Mixing
How to write the limits of triple integral $iiint f(x,y,z) dz dy dx$ over the annulus?
$begingroup$
How to write the limits of triple integral $iiint f(x,y,z) dz dy dx$ over an annulus which lies between the circle of radii $r$ and $R$, $r<R$? I am confused. I don't want to change into polar coordinates.
EDIT: I made a mistake in asking my question. It should be the spherical shell region lying between two spheres of radius $r$ and $R$. Sorry!
real-analysis calculus multiple-integral
edited Jan 9 at 3:50
David G. Stork
10.9k31432
asked Jan 9 at 3:00
ProblemBookProblemBook
32
$endgroup$
add a comment |
$begingroup$
How to write the limits of triple integral $iiint f(x,y,z) dz dy dx$ over an annulus which lies between the circle of radii $r$ and $R$, $r<R$? I am confused. I don't want to change into polar coordinates.
EDIT: I made a mistake in asking my question. It should be the spherical shell region lying between two spheres of radius $r$ and $R$. Sorry!
real-analysis calculus multiple-integral
edited Jan 9 at 3:50
David G. Stork
10.9k31432
asked Jan 9 at 3:00
ProblemBookProblemBook
32
$endgroup$
1
$begingroup$
$LaTeX$ note: You can get a triple integral by writingiiintrather thanintintint(visually you get $iiint$ instead of $intintint$).
$endgroup$
– JavaMan
Jan 9 at 3:07
add a comment |
$begingroup$
How to write the limits of triple integral $iiint f(x,y,z) dz dy dx$ over an annulus which lies between the circle of radii $r$ and $R$, $r<R$? I am confused. I don't want to change into polar coordinates.
EDIT: I made a mistake in asking my question. It should be the spherical shell region lying between two spheres of radius $r$ and $R$. Sorry!
real-analysis calculus multiple-integral
edited Jan 9 at 3:50
David G. Stork
10.9k31432
asked Jan 9 at 3:00
ProblemBookProblemBook
32
$endgroup$
How to write the limits of triple integral $iiint f(x,y,z) dz dy dx$ over an annulus which lies between the circle of radii $r$ and $R$, $r<R$? I am confused. I don't want to change into polar coordinates.
EDIT: I made a mistake in asking my question. It should be the spherical shell region lying between two spheres of radius $r$ and $R$. Sorry!
real-analysis calculus multiple-integral
real-analysis calculus multiple-integral
edited Jan 9 at 3:50
David G. Stork
10.9k31432
asked Jan 9 at 3:00
ProblemBookProblemBook
32
edited Jan 9 at 3:50
David G. Stork
10.9k31432
asked Jan 9 at 3:00
ProblemBookProblemBook
32
edited Jan 9 at 3:50
David G. Stork
10.9k31432
edited Jan 9 at 3:50
David G. Stork
10.9k31432
edited Jan 9 at 3:50
David G. Stork
10.9k31432
10.9k31432
asked Jan 9 at 3:00
ProblemBookProblemBook
32
asked Jan 9 at 3:00
ProblemBookProblemBook
32
asked Jan 9 at 3:00
ProblemBookProblemBook
32
32
1
$begingroup$
$LaTeX$ note: You can get a triple integral by writingiiintrather thanintintint(visually you get $iiint$ instead of $intintint$).
$endgroup$
– JavaMan
Jan 9 at 3:07
add a comment |
1
$begingroup$
$LaTeX$ note: You can get a triple integral by writingiiintrather thanintintint(visually you get $iiint$ instead of $intintint$).
$endgroup$
– JavaMan
Jan 9 at 3:07
1
1
$begingroup$
$LaTeX$ note: You can get a triple integral by writing
iiint rather than intintint (visually you get $iiint$ instead of $intintint$).$endgroup$
– JavaMan
Jan 9 at 3:07
$begingroup$
$LaTeX$ note: You can get a triple integral by writing
iiint rather than intintint (visually you get $iiint$ instead of $intintint$).$endgroup$
– JavaMan
Jan 9 at 3:07
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
In polar coordinates:
$$intlimits_{r=r_i}^R intlimits_{theta = 0}^{2 pi} intlimits_{z=z_i}^{z_f} f(r cos theta, r sin theta,z) r dr d theta dz$$
or in rectilinear coordinates...
$$intlimits_{x=-R}^R dx intlimits_{y = - sqrt{R^2 - x^2}}^{+ sqrt{R^2 - x^2}} dy intlimits_{z=z_i}^{z_f} dz f(x,y,z) - intlimits_{x=-r}^r dx intlimits_{y = - sqrt{r^2 - x^2}}^{+ sqrt{r^2 - x^2}} dy intlimits_{z=z_i}^{z_f} dz f(x,y,z)$$
Revised question:
$$intlimits_{x=-R}^R dx intlimits_{y = -sqrt{R^2 - x^2 - z^2}}^{+sqrt{R^2 - x^2 - z^2}} dy intlimits_{z=-sqrt{R^2 - x^2 - y^2}}^{+sqrt{R^2 - x^2 - y^2}} dz f(x,y,z) - intlimits_{x=-r}^r dx intlimits_{y = -sqrt{r^2 - x^2 - z^2}}^{+sqrt{r^2 - x^2 - z^2}} dy intlimits_{z=-sqrt{r^2 - x^2 - y^2}}^{+sqrt{r^2 -x^2 - y^2}} dz f(x,y,z)$$
answered Jan 9 at 3:05
David G. StorkDavid G. Stork
10.9k31432
$endgroup$
$begingroup$
G. Stroke, I made a mistake in asking my question. It should be the annular region lying between two spheres of radius $r$ and $R$. Sorry!. Can you please rewrite the limits for this spherical annulus.
$endgroup$
– ProblemBook
Jan 9 at 3:28
$begingroup$
I am not able to vote your comment as I am new here. Thank you!
$endgroup$
– ProblemBook
Jan 9 at 4:03
$begingroup$
I think there is some mistake. When I integrate with respect to z , shouldn't limits be independent of z? Similarly with other two integrals
$endgroup$
– ProblemBook
Jan 9 at 16:46
$begingroup$
Teeny typo: fixed. Thanks.
$endgroup$
– David G. Stork
Jan 9 at 18:51
$begingroup$
David G. Stroke, Sir, how can Change it into spherical coordinates. I that would be easy to evaluate. Also, the function $f(x,y,z)$ is inside the integrals? You have written it like this to make things look simpler, right?
$endgroup$
– ProblemBook
Jan 9 at 19:03
|
show 2 more comments
Your Answer
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1 Answer
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$begingroup$
In polar coordinates:
$$intlimits_{r=r_i}^R intlimits_{theta = 0}^{2 pi} intlimits_{z=z_i}^{z_f} f(r cos theta, r sin theta,z) r dr d theta dz$$
or in rectilinear coordinates...
$$intlimits_{x=-R}^R dx intlimits_{y = - sqrt{R^2 - x^2}}^{+ sqrt{R^2 - x^2}} dy intlimits_{z=z_i}^{z_f} dz f(x,y,z) - intlimits_{x=-r}^r dx intlimits_{y = - sqrt{r^2 - x^2}}^{+ sqrt{r^2 - x^2}} dy intlimits_{z=z_i}^{z_f} dz f(x,y,z)$$
Revised question:
$$intlimits_{x=-R}^R dx intlimits_{y = -sqrt{R^2 - x^2 - z^2}}^{+sqrt{R^2 - x^2 - z^2}} dy intlimits_{z=-sqrt{R^2 - x^2 - y^2}}^{+sqrt{R^2 - x^2 - y^2}} dz f(x,y,z) - intlimits_{x=-r}^r dx intlimits_{y = -sqrt{r^2 - x^2 - z^2}}^{+sqrt{r^2 - x^2 - z^2}} dy intlimits_{z=-sqrt{r^2 - x^2 - y^2}}^{+sqrt{r^2 -x^2 - y^2}} dz f(x,y,z)$$
answered Jan 9 at 3:05
David G. StorkDavid G. Stork
10.9k31432
$endgroup$
$begingroup$
G. Stroke, I made a mistake in asking my question. It should be the annular region lying between two spheres of radius $r$ and $R$. Sorry!. Can you please rewrite the limits for this spherical annulus.
$endgroup$
– ProblemBook
Jan 9 at 3:28
$begingroup$
I am not able to vote your comment as I am new here. Thank you!
$endgroup$
– ProblemBook
Jan 9 at 4:03
$begingroup$
I think there is some mistake. When I integrate with respect to z , shouldn't limits be independent of z? Similarly with other two integrals
$endgroup$
– ProblemBook
Jan 9 at 16:46
$begingroup$
Teeny typo: fixed. Thanks.
$endgroup$
– David G. Stork
Jan 9 at 18:51
$begingroup$
David G. Stroke, Sir, how can Change it into spherical coordinates. I that would be easy to evaluate. Also, the function $f(x,y,z)$ is inside the integrals? You have written it like this to make things look simpler, right?
$endgroup$
– ProblemBook
Jan 9 at 19:03
|
show 2 more comments
$begingroup$
In polar coordinates:
$$intlimits_{r=r_i}^R intlimits_{theta = 0}^{2 pi} intlimits_{z=z_i}^{z_f} f(r cos theta, r sin theta,z) r dr d theta dz$$
or in rectilinear coordinates...
$$intlimits_{x=-R}^R dx intlimits_{y = - sqrt{R^2 - x^2}}^{+ sqrt{R^2 - x^2}} dy intlimits_{z=z_i}^{z_f} dz f(x,y,z) - intlimits_{x=-r}^r dx intlimits_{y = - sqrt{r^2 - x^2}}^{+ sqrt{r^2 - x^2}} dy intlimits_{z=z_i}^{z_f} dz f(x,y,z)$$
Revised question:
$$intlimits_{x=-R}^R dx intlimits_{y = -sqrt{R^2 - x^2 - z^2}}^{+sqrt{R^2 - x^2 - z^2}} dy intlimits_{z=-sqrt{R^2 - x^2 - y^2}}^{+sqrt{R^2 - x^2 - y^2}} dz f(x,y,z) - intlimits_{x=-r}^r dx intlimits_{y = -sqrt{r^2 - x^2 - z^2}}^{+sqrt{r^2 - x^2 - z^2}} dy intlimits_{z=-sqrt{r^2 - x^2 - y^2}}^{+sqrt{r^2 -x^2 - y^2}} dz f(x,y,z)$$
answered Jan 9 at 3:05
David G. StorkDavid G. Stork
10.9k31432
$endgroup$
$begingroup$
G. Stroke, I made a mistake in asking my question. It should be the annular region lying between two spheres of radius $r$ and $R$. Sorry!. Can you please rewrite the limits for this spherical annulus.
$endgroup$
– ProblemBook
Jan 9 at 3:28
$begingroup$
I am not able to vote your comment as I am new here. Thank you!
$endgroup$
– ProblemBook
Jan 9 at 4:03
$begingroup$
I think there is some mistake. When I integrate with respect to z , shouldn't limits be independent of z? Similarly with other two integrals
$endgroup$
– ProblemBook
Jan 9 at 16:46
$begingroup$
Teeny typo: fixed. Thanks.
$endgroup$
– David G. Stork
Jan 9 at 18:51
$begingroup$
David G. Stroke, Sir, how can Change it into spherical coordinates. I that would be easy to evaluate. Also, the function $f(x,y,z)$ is inside the integrals? You have written it like this to make things look simpler, right?
$endgroup$
– ProblemBook
Jan 9 at 19:03
|
show 2 more comments
$begingroup$
In polar coordinates:
$$intlimits_{r=r_i}^R intlimits_{theta = 0}^{2 pi} intlimits_{z=z_i}^{z_f} f(r cos theta, r sin theta,z) r dr d theta dz$$
or in rectilinear coordinates...
$$intlimits_{x=-R}^R dx intlimits_{y = - sqrt{R^2 - x^2}}^{+ sqrt{R^2 - x^2}} dy intlimits_{z=z_i}^{z_f} dz f(x,y,z) - intlimits_{x=-r}^r dx intlimits_{y = - sqrt{r^2 - x^2}}^{+ sqrt{r^2 - x^2}} dy intlimits_{z=z_i}^{z_f} dz f(x,y,z)$$
Revised question:
$$intlimits_{x=-R}^R dx intlimits_{y = -sqrt{R^2 - x^2 - z^2}}^{+sqrt{R^2 - x^2 - z^2}} dy intlimits_{z=-sqrt{R^2 - x^2 - y^2}}^{+sqrt{R^2 - x^2 - y^2}} dz f(x,y,z) - intlimits_{x=-r}^r dx intlimits_{y = -sqrt{r^2 - x^2 - z^2}}^{+sqrt{r^2 - x^2 - z^2}} dy intlimits_{z=-sqrt{r^2 - x^2 - y^2}}^{+sqrt{r^2 -x^2 - y^2}} dz f(x,y,z)$$
answered Jan 9 at 3:05
David G. StorkDavid G. Stork
10.9k31432
$endgroup$
In polar coordinates:
$$intlimits_{r=r_i}^R intlimits_{theta = 0}^{2 pi} intlimits_{z=z_i}^{z_f} f(r cos theta, r sin theta,z) r dr d theta dz$$
or in rectilinear coordinates...
$$intlimits_{x=-R}^R dx intlimits_{y = - sqrt{R^2 - x^2}}^{+ sqrt{R^2 - x^2}} dy intlimits_{z=z_i}^{z_f} dz f(x,y,z) - intlimits_{x=-r}^r dx intlimits_{y = - sqrt{r^2 - x^2}}^{+ sqrt{r^2 - x^2}} dy intlimits_{z=z_i}^{z_f} dz f(x,y,z)$$
Revised question:
$$intlimits_{x=-R}^R dx intlimits_{y = -sqrt{R^2 - x^2 - z^2}}^{+sqrt{R^2 - x^2 - z^2}} dy intlimits_{z=-sqrt{R^2 - x^2 - y^2}}^{+sqrt{R^2 - x^2 - y^2}} dz f(x,y,z) - intlimits_{x=-r}^r dx intlimits_{y = -sqrt{r^2 - x^2 - z^2}}^{+sqrt{r^2 - x^2 - z^2}} dy intlimits_{z=-sqrt{r^2 - x^2 - y^2}}^{+sqrt{r^2 -x^2 - y^2}} dz f(x,y,z)$$
answered Jan 9 at 3:05
David G. StorkDavid G. Stork
10.9k31432
edited Jan 9 at 19:08
answered Jan 9 at 3:05
David G. StorkDavid G. Stork
10.9k31432
answered Jan 9 at 3:05
David G. StorkDavid G. Stork
10.9k31432
answered Jan 9 at 3:05
David G. StorkDavid G. Stork
10.9k31432
10.9k31432
$begingroup$
G. Stroke, I made a mistake in asking my question. It should be the annular region lying between two spheres of radius $r$ and $R$. Sorry!. Can you please rewrite the limits for this spherical annulus.
$endgroup$
– ProblemBook
Jan 9 at 3:28
$begingroup$
I am not able to vote your comment as I am new here. Thank you!
$endgroup$
– ProblemBook
Jan 9 at 4:03
$begingroup$
I think there is some mistake. When I integrate with respect to z , shouldn't limits be independent of z? Similarly with other two integrals
$endgroup$
– ProblemBook
Jan 9 at 16:46
$begingroup$
Teeny typo: fixed. Thanks.
$endgroup$
– David G. Stork
Jan 9 at 18:51
$begingroup$
David G. Stroke, Sir, how can Change it into spherical coordinates. I that would be easy to evaluate. Also, the function $f(x,y,z)$ is inside the integrals? You have written it like this to make things look simpler, right?
$endgroup$
– ProblemBook
Jan 9 at 19:03
|
show 2 more comments
$begingroup$
G. Stroke, I made a mistake in asking my question. It should be the annular region lying between two spheres of radius $r$ and $R$. Sorry!. Can you please rewrite the limits for this spherical annulus.
$endgroup$
– ProblemBook
Jan 9 at 3:28
$begingroup$
I am not able to vote your comment as I am new here. Thank you!
$endgroup$
– ProblemBook
Jan 9 at 4:03
$begingroup$
I think there is some mistake. When I integrate with respect to z , shouldn't limits be independent of z? Similarly with other two integrals
$endgroup$
– ProblemBook
Jan 9 at 16:46
$begingroup$
Teeny typo: fixed. Thanks.
$endgroup$
– David G. Stork
Jan 9 at 18:51
$begingroup$
David G. Stroke, Sir, how can Change it into spherical coordinates. I that would be easy to evaluate. Also, the function $f(x,y,z)$ is inside the integrals? You have written it like this to make things look simpler, right?
$endgroup$
– ProblemBook
Jan 9 at 19:03
$begingroup$
G. Stroke, I made a mistake in asking my question. It should be the annular region lying between two spheres of radius $r$ and $R$. Sorry!. Can you please rewrite the limits for this spherical annulus.
$endgroup$
– ProblemBook
Jan 9 at 3:28
$begingroup$
G. Stroke, I made a mistake in asking my question. It should be the annular region lying between two spheres of radius $r$ and $R$. Sorry!. Can you please rewrite the limits for this spherical annulus.
$endgroup$
– ProblemBook
Jan 9 at 3:28
$begingroup$
I am not able to vote your comment as I am new here. Thank you!
$endgroup$
– ProblemBook
Jan 9 at 4:03
$begingroup$
I am not able to vote your comment as I am new here. Thank you!
$endgroup$
– ProblemBook
Jan 9 at 4:03
$begingroup$
I think there is some mistake. When I integrate with respect to z , shouldn't limits be independent of z? Similarly with other two integrals
$endgroup$
– ProblemBook
Jan 9 at 16:46
$begingroup$
I think there is some mistake. When I integrate with respect to z , shouldn't limits be independent of z? Similarly with other two integrals
$endgroup$
– ProblemBook
Jan 9 at 16:46
$begingroup$
Teeny typo: fixed. Thanks.
$endgroup$
– David G. Stork
Jan 9 at 18:51
$begingroup$
Teeny typo: fixed. Thanks.
$endgroup$
– David G. Stork
Jan 9 at 18:51
$begingroup$
David G. Stroke, Sir, how can Change it into spherical coordinates. I that would be easy to evaluate. Also, the function $f(x,y,z)$ is inside the integrals? You have written it like this to make things look simpler, right?
$endgroup$
– ProblemBook
Jan 9 at 19:03
$begingroup$
David G. Stroke, Sir, how can Change it into spherical coordinates. I that would be easy to evaluate. Also, the function $f(x,y,z)$ is inside the integrals? You have written it like this to make things look simpler, right?
$endgroup$
– ProblemBook
Jan 9 at 19:03
|
show 2 more comments
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$begingroup$
$LaTeX$ note: You can get a triple integral by writing
iiintrather thanintintint(visually you get $iiint$ instead of $intintint$).$endgroup$
– JavaMan
Jan 9 at 3:07