y. That means for any z, the value is the same for the same x and y. So the domain for real values of the function in terms of x, y, and z is not just the unit circle in the x-y plane but the unit circle for every z. Think of the function f(x,y,z) as a density function in 3D space. For example,

f(1,0,0)=0
f(0,0,0)=1

Now suppose you colored the values of the function for every value of x, y, and z in it's domain, the smaller the value of f, the lighter the color. Say the colors go from white for small values to black when f(x,y,z)=1. Now consider that "natural" domain in the x, y, z plane, a cylinder right with all those values of f(x,y,z) colored according to their values. Wouldn't that figure look like a density plot with white on the perimeter where f(x,y,z)=0 and as you get closer to the z-axis, the colors get darker and darker until at the z-axis, they are black.

Ok, now plot that in Mathematica.
 
Ah perfect. I understand the circle at every value of z. I don't understand the 'density function' part though. PErhaps that one will come with time.

Thank you, jackmell.
 
Assignment problem. I have a new issue regarding visualizing graphs of multiple variables.

How do you visualize the two of these:

f(x,y)=yln(1+x)
f(x,y)=sin-1(xy)


These completely baffle me. I can see z=ln(1+x), but when you multiply it by y, how does that change it? Does it extend infinitely in the +/- y direction? As for the second one... I can graph z=sin-1(x), but I'm not sure how the newly introduced y changes things...

The answer is in the back of the book for both questions. Both are sketched in the xy-plane only. The first has an open boundary at y=-1, and is colored for all values in the positive direction from this line. The second graph is much like a star, with parabolas in each quadrant, dipping toward the origin (opening outwards), and the spikes of the star along each of the two axes (four directions of x-, x+, y-, y+)...

I'd really like to get better at visualizing these myself. But not sure where to start.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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