Evaluate $f(x,y,z) = \frac{y}{\sqrt{z}}$ in $y \geq 0 \ $ , $0 \leq z \leq x^2$ and $(x-2)^2 + y^2 \leq 4$

Question

Evaluate $f(x,y,z) = \frac{y}{\sqrt{z}}$ in $y \geq 0 \ $ , $0 \leq z \leq x^2$ and $(x-2)^2 + y^2 \leq 4$

The integral set up I got is

$$\int_{0}^{\frac{\pi}{2}} \int_{0}^{4 \cos(\theta)} \int_{0}^{r^2\cos^2(\theta)} \frac{r \sin(\theta)}{\sqrt{z}} \ dzdrd\theta$$

but the answer I get, $\dfrac{128}{5}$, is different that this textbook's answer (which is $\dfrac{64}{3}$).

Am I missing something?

Note: This wouldn't be the first time this textbook contains an error.

Answer 1

Your integrand should be $\frac{r \sin(\theta)}{\sqrt{z}} \ \color{red}r dzdrd\theta.$

In fact, you have forgotten the Jacobian $\color{red}r$ of the transformation from cartesian to cylindrical coordinates (http://mathworld.wolfram.com/CylindricalCoordinates.html).

All the rest is correct; in particular, the boundary of the half circle (at the base of the integration domain) has polar equation $r=4\cos(\theta)$ for $0 \leq \theta \leq \tfrac{\pi}{2}$ (and not $\tfrac{\pi}{2}$ as I thought at first).

It can be of interest to know how it is possible to check this result by a numerical simulation.

Here is a way to do this (Matlab program below), with results at $\pm 1/\sqrt{N}=1/1000$ of the true result.

clear all;close all;
N=10^6;S=0;
for k=1:N;
    % unif. distributed random point in enclosing brick 
    % [0,4]x[0,2]x[0,16] with volume 128:
    x=4*rand;y=2*rand;z=16*rand;
    if z<x^2 && (x-2)^2+y^2<4
        S=S+y/sqrt(z);
    end
end
128*S/N,%=vol. of integrated zone times mean value
% good approximation of exact value 64/3

Remark: There are different ways to obtain an approximate value of an integral by simulation. This one could be considered as a "brute force" version.