How to draw the domain of a three variable function?
$begingroup$
I have this function
$$f(x,y,z)= sqrt x+sqrt y+sqrt z+ln(4-x²-y²-z²)$$
I know the domain is
$$Df={(x,y,z);x≥0,y≥0,z≥0,x²+y²+z²<4}$$
I need to draw the domain under the form of a sphere, as represented below:
Now my question is , the equation which constructed the general sphere here is x²+y²+z²=4.
But how do i know which parts of the sphere to highlight using the constraints, or else that all x y and z ≥0?
Meaning, which parts of the sphere may correspond to this restrictions? How do i know what to highlight to show the domain ?
EDIT: THIS IS NOT A DUPLICATE. Please do not mark it as such.
integration multivariable-calculus
edited Dec 15 '18 at 11:14
gimusi
92.9k84494
asked Dec 15 '18 at 10:59
BM97BM97
758
$endgroup$
add a comment |
$begingroup$
I have this function
$$f(x,y,z)= sqrt x+sqrt y+sqrt z+ln(4-x²-y²-z²)$$
I know the domain is
$$Df={(x,y,z);x≥0,y≥0,z≥0,x²+y²+z²<4}$$
I need to draw the domain under the form of a sphere, as represented below:
Now my question is , the equation which constructed the general sphere here is x²+y²+z²=4.
But how do i know which parts of the sphere to highlight using the constraints, or else that all x y and z ≥0?
Meaning, which parts of the sphere may correspond to this restrictions? How do i know what to highlight to show the domain ?
EDIT: THIS IS NOT A DUPLICATE. Please do not mark it as such.
integration multivariable-calculus
edited Dec 15 '18 at 11:14
gimusi
92.9k84494
asked Dec 15 '18 at 10:59
BM97BM97
758
$endgroup$
1
$begingroup$
The domain is the intersection of the open ball with the first octant. Isn’t that what you are showing?
$endgroup$
– MPW
Dec 15 '18 at 11:03
add a comment |
$begingroup$
I have this function
$$f(x,y,z)= sqrt x+sqrt y+sqrt z+ln(4-x²-y²-z²)$$
I know the domain is
$$Df={(x,y,z);x≥0,y≥0,z≥0,x²+y²+z²<4}$$
I need to draw the domain under the form of a sphere, as represented below:
Now my question is , the equation which constructed the general sphere here is x²+y²+z²=4.
But how do i know which parts of the sphere to highlight using the constraints, or else that all x y and z ≥0?
Meaning, which parts of the sphere may correspond to this restrictions? How do i know what to highlight to show the domain ?
EDIT: THIS IS NOT A DUPLICATE. Please do not mark it as such.
integration multivariable-calculus
edited Dec 15 '18 at 11:14
gimusi
92.9k84494
asked Dec 15 '18 at 10:59
BM97BM97
758
$endgroup$
I have this function
$$f(x,y,z)= sqrt x+sqrt y+sqrt z+ln(4-x²-y²-z²)$$
I know the domain is
$$Df={(x,y,z);x≥0,y≥0,z≥0,x²+y²+z²<4}$$
I need to draw the domain under the form of a sphere, as represented below:
Now my question is , the equation which constructed the general sphere here is x²+y²+z²=4.
But how do i know which parts of the sphere to highlight using the constraints, or else that all x y and z ≥0?
Meaning, which parts of the sphere may correspond to this restrictions? How do i know what to highlight to show the domain ?
EDIT: THIS IS NOT A DUPLICATE. Please do not mark it as such.
integration multivariable-calculus
integration multivariable-calculus
edited Dec 15 '18 at 11:14
gimusi
92.9k84494
asked Dec 15 '18 at 10:59
BM97BM97
758
edited Dec 15 '18 at 11:14
gimusi
92.9k84494
asked Dec 15 '18 at 10:59
BM97BM97
758
edited Dec 15 '18 at 11:14
gimusi
92.9k84494
edited Dec 15 '18 at 11:14
gimusi
92.9k84494
edited Dec 15 '18 at 11:14
gimusi
92.9k84494
92.9k84494
asked Dec 15 '18 at 10:59
BM97BM97
758
asked Dec 15 '18 at 10:59
BM97BM97
758
asked Dec 15 '18 at 10:59
BM97BM97
758
758
1
$begingroup$
The domain is the intersection of the open ball with the first octant. Isn’t that what you are showing?
$endgroup$
– MPW
Dec 15 '18 at 11:03
add a comment |
1
$begingroup$
The domain is the intersection of the open ball with the first octant. Isn’t that what you are showing?
$endgroup$
– MPW
Dec 15 '18 at 11:03
1
1
$begingroup$
The domain is the intersection of the open ball with the first octant. Isn’t that what you are showing?
$endgroup$
– MPW
Dec 15 '18 at 11:03
$begingroup$
The domain is the intersection of the open ball with the first octant. Isn’t that what you are showing?
$endgroup$
– MPW
Dec 15 '18 at 11:03
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
The points
$x^2+y^2+z^2 < 4$ are inside the sphere (excluding the points on the surface)
$x,y,zge 0$ are those in the first octant
therefore the domain are points in the first octant and inside the sphere.
answered Dec 15 '18 at 11:06
gimusigimusi
92.9k84494
$endgroup$
add a comment |
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$begingroup$
The points
$x^2+y^2+z^2 < 4$ are inside the sphere (excluding the points on the surface)
$x,y,zge 0$ are those in the first octant
therefore the domain are points in the first octant and inside the sphere.
answered Dec 15 '18 at 11:06
gimusigimusi
92.9k84494
$endgroup$
add a comment |
$begingroup$
The points
$x^2+y^2+z^2 < 4$ are inside the sphere (excluding the points on the surface)
$x,y,zge 0$ are those in the first octant
therefore the domain are points in the first octant and inside the sphere.
answered Dec 15 '18 at 11:06
gimusigimusi
92.9k84494
$endgroup$
add a comment |
$begingroup$
The points
$x^2+y^2+z^2 < 4$ are inside the sphere (excluding the points on the surface)
$x,y,zge 0$ are those in the first octant
therefore the domain are points in the first octant and inside the sphere.
answered Dec 15 '18 at 11:06
gimusigimusi
92.9k84494
$endgroup$
The points
$x^2+y^2+z^2 < 4$ are inside the sphere (excluding the points on the surface)
$x,y,zge 0$ are those in the first octant
therefore the domain are points in the first octant and inside the sphere.
answered Dec 15 '18 at 11:06
gimusigimusi
92.9k84494
answered Dec 15 '18 at 11:06
gimusigimusi
92.9k84494
answered Dec 15 '18 at 11:06
gimusigimusi
92.9k84494
answered Dec 15 '18 at 11:06
gimusigimusi
92.9k84494
92.9k84494
add a comment |
add a comment |
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$begingroup$
The domain is the intersection of the open ball with the first octant. Isn’t that what you are showing?
$endgroup$
– MPW
Dec 15 '18 at 11:03