Rewrite Cartesian in Cylindrical form

[nysnacc]

Homework Statement

Homework Equations

Cartesian to Cylindrial

The Attempt at a Solution

What I was doing is that, I changed the limits of z, and the function.

 

[Ssnow]

you must be precise where $\rho$ varies. I give you an hint. Start from $-\sqrt{4-x^2-y^2}\leq z\leq \sqrt{4-x^2-y^2}$ and prove that $0\leq \rho \leq 2$...

 

[nysnacc]

So the whole volume is above and below the xy-plane... Can I divide the whole volume into 2 symmetric sub volume?

0≤ρ≤20≤ρ≤2 0\leq \rho \leq 2 as Φ is 0 where r = 0 :)

 

[Mark44]

Minor point, but your title (Rewrite Cartesian in Cylindrial form) and what you have in Relevant Equations, are incorrect. The problem asks you to transform the integral to spherical form, which is actually what you're doing. It might indicate that you don't have a clear understanding of the difference between cylindrical (not cylindrial, which I don't think is a word) coordinates and spherical coordinates.

 

[nysnacc]

θ

Minor point, but your title (Rewrite Cartesian in Cylindrial form) and what you have in Relevant Equations, are incorrect. The problem asks you to transform the integral to spherical form, which is actually what you're doing. It might indicate that you don't have a clear understanding of the difference between cylindrical (not cylindrial, which I don't think is a word) coordinates and spherical coordinates.

Okay, so am I setting the limits correctly in my work? thanks

 

[Mark44]

Okay, so am I setting the limits correctly in my work? thanks

I don't think you are, but I haven't looked that closely at your work. Can you describe, in words, what the region of integration looks like?

 

[nysnacc]

In xy plane, the shape is a half circle in +x then z is bounded between -√ to +√ (so above and below xy plane)

Then r = x^2 + y^2 and θ goes from 0 to π, Φ goes from 0 to π as the volume is above and below the plane

 

[Mark44]

In xy plane, the shape is a half circle in +x then z is bounded between -√ to +√ (so above and below xy plane)

Then r = x^2 + y^2 and θ goes from 0 to π, Φ goes from 0 to π as the volume is above and below the plane

But what is the shape of the three-dimensional object that is the region of integration? That's what I'd like you to tell me. A very important aspect of being able to transform iterated integrals from one form to another is being able to correctly describe the region of integration. Once you understand this, evaluating the integrals is more-or-less mechanical.

 

[nysnacc]

A half sphere.

 

[LCKurtz]

You need to describe where the half sphere is located in the xyz coordinate system. You need to have the picture to get the limits correct. And, no, your limits in post #1 are not correct. For one thing, if the integration variable is $d\rho$, there cannot be any $\rho$ in the limits. And there is at least one other error.

 

[nysnacc]

Thanks.

By sketching, I come up with something like this:

θ: 0 to π
Φ: 0 to π
rho: 0 to 2 (the bound of the sphere)

And then I replace the function y^2 √ x^2+y^2+z^2 in terms of rho, θ and Φ

 

[nysnacc]

p is rho

∫ ∫ ∫ (p sinΦ sinθ)² (p) p²sinΦ dp dΦ dθ ! ! !

 

[LCKurtz]

You haven't given us a description or picture of where the half sphere is located in the xyz coordinate system, and as a consequence your limits on $\theta$ are incorrect.