How Do I Properly Convert a Triple Integral to Cylindrical Coordinates?

[MaximumTaco]

Firstly, can someone please demonstrate the proper Latex code for the terminals on a multiple integral? Thanks!

Anyway, as you can probably see, I'm calculating the volume enclosed by x^2+y^2+z^2 = 2 and z = x^2+y^2 using a change to cylindrical coordinates.

<br /> <br /> V = \iiint_{0}^{2\pi}_{0}^{\sqrt{2}}_{\rho^2}^{\sqrt{2-\rho^2}} \rho dz\,d\rho\,d\phi<br /> <br />

Is that integral right?

The answer i get when evaluating the integral is (pi/3)(4sqrt(2) - 6), which is the same doing it by hand or using Mathematica. But it doesn't agree with the expected answer.

 

[dextercioby]

V=\int_{0}^{1}\rho \ d\rho \int_{0}^{2\pi} d\varphi \int_{0}^{\rho^{2}} dz

Daniel.

 

[M98Ranger]

To properly convert a triple integral to cylindrical coordinates, follow these steps:

1. Identify the limits of integration for each variable in the Cartesian coordinates. In this case, the limits are x = 0 to x = sqrt(2), y = 0 to y = sqrt(2), and z = 0 to z = x^2 + y^2.

2. Convert the Cartesian coordinates to cylindrical coordinates using the equations x = \rho \cos{\phi}, y = \rho \sin{\phi}, and z = z.

3. Substitute the converted limits into the integral. The integral should now be in terms of \rho, \phi, and z.

4. Change the order of integration if necessary. In this case, since the limits for z depend on \rho and \phi, it is best to integrate with respect to z first.

5. Evaluate the integral by integrating with respect to z, then \rho, and finally \phi.

The integral you have written in LaTeX looks correct. However, the answer you obtained may not be the expected answer due to errors in the evaluation or a typo in the original problem. It is always a good idea to double check your work and use a calculator or computer program to verify your answer.