Find a function using cartesian coordinates of a given sphere The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Find the volume under the surfaceConverting an integral from Cartesian to Polar coordinates.Using triple integral to find the volume of a sphere with cylindrical coordinatesTriple integral over a sphere with parameter $2n$?Find volume between two spheres using cylindrical & spherical coordinatesVerify Divergence Theorem (using Spherical Coordinates)Set up integral in spherical coordinates outside cylinder but inside sphereCartesian to Spherical Coordinate Conversion for Triple IntegralVolume of Region using Spherical Polar CoordinatesOrbiting spherical coordinates around sphereFind a point of tangency in a plane from a point to a sphere using spherical coordinates
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Find a function using cartesian coordinates of a given sphere
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Find the volume under the surfaceConverting an integral from Cartesian to Polar coordinates.Using triple integral to find the volume of a sphere with cylindrical coordinatesTriple integral over a sphere with parameter $2n$?Find volume between two spheres using cylindrical & spherical coordinatesVerify Divergence Theorem (using Spherical Coordinates)Set up integral in spherical coordinates outside cylinder but inside sphereCartesian to Spherical Coordinate Conversion for Triple IntegralVolume of Region using Spherical Polar CoordinatesOrbiting spherical coordinates around sphereFind a point of tangency in a plane from a point to a sphere using spherical coordinates
$begingroup$
I'm having major difficulty with my maths problem, and any help with understanding and moving forward with the problem would be most appreciated.
My Problem:
Consider the sphere $$S_R=(x,y,z)inmathbbR^3x^2+y^2+z^2=R^2$$
where $R>0$ is the radius of the sphere.
- Using cartesian coordinates, find a function $f(x,y)$ whose graph is the upper hemisphere of $S_R$, and use it to set up a repeated integral for the volume of the sphere.
Solution:
I've had an attempt at finding the function $f(x,y)$, where I have found that:$$f(x,y)=sqrtR^2-z^2$$
I'm unsure whether this is correct as I didn't know whether, as I'm trying to find the function of $(x,y)$ do I just rearrange the equation of the sphere for $x$ and $y$, or am I meant to be using $z$ as my function, so I would have:
$$z(x,y)=sqrtR^2-x^2-y^2$$.
That's all I've managed to do so far, as I'm unsure whether the function $f(x,y)$ is correct and I don't know how to set up the function as a repeated integral.
So any help on this problem would be amazing.
Thank you in advance.
calculus functions spherical-coordinates
edited Apr 8 at 13:02
eranreches
3,846925
asked Apr 8 at 12:59
The StatisticianThe Statistician
117112
$endgroup$
add a comment |
$begingroup$
I'm having major difficulty with my maths problem, and any help with understanding and moving forward with the problem would be most appreciated.
My Problem:
Consider the sphere $$S_R=(x,y,z)inmathbbR^3x^2+y^2+z^2=R^2$$
where $R>0$ is the radius of the sphere.
- Using cartesian coordinates, find a function $f(x,y)$ whose graph is the upper hemisphere of $S_R$, and use it to set up a repeated integral for the volume of the sphere.
Solution:
I've had an attempt at finding the function $f(x,y)$, where I have found that:$$f(x,y)=sqrtR^2-z^2$$
I'm unsure whether this is correct as I didn't know whether, as I'm trying to find the function of $(x,y)$ do I just rearrange the equation of the sphere for $x$ and $y$, or am I meant to be using $z$ as my function, so I would have:
$$z(x,y)=sqrtR^2-x^2-y^2$$.
That's all I've managed to do so far, as I'm unsure whether the function $f(x,y)$ is correct and I don't know how to set up the function as a repeated integral.
So any help on this problem would be amazing.
Thank you in advance.
calculus functions spherical-coordinates
edited Apr 8 at 13:02
eranreches
3,846925
asked Apr 8 at 12:59
The StatisticianThe Statistician
117112
$endgroup$
add a comment |
$begingroup$
I'm having major difficulty with my maths problem, and any help with understanding and moving forward with the problem would be most appreciated.
My Problem:
Consider the sphere $$S_R=(x,y,z)inmathbbR^3x^2+y^2+z^2=R^2$$
where $R>0$ is the radius of the sphere.
- Using cartesian coordinates, find a function $f(x,y)$ whose graph is the upper hemisphere of $S_R$, and use it to set up a repeated integral for the volume of the sphere.
Solution:
I've had an attempt at finding the function $f(x,y)$, where I have found that:$$f(x,y)=sqrtR^2-z^2$$
I'm unsure whether this is correct as I didn't know whether, as I'm trying to find the function of $(x,y)$ do I just rearrange the equation of the sphere for $x$ and $y$, or am I meant to be using $z$ as my function, so I would have:
$$z(x,y)=sqrtR^2-x^2-y^2$$.
That's all I've managed to do so far, as I'm unsure whether the function $f(x,y)$ is correct and I don't know how to set up the function as a repeated integral.
So any help on this problem would be amazing.
Thank you in advance.
calculus functions spherical-coordinates
edited Apr 8 at 13:02
eranreches
3,846925
asked Apr 8 at 12:59
The StatisticianThe Statistician
117112
$endgroup$
I'm having major difficulty with my maths problem, and any help with understanding and moving forward with the problem would be most appreciated.
My Problem:
Consider the sphere $$S_R=(x,y,z)inmathbbR^3x^2+y^2+z^2=R^2$$
where $R>0$ is the radius of the sphere.
- Using cartesian coordinates, find a function $f(x,y)$ whose graph is the upper hemisphere of $S_R$, and use it to set up a repeated integral for the volume of the sphere.
Solution:
I've had an attempt at finding the function $f(x,y)$, where I have found that:$$f(x,y)=sqrtR^2-z^2$$
I'm unsure whether this is correct as I didn't know whether, as I'm trying to find the function of $(x,y)$ do I just rearrange the equation of the sphere for $x$ and $y$, or am I meant to be using $z$ as my function, so I would have:
$$z(x,y)=sqrtR^2-x^2-y^2$$.
That's all I've managed to do so far, as I'm unsure whether the function $f(x,y)$ is correct and I don't know how to set up the function as a repeated integral.
So any help on this problem would be amazing.
Thank you in advance.
calculus functions spherical-coordinates
calculus functions spherical-coordinates
edited Apr 8 at 13:02
eranreches
3,846925
asked Apr 8 at 12:59
The StatisticianThe Statistician
117112
edited Apr 8 at 13:02
eranreches
3,846925
asked Apr 8 at 12:59
The StatisticianThe Statistician
117112
edited Apr 8 at 13:02
eranreches
3,846925
edited Apr 8 at 13:02
eranreches
3,846925
edited Apr 8 at 13:02
eranreches
3,846925
3,846925
asked Apr 8 at 12:59
The StatisticianThe Statistician
117112
asked Apr 8 at 12:59
The StatisticianThe Statistician
117112
asked Apr 8 at 12:59
The StatisticianThe Statistician
117112
117112
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If I give you $x$ and $y$, the only remaining value to fully specifies the point in the 3D cartesian coordinate system is $z$. So, the function you are looking for is $z=f(x,y)$. This can be done by rearranging the terms in the equation of the sphere, exactly as you did. The solution for $z$, given $x,y$, is
$$z=f(x,y)=pmsqrtR^2-x^2-y^2$$
There are two solutions, one is the upper hemisphere ($+$) and one is the lower hemisphere ($-$). You chose the correct one. You can think about this function as representing a surface - you give me a point in the $rm XY$ plane $(x,y)$, and in return I give you the height $f(x,y)$ of the surface above this point.
answered Apr 8 at 13:09
eranrecheseranreches
3,846925
$endgroup$
$begingroup$
Ok thank you so much for this. So the second part, setting up the repeated integral, would I go about this in doing a triple integral with respect to R, x and y? Using only the upper hemisphere or both?
$endgroup$
– The Statistician
Apr 8 at 14:39
$begingroup$
You need to calculate the volume under the upper hemisphere. This volume would be half of the volume of the sphere. See this question for example math.stackexchange.com/questions/433326/….
$endgroup$
– eranreches
Apr 8 at 14:48
$begingroup$
Ok, thank you so much for your help.
$endgroup$
– The Statistician
Apr 8 at 15:08
$begingroup$
I was wondering how I would go about doing the same problem but instead of using cartesian coordinates, use cylindrical coordinates to find a function $f(r,theta)$ ?
$endgroup$
– The Statistician
Apr 8 at 15:14
$begingroup$
Same. You now that $r^2=x^2+y^2$ so your function is $f(r,theta)=sqrtR^2-r^2$.
$endgroup$
– eranreches
Apr 8 at 15:18
|
show 1 more comment
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If I give you $x$ and $y$, the only remaining value to fully specifies the point in the 3D cartesian coordinate system is $z$. So, the function you are looking for is $z=f(x,y)$. This can be done by rearranging the terms in the equation of the sphere, exactly as you did. The solution for $z$, given $x,y$, is
$$z=f(x,y)=pmsqrtR^2-x^2-y^2$$
There are two solutions, one is the upper hemisphere ($+$) and one is the lower hemisphere ($-$). You chose the correct one. You can think about this function as representing a surface - you give me a point in the $rm XY$ plane $(x,y)$, and in return I give you the height $f(x,y)$ of the surface above this point.
answered Apr 8 at 13:09
eranrecheseranreches
3,846925
$endgroup$
$begingroup$
Ok thank you so much for this. So the second part, setting up the repeated integral, would I go about this in doing a triple integral with respect to R, x and y? Using only the upper hemisphere or both?
$endgroup$
– The Statistician
Apr 8 at 14:39
$begingroup$
You need to calculate the volume under the upper hemisphere. This volume would be half of the volume of the sphere. See this question for example math.stackexchange.com/questions/433326/….
$endgroup$
– eranreches
Apr 8 at 14:48
$begingroup$
Ok, thank you so much for your help.
$endgroup$
– The Statistician
Apr 8 at 15:08
$begingroup$
I was wondering how I would go about doing the same problem but instead of using cartesian coordinates, use cylindrical coordinates to find a function $f(r,theta)$ ?
$endgroup$
– The Statistician
Apr 8 at 15:14
$begingroup$
Same. You now that $r^2=x^2+y^2$ so your function is $f(r,theta)=sqrtR^2-r^2$.
$endgroup$
– eranreches
Apr 8 at 15:18
|
show 1 more comment
$begingroup$
If I give you $x$ and $y$, the only remaining value to fully specifies the point in the 3D cartesian coordinate system is $z$. So, the function you are looking for is $z=f(x,y)$. This can be done by rearranging the terms in the equation of the sphere, exactly as you did. The solution for $z$, given $x,y$, is
$$z=f(x,y)=pmsqrtR^2-x^2-y^2$$
There are two solutions, one is the upper hemisphere ($+$) and one is the lower hemisphere ($-$). You chose the correct one. You can think about this function as representing a surface - you give me a point in the $rm XY$ plane $(x,y)$, and in return I give you the height $f(x,y)$ of the surface above this point.
answered Apr 8 at 13:09
eranrecheseranreches
3,846925
$endgroup$
$begingroup$
Ok thank you so much for this. So the second part, setting up the repeated integral, would I go about this in doing a triple integral with respect to R, x and y? Using only the upper hemisphere or both?
$endgroup$
– The Statistician
Apr 8 at 14:39
$begingroup$
You need to calculate the volume under the upper hemisphere. This volume would be half of the volume of the sphere. See this question for example math.stackexchange.com/questions/433326/….
$endgroup$
– eranreches
Apr 8 at 14:48
$begingroup$
Ok, thank you so much for your help.
$endgroup$
– The Statistician
Apr 8 at 15:08
$begingroup$
I was wondering how I would go about doing the same problem but instead of using cartesian coordinates, use cylindrical coordinates to find a function $f(r,theta)$ ?
$endgroup$
– The Statistician
Apr 8 at 15:14
$begingroup$
Same. You now that $r^2=x^2+y^2$ so your function is $f(r,theta)=sqrtR^2-r^2$.
$endgroup$
– eranreches
Apr 8 at 15:18
|
show 1 more comment
$begingroup$
If I give you $x$ and $y$, the only remaining value to fully specifies the point in the 3D cartesian coordinate system is $z$. So, the function you are looking for is $z=f(x,y)$. This can be done by rearranging the terms in the equation of the sphere, exactly as you did. The solution for $z$, given $x,y$, is
$$z=f(x,y)=pmsqrtR^2-x^2-y^2$$
There are two solutions, one is the upper hemisphere ($+$) and one is the lower hemisphere ($-$). You chose the correct one. You can think about this function as representing a surface - you give me a point in the $rm XY$ plane $(x,y)$, and in return I give you the height $f(x,y)$ of the surface above this point.
answered Apr 8 at 13:09
eranrecheseranreches
3,846925
$endgroup$
If I give you $x$ and $y$, the only remaining value to fully specifies the point in the 3D cartesian coordinate system is $z$. So, the function you are looking for is $z=f(x,y)$. This can be done by rearranging the terms in the equation of the sphere, exactly as you did. The solution for $z$, given $x,y$, is
$$z=f(x,y)=pmsqrtR^2-x^2-y^2$$
There are two solutions, one is the upper hemisphere ($+$) and one is the lower hemisphere ($-$). You chose the correct one. You can think about this function as representing a surface - you give me a point in the $rm XY$ plane $(x,y)$, and in return I give you the height $f(x,y)$ of the surface above this point.
answered Apr 8 at 13:09
eranrecheseranreches
3,846925
answered Apr 8 at 13:09
eranrecheseranreches
3,846925
answered Apr 8 at 13:09
eranrecheseranreches
3,846925
answered Apr 8 at 13:09
eranrecheseranreches
3,846925
3,846925
$begingroup$
Ok thank you so much for this. So the second part, setting up the repeated integral, would I go about this in doing a triple integral with respect to R, x and y? Using only the upper hemisphere or both?
$endgroup$
– The Statistician
Apr 8 at 14:39
$begingroup$
You need to calculate the volume under the upper hemisphere. This volume would be half of the volume of the sphere. See this question for example math.stackexchange.com/questions/433326/….
$endgroup$
– eranreches
Apr 8 at 14:48
$begingroup$
Ok, thank you so much for your help.
$endgroup$
– The Statistician
Apr 8 at 15:08
$begingroup$
I was wondering how I would go about doing the same problem but instead of using cartesian coordinates, use cylindrical coordinates to find a function $f(r,theta)$ ?
$endgroup$
– The Statistician
Apr 8 at 15:14
$begingroup$
Same. You now that $r^2=x^2+y^2$ so your function is $f(r,theta)=sqrtR^2-r^2$.
$endgroup$
– eranreches
Apr 8 at 15:18
|
show 1 more comment
$begingroup$
Ok thank you so much for this. So the second part, setting up the repeated integral, would I go about this in doing a triple integral with respect to R, x and y? Using only the upper hemisphere or both?
$endgroup$
– The Statistician
Apr 8 at 14:39
$begingroup$
You need to calculate the volume under the upper hemisphere. This volume would be half of the volume of the sphere. See this question for example math.stackexchange.com/questions/433326/….
$endgroup$
– eranreches
Apr 8 at 14:48
$begingroup$
Ok, thank you so much for your help.
$endgroup$
– The Statistician
Apr 8 at 15:08
$begingroup$
I was wondering how I would go about doing the same problem but instead of using cartesian coordinates, use cylindrical coordinates to find a function $f(r,theta)$ ?
$endgroup$
– The Statistician
Apr 8 at 15:14
$begingroup$
Same. You now that $r^2=x^2+y^2$ so your function is $f(r,theta)=sqrtR^2-r^2$.
$endgroup$
– eranreches
Apr 8 at 15:18
$begingroup$
Ok thank you so much for this. So the second part, setting up the repeated integral, would I go about this in doing a triple integral with respect to R, x and y? Using only the upper hemisphere or both?
$endgroup$
– The Statistician
Apr 8 at 14:39
$begingroup$
Ok thank you so much for this. So the second part, setting up the repeated integral, would I go about this in doing a triple integral with respect to R, x and y? Using only the upper hemisphere or both?
$endgroup$
– The Statistician
Apr 8 at 14:39
$begingroup$
You need to calculate the volume under the upper hemisphere. This volume would be half of the volume of the sphere. See this question for example math.stackexchange.com/questions/433326/….
$endgroup$
– eranreches
Apr 8 at 14:48
$begingroup$
You need to calculate the volume under the upper hemisphere. This volume would be half of the volume of the sphere. See this question for example math.stackexchange.com/questions/433326/….
$endgroup$
– eranreches
Apr 8 at 14:48
$begingroup$
Ok, thank you so much for your help.
$endgroup$
– The Statistician
Apr 8 at 15:08
$begingroup$
Ok, thank you so much for your help.
$endgroup$
– The Statistician
Apr 8 at 15:08
$begingroup$
I was wondering how I would go about doing the same problem but instead of using cartesian coordinates, use cylindrical coordinates to find a function $f(r,theta)$ ?
$endgroup$
– The Statistician
Apr 8 at 15:14
$begingroup$
I was wondering how I would go about doing the same problem but instead of using cartesian coordinates, use cylindrical coordinates to find a function $f(r,theta)$ ?
$endgroup$
– The Statistician
Apr 8 at 15:14
$begingroup$
Same. You now that $r^2=x^2+y^2$ so your function is $f(r,theta)=sqrtR^2-r^2$.
$endgroup$
– eranreches
Apr 8 at 15:18
$begingroup$
Same. You now that $r^2=x^2+y^2$ so your function is $f(r,theta)=sqrtR^2-r^2$.
$endgroup$
– eranreches
Apr 8 at 15:18
|
show 1 more comment
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