Converting Polar Triple Integral to Spherical One

[harrietstowe]

Homework Statement

Evaluate the following triple integral by switching it to spherical coordinates?
The integrand is r dzdrdθ
The limits for the inner integral are 0 to r
The limits for the middle integral are 0 to 3
The limits for the outer integral are 0 to 2π

Homework Equations

The Attempt at a Solution

I know that the limits for θ will stay the same but I need help finding the limits for ρ and Φ.
Also I believe the integrand should be (ρ^3)(sin(Φ))^2 dρdΦdθ

 

[Char. Limit]

Homework Statement

Evaluate the following triple integral by switching it to spherical coordinates?
The integrand is r dzdrdθ
The limits for the inner integral are 0 to r
The limits for the middle integral are 0 to 3
The limits for the outer integral are 0 to 2π

Homework Equations

The Attempt at a Solution

I know that the limits for θ will stay the same but I need help finding the limits for ρ and Φ.
Also I believe the integrand should be (ρ^3)(sin(Φ))^2 dρdΦdθ

So, the integral looks like this?

$$\int_0^{2\pi} \int_0^3 \int_0^r r dz dr d\theta$$

And you want to convert it from cylindrical (not polar) to spherical co-ordinates, right?

Ah, let me check my book, I'll be right back.

 

[harrietstowe]

that is correct

 

[hunt_mat]

You're asked to convert from spherical co-ordinates to cylindrical ones?

 

[Char. Limit]

All right, I can't seem to find my book... I'm sorry I couldn't help you out!

 

[harrietstowe]

its cylindrical to spherical and I figured it out

 

[mege]

Take note:

(ρ^2)(sin(Φ)) dρdΦdθ is the correct 'basic' integrand for spherical. You don't need to multiply by 'r' (or it's spherical equivalent) from cylindrical again as the above is already taking the conversion into account.