Write a triple integral in spherical coordinates

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Homework Help Overview

The discussion revolves around writing a triple integral in spherical coordinates to represent the volume of a sphere defined by the equation X² + Y² + Z² = 16, specifically focusing on the portion that lies in the first octant where x, y, and z are all non-negative.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the transformation from rectangular to spherical coordinates, noting the need for proper limits for the integral. There is confusion regarding the correct representation of the volume element in spherical coordinates and the limits for the angles involved.

Discussion Status

Some participants have pointed out the importance of understanding the volume element in spherical coordinates and have suggested visualizing the sphere to clarify the limits for the angles. There is ongoing exploration of the limits for the angle ø, with some participants expressing uncertainty about how to derive them without specific values for z.

Contextual Notes

Participants mention the first octant and the constraints of the problem, including the requirement to find limits for the spherical coordinates without using a Jacobian, as they are focusing on volume calculation rather than a change of variables.

Jtechguy21
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Homework Statement



Write a triple integral in spherical coordinates that represents the volume of the part of the sphere
X^2+Y^2+Z^2=16 that lies in the first octant(where x,y, and z are coordinates are all greater than or equal to zero)


Homework Equations



So i know this is in rectangular form (x,y,z) trying to get it into (p,Θ,ø)

The Attempt at a Solution




X^2+y^2+z^2=16
p^2=16
p=4

To get Θ the formula is
arccos z/(square root of x^2+y^2+z^2)

when i solve for z I get z=√(16-r^2)

These are the limits i know
The limits for Dz are from z=0 to z=√(16-r^2)
The limits for Dr are from r=0 to r= pi/2

I do not know how to find the limits for DΘ (theta)

since the arccos √(16-r^2)/(square root of x^2+y^2+z^2)
should give me my theta. Θ
but i have no real values for x y and z. so i don't know how to approach this.
thank you

∫ ∫ ∫
rDzDrDΘ
 
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You don't seem to understand the problem. You should figure out how a differential element of volume dV is represented in spherical coordinates. In cartesian coordinates, dV = dx dy dz. You appear to have used cylindrical coordinates, r, z, and Θ.

http://en.wikipedia.org/wiki/Spherical_coordinate_system
 
Indeed, and in applying that transformation properly, you will wind up needing to solve:

##r sin(\theta) cos(\phi) ≥ 0##
##r sin(\theta) sin(\phi) ≥ 0##
##r cos(\theta) ≥ 0##

Don't forget the Jacobian.
 
@JtechGuy21: Don't set that problem up in rectangular coordinates and try to change it to spherical coordinates. You don't need to change variables and mess with Jacobians. Just remember$$
\text{Volume} =\iiint_V 1~dV$$All you need is the ##dV## formula in spherical coordinates and then to put the ##\rho,\phi,\theta## limits on. They are very easy and can be easily seen geometrically.
 
Last edited:
aaaaaah okay.
yes i was turning them into cylindrical for some reason...

Thanks for pointing that out.

p=4
x=4SinøCosΘ
y=4SinøSinΘ
Z=4Cosø

1Dv=p^2sin dp dø dΘ

∭ p^2 Sinø dp dø dΘ

So that means my limits for;
p is 0 to 4
Θ is 0 to pi/2 (because its quadrant one)
and i have no idea how to find my ø limits?
Since i don't have a real value for z. I can't take the arc cos of z to get my ø?
 
It would be good for you to quote who you are replying to.

Jtechguy21 said:
aaaaaah okay.
yes i was turning them into cylindrical for some reason...

Thanks for pointing that out.

p=4
x=4SinøCosΘ
y=4SinøSinΘ
Z=4Cosø

1Dv=p^2sin dp dø dΘ

∭ p^2 Sinø dp dø dΘ

So that means my limits for;
p is 0 to 4
Θ is 0 to pi/2 (because its quadrant one)
and i have no idea how to find my ø limits?
Since i don't have a real value for z. I can't take the arc cos of z to get my ø?


Look at the picture of the sphere in the first octant. Draw a radius and label ##\rho##, ##\phi## and ##\theta##. It is obvious from the picture what their ranges are. You have ##\rho## and ##\theta## correct so far.
 
Zondrina said:
Indeed, and in applying that transformation properly, you will wind up needing to solve:

##r sin(\theta) cos(\phi) ≥ 0##
##r sin(\theta) sin(\phi) ≥ 0##
##r cos(\theta) ≥ 0##

Don't forget the Jacobian.

hello.
i don't think i need to use a jacobian here. I am in the section that involves using triple integrals in cylindrical and spherical coordinates.(not saying its not possible )

The section afterwards is change of variables:jacobians.
Which I've done before, but since we are finding volume in this case, i don't know how jacobians would help since jacobians deal with area in the uv plane in comparison to the xy plane
 
LCKurtz said:
It would be good for you to quote who you are replying to.



Look at the picture of the sphere in the first octant. Draw a radius and label ##\rho##, ##\phi## and ##\theta##. It is obvious from the picture what their ranges are. You have ##\rho## and ##\theta## correct so far.

This might be a dumb question, but how do i draw this picture so i could visually see what's happening.
Do i have to put it into z= format?
and use wolfgram alpha
 
Jtechguy21 said:
This might be a dumb question, but how do i draw this picture so i could visually see what's happening.
Do i have to put it into z= format?
and use wolfgram alpha

Just draw it by hand. Draw a section of a sphere that is in the first octant. Surely your book has such pictures. Probably right where it introduces spherical coordinates.

Or look here: http://mathinsight.org/spherical_coordinates

It isn't hard to see the limits for the variables in the first octant. Be aware that physics and math books have the ##\phi## and ##\theta## switched. That link uses the math convention.
 

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