Triple Integral in spherical coordinate The Next CEO of Stack OverflowTriple Integral in...
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Triple Integral in spherical coordinate
The Next CEO of Stack OverflowTriple Integral in cylindrical coordinatelimits of integration in spherical coordinates.Solving a triple integral with Spherical CoordinatesTwo spheres, triple integration, not their intersectionVolumes using triple integrationInterchange $phi$ and $theta$ in spherical coordinatesFinding volume of region in first octant underneath paraboloidHow to write six possible iterated integral?How to find volume of the region ${(x,y,z)|,0 le (x-1)^2+y^2 le z(1-z)}$?Triple integrals in spherical coordinates, volume of octant
$begingroup$
$displaystyleiiint_R (x^2+y^2+z^2)^{-2},dx,dy,dz$ where $R$ is in the region in the first octant outside the sphere $x^2+y^2+z^2 = 1$;
Hi guys, I don't quite get which region is this, is it that I find the volume of this sphere and minus by the volume of the first octant? Thanks in advance.
integration volume spheres multiple-integral
edited Mar 16 at 13:36
MathIsFun
1915
asked Mar 16 at 8:56
joshjosh
177
$endgroup$
add a comment |
$begingroup$
$displaystyleiiint_R (x^2+y^2+z^2)^{-2},dx,dy,dz$ where $R$ is in the region in the first octant outside the sphere $x^2+y^2+z^2 = 1$;
Hi guys, I don't quite get which region is this, is it that I find the volume of this sphere and minus by the volume of the first octant? Thanks in advance.
integration volume spheres multiple-integral
edited Mar 16 at 13:36
MathIsFun
1915
asked Mar 16 at 8:56
joshjosh
177
$endgroup$
$begingroup$
If you want to use spherical coordinates, do you know the transformations for your variables?
$endgroup$
– Boots
Mar 16 at 9:03
$begingroup$
hi boots, yea i know, is just that i dont quite understand the question. I am unsure what region is it that i required to calculate
$endgroup$
– josh
Mar 16 at 9:05
$begingroup$
I tried using the limit for the first octant of the sphere and it gives me the same answer as the answer sheet but from the question, it doesn't seem like it is asking for the first octant.
$endgroup$
– josh
Mar 16 at 9:07
add a comment |
$begingroup$
$displaystyleiiint_R (x^2+y^2+z^2)^{-2},dx,dy,dz$ where $R$ is in the region in the first octant outside the sphere $x^2+y^2+z^2 = 1$;
Hi guys, I don't quite get which region is this, is it that I find the volume of this sphere and minus by the volume of the first octant? Thanks in advance.
integration volume spheres multiple-integral
edited Mar 16 at 13:36
MathIsFun
1915
asked Mar 16 at 8:56
joshjosh
177
$endgroup$
$displaystyleiiint_R (x^2+y^2+z^2)^{-2},dx,dy,dz$ where $R$ is in the region in the first octant outside the sphere $x^2+y^2+z^2 = 1$;
Hi guys, I don't quite get which region is this, is it that I find the volume of this sphere and minus by the volume of the first octant? Thanks in advance.
integration volume spheres multiple-integral
integration volume spheres multiple-integral
edited Mar 16 at 13:36
MathIsFun
1915
asked Mar 16 at 8:56
joshjosh
177
edited Mar 16 at 13:36
MathIsFun
1915
asked Mar 16 at 8:56
joshjosh
177
edited Mar 16 at 13:36
MathIsFun
1915
edited Mar 16 at 13:36
MathIsFun
1915
edited Mar 16 at 13:36
MathIsFun
1915
1915
asked Mar 16 at 8:56
joshjosh
177
asked Mar 16 at 8:56
joshjosh
177
asked Mar 16 at 8:56
joshjosh
177
177
$begingroup$
If you want to use spherical coordinates, do you know the transformations for your variables?
$endgroup$
– Boots
Mar 16 at 9:03
$begingroup$
hi boots, yea i know, is just that i dont quite understand the question. I am unsure what region is it that i required to calculate
$endgroup$
– josh
Mar 16 at 9:05
$begingroup$
I tried using the limit for the first octant of the sphere and it gives me the same answer as the answer sheet but from the question, it doesn't seem like it is asking for the first octant.
$endgroup$
– josh
Mar 16 at 9:07
add a comment |
$begingroup$
If you want to use spherical coordinates, do you know the transformations for your variables?
$endgroup$
– Boots
Mar 16 at 9:03
$begingroup$
hi boots, yea i know, is just that i dont quite understand the question. I am unsure what region is it that i required to calculate
$endgroup$
– josh
Mar 16 at 9:05
$begingroup$
I tried using the limit for the first octant of the sphere and it gives me the same answer as the answer sheet but from the question, it doesn't seem like it is asking for the first octant.
$endgroup$
– josh
Mar 16 at 9:07
$begingroup$
If you want to use spherical coordinates, do you know the transformations for your variables?
$endgroup$
– Boots
Mar 16 at 9:03
$begingroup$
If you want to use spherical coordinates, do you know the transformations for your variables?
$endgroup$
– Boots
Mar 16 at 9:03
$begingroup$
hi boots, yea i know, is just that i dont quite understand the question. I am unsure what region is it that i required to calculate
$endgroup$
– josh
Mar 16 at 9:05
$begingroup$
hi boots, yea i know, is just that i dont quite understand the question. I am unsure what region is it that i required to calculate
$endgroup$
– josh
Mar 16 at 9:05
$begingroup$
I tried using the limit for the first octant of the sphere and it gives me the same answer as the answer sheet but from the question, it doesn't seem like it is asking for the first octant.
$endgroup$
– josh
Mar 16 at 9:07
$begingroup$
I tried using the limit for the first octant of the sphere and it gives me the same answer as the answer sheet but from the question, it doesn't seem like it is asking for the first octant.
$endgroup$
– josh
Mar 16 at 9:07
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
No. This has nothing to do with computing volumes. The idea is to compute the integral of $(x^2+y^2+z^2)^{-2}$ in the region of the first octant outside the unit sphre. Turning this into spherical coordinates, this becomes$$int_1^inftyint_0^{fracpi2}int_0^{fracpi2}frac{rho^2sinvarphi}{rho^4},mathrm dtheta,mathrm dvarphi,mathrm drho.$$Can you take it from here?
answered Mar 16 at 9:09
José Carlos SantosJosé Carlos Santos
171k23132240
$endgroup$
$begingroup$
Thank you so much, this helps. I understand and got the answer as well.
$endgroup$
– josh
Mar 16 at 14:19
$begingroup$
I'm glad I could help.
$endgroup$
– José Carlos Santos
Mar 16 at 14:24
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
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active
oldest
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active
oldest
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$begingroup$
No. This has nothing to do with computing volumes. The idea is to compute the integral of $(x^2+y^2+z^2)^{-2}$ in the region of the first octant outside the unit sphre. Turning this into spherical coordinates, this becomes$$int_1^inftyint_0^{fracpi2}int_0^{fracpi2}frac{rho^2sinvarphi}{rho^4},mathrm dtheta,mathrm dvarphi,mathrm drho.$$Can you take it from here?
answered Mar 16 at 9:09
José Carlos SantosJosé Carlos Santos
171k23132240
$endgroup$
$begingroup$
Thank you so much, this helps. I understand and got the answer as well.
$endgroup$
– josh
Mar 16 at 14:19
$begingroup$
I'm glad I could help.
$endgroup$
– José Carlos Santos
Mar 16 at 14:24
add a comment |
$begingroup$
No. This has nothing to do with computing volumes. The idea is to compute the integral of $(x^2+y^2+z^2)^{-2}$ in the region of the first octant outside the unit sphre. Turning this into spherical coordinates, this becomes$$int_1^inftyint_0^{fracpi2}int_0^{fracpi2}frac{rho^2sinvarphi}{rho^4},mathrm dtheta,mathrm dvarphi,mathrm drho.$$Can you take it from here?
answered Mar 16 at 9:09
José Carlos SantosJosé Carlos Santos
171k23132240
$endgroup$
$begingroup$
Thank you so much, this helps. I understand and got the answer as well.
$endgroup$
– josh
Mar 16 at 14:19
$begingroup$
I'm glad I could help.
$endgroup$
– José Carlos Santos
Mar 16 at 14:24
add a comment |
$begingroup$
No. This has nothing to do with computing volumes. The idea is to compute the integral of $(x^2+y^2+z^2)^{-2}$ in the region of the first octant outside the unit sphre. Turning this into spherical coordinates, this becomes$$int_1^inftyint_0^{fracpi2}int_0^{fracpi2}frac{rho^2sinvarphi}{rho^4},mathrm dtheta,mathrm dvarphi,mathrm drho.$$Can you take it from here?
answered Mar 16 at 9:09
José Carlos SantosJosé Carlos Santos
171k23132240
$endgroup$
No. This has nothing to do with computing volumes. The idea is to compute the integral of $(x^2+y^2+z^2)^{-2}$ in the region of the first octant outside the unit sphre. Turning this into spherical coordinates, this becomes$$int_1^inftyint_0^{fracpi2}int_0^{fracpi2}frac{rho^2sinvarphi}{rho^4},mathrm dtheta,mathrm dvarphi,mathrm drho.$$Can you take it from here?
answered Mar 16 at 9:09
José Carlos SantosJosé Carlos Santos
171k23132240
answered Mar 16 at 9:09
José Carlos SantosJosé Carlos Santos
171k23132240
answered Mar 16 at 9:09
José Carlos SantosJosé Carlos Santos
171k23132240
answered Mar 16 at 9:09
José Carlos SantosJosé Carlos Santos
171k23132240
171k23132240
$begingroup$
Thank you so much, this helps. I understand and got the answer as well.
$endgroup$
– josh
Mar 16 at 14:19
$begingroup$
I'm glad I could help.
$endgroup$
– José Carlos Santos
Mar 16 at 14:24
add a comment |
$begingroup$
Thank you so much, this helps. I understand and got the answer as well.
$endgroup$
– josh
Mar 16 at 14:19
$begingroup$
I'm glad I could help.
$endgroup$
– José Carlos Santos
Mar 16 at 14:24
$begingroup$
Thank you so much, this helps. I understand and got the answer as well.
$endgroup$
– josh
Mar 16 at 14:19
$begingroup$
Thank you so much, this helps. I understand and got the answer as well.
$endgroup$
– josh
Mar 16 at 14:19
$begingroup$
I'm glad I could help.
$endgroup$
– José Carlos Santos
Mar 16 at 14:24
$begingroup$
I'm glad I could help.
$endgroup$
– José Carlos Santos
Mar 16 at 14:24
add a comment |
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$begingroup$
If you want to use spherical coordinates, do you know the transformations for your variables?
$endgroup$
– Boots
Mar 16 at 9:03
$begingroup$
hi boots, yea i know, is just that i dont quite understand the question. I am unsure what region is it that i required to calculate
$endgroup$
– josh
Mar 16 at 9:05
$begingroup$
I tried using the limit for the first octant of the sphere and it gives me the same answer as the answer sheet but from the question, it doesn't seem like it is asking for the first octant.
$endgroup$
– josh
Mar 16 at 9:07