Determine and sketch the domain of the function f(x,y)

[kan222]

I know what's under the root must be equal or larger than zero and writing that I get:
$$x^2+y^2<=16$$$$x^2+y^2>=4$$
The graph is then an elipse? Is it with between +-1 of x and y and if so, why is that?

 

[Steven G]

If x^2+y^2<=16 & x^2+y^2>=4, then what is the domain. That is what is the intersection of x^2+y^2<=16 & x^2+y^2>=4???

 

[JeffM]

You are not asked to sketch the surface represented by f(x, y). You are asked to sketch the domain in the xy plane.

You are 90% there.

Thank you for giving the complete problem.

 

[Steven G]

View attachment 33883

I know what's under the root must be equal or larger than zero and writing that I get:
$$x^2+y^2<=16$$$$x^2+y^2>=4$$
The graph is then an elipse? Is it with between +-1 of x and y and if so, why is that?

The graph is definitely NOT an ellipse because an ellipse is 2-D and your graph is 3-D. Look up what a 3-D 'ellipse' is called.

 

[JeffM]

Hint

The way I like to deal with inequalities is to start by solving the related equality.

To cope with $x^2 + y^2 \ge 4$, I first look at $x^2 + y^2 = 4$.

 

[kan222]

Hint

The way I like to deal with inequalities is to start by solving the related equality.

To cope with $x^2 + y^2 \ge 4$, I first look at $x^2 + y^2 = 4$.

Well yes, but should I then get $$x^2=4$$ for the domain, x=-2,+2 and y^2=4, y=-2,+2 and that would be the range?

 

[Dr.Peterson]

Well yes, but should I then get $$x^2=4$$ for the domain, x=-2,+2 and y^2=4, y=-2,+2 and that would be the range?

Are you finding the intercepts (setting x or y to zero)? That's not what you need here.

If you graph the equation $x^2+y^2=4$, what will that graph look like? (You almost had that at the start.)

Then, what region would you shade in for the graph of $x^2+y^2\ge4$?

That will be half of what you need to do to show the domain of f.

 

[Deleted member 4993]

Well yes, but should I then get $$x^2=4$$ for the domain, x=-2,+2 and y^2=4, y=-2,+2 and that would be the range?

Graph x² + y² = 4

The domain and range of this graph would be obvious to you. If not, show us the graph and we can talk you throught.

 

[JeffM]

Well yes, but should I then get $$x^2=4$$ for the domain, x=-2,+2 and y^2=4, y=-2,+2 and that would be the range?

Yes, (2, 2) is in the domain, but there are an infinite number of points in the domain.

Every pair of points such that $x^2 + y^2 = 4$ is in the domain. That is an infinite number of real numbers right there. But that does not begin to exhaust the points in the domain.

I suspect you are having trouble here because we are dealing with a MULTIVARIATE function. There are two independent variables, x and y. There is a dependent variable, which is typically called z.

$$z = f(x, y) = \sqrt{16 - x^2 - y^2} + \sqrt{x^2 + y^2 - 4}$$
Questions about range in this problem apply to z. Questions about domain in this problem apply to x-y pairs..

You have in your first post answered analytically the question about the domain, which lies in the xy plane.

$$4 \le x^2 + y^2 \le 16$$
All you need to do further is to sketch that.

 

[kan222]

Would it be like this? for x there is -2,+2 and -4,+4

 

[Steven G]

Yes and no. Yes, the domain would look like that beautiful pink area that is highlighted in your graph.
No to for x there is -2,+2 and -4,+4. Can you please tell us what that means?

 

[JeffM]

That is indeed a picture of the domain (not the function).

Do you understand the question and answer now? The domain specifies the allowed PAIRS of x’s and y’s. You keep making this about just x as though there is only one independent variable. Please read my previous answer and ask questions.

 

[Steven G]

Well yes, but should I then get $$x^2=4$$ for the domain, x=-2,+2 and y^2=4, y=-2,+2 and that would be the range?

The domain is what the x and y values can be. The range is what the z values can be.

 

[kan222]

I

That is indeed a picture of the domain (not the function).

Do you understand the question and answer now? The domain specifies the allowed PAIRS of x’s and y’s. You keep making this about just x as though there is only one independent variable. Please read my previous answer and ask questions

Think I got it, there's two cases; x=+-2 and y=+-4 but also x=+-4 and y=+-2, which is all according to the graph

 

[JeffM]

I

Think I got it, there's two cases; x=+-2 and y=+-4 but also x=+-4 and y=+-2, which is all according to the graph

No, you are way off base. You keep looking for some finite number of pairs. ANY pair of x, y values in the pink area (including the boundaries) is a valid set of inputs for this function. That is what we mean by a domain of a function: it describes every input for which the function is defined. This function has two inputs. Therefore, the domain is specified by pairs of numbers.

The word “graph“ may be confusing you. Technically, it is called a locus of points.

You gave the answer in your very first post

$$\text {The domain of } f(x, \ y) \text { is any } x, \ y \text { pair such that} \\ 4 \le x^2 + y^2 \le 16.$$
All that remained to do was to make a visual representation of the infinite number of pairs that satisfy that condition.

You were never asked to draw the graph of the function itself (at least not as part of what you wrote in your first post).

You have made this question so much harder than it is. I really want to help you here because there is some confusion in your mind about functions or domains or graphs or loci of points. If we do not clear it up, you will continue to flail about on questions about basic concepts.

 

[kan222]

No, you are way off base. You keep looking for some finite number of pairs. ANY pair of x, y values in the pink area (including the boundaries) is a valid set of inputs for this function. That is what we mean by a domain of a function: it describes every input for which the function is defined. This function has two inputs. Therefore, the domain is specified by pairs of numbers.

The word “graph“ may be confusing you. Technically, it is called a locus of points.

You gave the answer in your very first post

$$\text {The domain of } f(x, \ y) \text { is any } x, \ y \text { pair such that} \\ 4 \le x^2 + y^2 \le 16.$$
All that remained to do was to make a visual representation of the infinite number of pairs that satisfy that condition.

You were never asked to draw the graph of the function itself (at least not as part of what you wrote in your first post).

You have made this question so much harder than it is. I really want to help you here because there is some confusion in your mind about functions or domains or graphs or loci of points. If we do not clear it up, you will continue to flail about on questions about basic concepts.

So i only use the finite pairs to help me graph the boundaries of the domain? Any value of x or y in that area will satisfy it's conditions, $$4 \le x^2 + y^2 \le 16$$

 

[JeffM]

So i only use the finite pairs to help me graph the boundaries of the domain? Any value of x or y in that area will satisfy it's conditions, $$4 \le x^2 + y^2 \le 16$$

YES. Now you are right on target.

A function is defined for certain inputs. The domain of a function is simply the entire set of inputs for which the function is defined.

The functions that calculus works with have infinite domains. If a real function has one independent variable, then the domain is an infinite set of real numbers and can be represented as part or all of the real number line. If a real function has two independent variables, then the domain is an infinite set of ordered pairs of real numbers and can be represented as part or all of the real number plane. If a real function has three independent variables, then the domain is an infinite set of ordered triplets of real numbers and can be represented in three-dimensional space.

If the domain is restricted, then we usually describe it by describing its boundaries. In the specific problem you worked on, the domain included every point on the circumference of two circles and every point between those two circles.

You good now?

 

[kan222]

YES. Now you are right on target.

A function is defined for certain inputs. The domain of a function is simply the entire set of inputs for which the function is defined.

The functions that calculus works with have infinite domains. If a real function has one independent variable, then the domain is an infinite set of real numbers and can be represented as part or all of the real number line. If a real function has two independent variables, then the domain is an infinite set of ordered pairs of real numbers and can be represented as part or all of the real number plane. If a real function has three independent variables, then the domain is an infinite set of ordered triplets of real numbers and can be represented in three-dimensional space.

If the domain is restricted, then we usually describe it by describing its boundaries. In the specific problem you worked on, the domain included every point on the circumference of two circles and every point between those two circles.

You good now?

YES, I got it! Did a few more problems like this and I've done them correctly. I understand now. Thank you so much for having patience with me, this means a lot.

 

[Steven G]

So i only use the finite pairs to help me graph the boundaries of the domain?

No, that is not correct. Just look at the graph you came up with. The boundary points all lie on the circles x^2+y^2 = 4 and x^2 + y^2 =16. There are NOT a finite number of points around the circles! Every circle, whose radius is positive, will have an infinite number of points.

 

[Deleted member 4993]

only use the finite pairs

What exactly do you mean by finite pairs?