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True or False? The integral from 0 to 2pi?

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2 Answers

It's false.

Whenever you convert an n-fold integral from x1-....-xn coordinate system to the u1-...-un coordinate system, you get a term called the Jacobian.

I'm pretty sure you must have already covered Jacobians in your class, but here is a review.

It is an n by n matrix determinant. The ij entry in the matrix is the partial derivative ∂x/ ∂uj

Of course to evaluate this determinant you need to know the conversion formulas from the old system to the new system.

The Jacobian is denoted by ∂(x1,...,xn) / ∂(u1,...,un).

Once you have the Jacobian, then the integral transformation is

∫...∫ f dx1 ... dxn = ∫...∫ f  * Jacobian * du1 ... dun.

For example, from cartesian coordinates (x,y) to polar coordinates (r,θ), we have

x = r cos θ and y = r sin θ so that

                ∂ ( x , y )       |  ∂x/∂r       ∂x/∂θ  |        |    cos θ      -r sin θ    |

Jacobian = ----------- =   |                          |   =   |                                |

                 ∂ ( r , θ )      |  ∂y/∂r       ∂y/∂θ  |        |     sin θ      r cos θ    |


= (cos θ)(r cos θ) - (sin θ)(-rsin θ) = r(sin2 θ + cos2 θ) = r

Thus we have that when integrating over a region R,

∫∫R f(x , y) dx dy = ∫∫R f(r cos θ, r sin θ) r dr dθ

The jacobian for cartesian to cylindrical coordinates is also r, so we get

∫∫∫R f(x , y , z) dx dy dz = ∫∫∫R f(r cos θ , r sin θ , z) r dr dθ dz.

In your question, you said ∫0 ∫04 ∫r4 dz dr dθ which is missing the Jacobian.

The correct expression for the volume would be ∫004r4 r dz dr dθ.