Triple integral and cartesian coordinates

1. Dec 28, 2011

queenstudy

we all know that triple integral can be solved by either cartesian coordinates , spherical ,or cylindrical coordinates
i just need like some advice in knowing when the variable used is constant and when it is not
for example : r in cylindrical coordinates can it be constant or not?? because i always see it in terms of y and z like that

2. Dec 28, 2011

I like Serena

Hi queenstudy!

Can you give a couple of examples of the integrals you mean?

Taken literally, if you keep r in cylindrical coordinates constant, you have a double integral, meaning you integrate over a cylindrical surface instead of a cylindrical volume.

3. Jan 1, 2012

queenstudy

by fixing the plane Z=z change the domain of definitoin of the triple integral f(x,y,z) where D is
D=(x,y,z) belongs to R cube and it is the intersection of the sphere and cone (icecream cone) sphere xsquare +y square + z square =2 and z=xsquare + y square
z is positive
in cartesian coordinates
my question here is when i do my projection i will have a circle , and if i want it in cartesian i need like 8 integrals because the plane Z=z should be cut twice and for each domain the intersection is a circle and in each circle we have y and x positive x and y negative and one positive and one negative so can i replace those with r and theta and save time and work and still in cartesian or not??
do you have any extra double and triple integral because i solved on google and want some more and thank you thank you twice serena for helping me out

4. Jan 1, 2012

I like Serena

Hmm, as yet I'm having trouble understanding what you write...?

You appear to refer to the integral (polar):
$$\int_{r=0}^{r=R} \int_{\theta=0}^{\theta=2\pi} \int_{\phi=0}^{\phi={\pi \over 4}} ... d\phi d\theta dr$$

Or cartesian:
$$\int_{x=-\frac R 2 \sqrt 2}^{x=+\frac R 2 \sqrt 2} \int_{y=-\sqrt{R^2-x^2}}^{y=+\sqrt{R^2-x^2}} \int_{z=\sqrt{x^2+y^2}}^{z=\sqrt{R^2-x^2-y^2}} ... dz dy dx$$

Does this look like what you mean?

Last edited: Jan 1, 2012
5. Jan 1, 2012

queenstudy

what you wrote is true but i am supposed to write z as two constants, and when you study x and y , you must write in terms of constant K , thus in both answers you wrote they arent correct ,
let me ask a simpler question:
i should write the domain in cartesian may i write x and y in polar coordinates and still call them cartesian because if i keep them in cartesian , x and y will have 4 domains when:
1) x and y postive ; 2) x and y negative; 3) x positive and y negative ; 4) x negative and y postive
i hope i explained my question better

6. Jan 1, 2012

I like Serena

Well, you can rearrange the cartesian integral so z is integrated last.
In that case you would need to split the integral into 2 integrals.
One integral for z from 0 to $\frac R 2 \sqrt 2$ and one from $\frac R 2 \sqrt 2$ to R.
You would not need to split the x or y integrations though.

Perhaps you can show what you think it might be?

Cartesian means not-polar, but why do you think you would have to?
There is no need to split x and y in the 4 domains you mention.

7. Jan 2, 2012

queenstudy

because according to our proffesor when x is negative y has the possibility to be positive or negative right? so we have two cases
now is x is positive then y can be either positive or negative right? so we also have two possibilities , thus we have 4 cases thats and that is only the case of the intersection of the plane z=constant with the cone
if we do it with the sphere inside the domain we will obtain the same 4 parts thus the total will be 8 parts which is long
that is why we choose polar here for a circle or to me be more specific a full disk because r is only positive and theta is between zero and two pie

8. Jan 2, 2012

I like Serena

I think you're mixing up a single integral with higher dimensional integrals.

If you have a single integral to find the area of (part of) a circle disk, you have the problem that y has a negative and positive value and you have to choose between one of them.

If you have a double integral over a circle disk, y runs from the negative value to the positive value, which covers the entire circle disk.
So there is no need to split up the integral.

9. Jan 2, 2012

queenstudy

but in class when i we projected the domain we got a circle , and it got 4 domains where x and y contain the domain of the circle
okay if we have a disk on the xoy axis and i want to make double integration with respect to x and y dont i get 4 parts of the circles thus 4 domains?? then we use polar coordinate and becomes a simple integration right????????
if this is true , MY question is that can r and theta be considered also as cartesian coordinates??????

10. Jan 2, 2012

I like Serena

Can you show me integrals of what you mean?

As for your question whether r and theta can be considered cartesian coordinates, I believe we covered that in your other thread.
All I can say is that the word "cartesian" means x and y, which is not r and theta.