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.
volume = ∫[0, 4] dr∫[r, 4] dz ∫[0, 2pi] r dθ = (64/3)pi
Answer: True
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.
volume = ∫[0, 4] dr∫[r, 4] dz ∫[0, 2pi] r dθ = (64/3)pi
Answer: True
It's false.
Whenever you convert an n-fold integral from x_{1}-....-x_{n} coordinate system to the u_{1}-...-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 ∂(x_{1},...,x_{n}) / ∂(u_{1},...,u_{n}).
Once you have the Jacobian, then the integral transformation is
∫...∫ f dx_{1} ... dx_{n} = ∫...∫ f * Jacobian * du_{1} ... 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(sin^{2} θ + cos^{2} 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} ∫_{0}^{4} ∫_{r}^{4} dz dr dθ which is missing the Jacobian.
The correct expression for the volume would be ∫_{0}^{2π} ∫_{0}^{4} ∫_{r}^{4} r dz dr dθ.
Comments
But the answer in the book says false.
This is a simple geometry problem. You can use geometry to verify the result.