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

Question

True or False? The integral from 0 to 2pi, from 0 to 4, from r to 4 dz dr dtheta represents the volume enclosed by the cone z=sqrt(x^2+y^2) and the plane z=4.

Answer 1

volume = ∫[0, 4] dr∫[r, 4] dz ∫[0, 2pi] r dθ = (64/3)pi

Answer: True

Answer 2

It's false.

Whenever you convert an n-fold integral from x₁-....-x_n coordinate system to the u₁-...-u_n 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_i / ∂u_j

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,...,x_n) / ∂(u₁,...,u_n).

Once you have the Jacobian, then the integral transformation is

∫...∫ f dx₁ ... dx_n = ∫...∫ f  * Jacobian * du₁ ... du_n.

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 ∫₀^2π ∫₀⁴ ∫_r⁴ dz dr dθ which is missing the Jacobian.

The correct expression for the volume would be ∫₀^2π ∫₀⁴ ∫_r⁴ r dz dr dθ.