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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!

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    $\begingroup$ $\LaTeX$ note: You can get a triple integral by writing \iiint rather than \int\int\int (visually you get $\iiint$ instead of $\int\int\int$). $\endgroup$ – JavaMan Jan 9 at 3:07
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In polar coordinates:

$$\int\limits_{r=r_i}^R \int\limits_{\theta = 0}^{2 \pi} \int\limits_{z=z_i}^{z_f} f(r \cos \theta, r \sin \theta,z)\ r\ dr\ d \theta\ dz$$

or in rectilinear coordinates...

$$\int\limits_{x=-R}^R\ dx \int\limits_{y = - \sqrt{R^2 - x^2}}^{+ \sqrt{R^2 - x^2}}\ dy \int\limits_{z=z_i}^{z_f}\ dz\ f(x,y,z) - \int\limits_{x=-r}^r\ dx\ \int\limits_{y = - \sqrt{r^2 - x^2}}^{+ \sqrt{r^2 - x^2}}\ dy\ \int\limits_{z=z_i}^{z_f}\ dz\ f(x,y,z)$$

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Revised question:

$$\int\limits_{x=-R}^R\ dx \int\limits_{y = -\sqrt{R^2 - x^2 - z^2}}^{+\sqrt{R^2 - x^2 - z^2}}\ dy \int\limits_{z=-\sqrt{R^2 - x^2 - y^2}}^{+\sqrt{R^2 - x^2 - y^2}}\ dz\ f(x,y,z) - \int\limits_{x=-r}^r\ dx \int\limits_{y = -\sqrt{r^2 - x^2 - z^2}}^{+\sqrt{r^2 - x^2 - z^2}}\ dy \int\limits_{z=-\sqrt{r^2 - x^2 - y^2}}^{+\sqrt{r^2 -x^2 - y^2}}\ dz\ f(x,y,z)$$

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  • $\begingroup$ 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. $\endgroup$ – user421818 Jan 9 at 3:28
  • $\begingroup$ I am not able to vote your comment as I am new here. Thank you! $\endgroup$ – user421818 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 $\endgroup$ – user421818 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? $\endgroup$ – user421818 Jan 9 at 19:03

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