I have to solve this integral: $$ \int_D \frac {4x 4y 4z}{x^2}dxdydz, \quad \text{where} \quad D=\left\{(x, y,z) \in \mathbb R^3 \mid x,z \in [1,e], \ 0\le y\le\sqrt{\ln z}\right\}$$
I started with $$ \int_1^e \int_0^{\sqrt{\ln z}} \int_1^e \frac {4x4y4z}{x^2}dxdydz $$
but I'm stuck and can't calculate the first integral right because its hard for me to integrate a quotient. It would be nice if you would show me the first step with an explantation so I can try to go on.
Just separate $x$ and change the order of $y \ \& \ z$:
$$\int_D \frac{4x4y4z}{x^2}dx dy dz = 64 \left( \int_1^e\frac{1}{x}dx \right) \left( \int_0^1 \int_{e^{y^2}}^e yz dz dy \right) = \dots $$
For boundary of $z$, just inverse the inequality $y \leq \sqrt{\ln z}$.
You can do it $$\begin{align} & \int_1^e \int_0^{\sqrt{\ln z}} \int_1^e \frac {4x4y4z}{x^2}dx dy dz \\ = & \int_1^e\int_0^{\sqrt{\ln z}} \left( 64yz\ln x \right) _1^e dy dz\\ = & \int_1^e\int_0^{\sqrt{\ln z}} 64yz dy dz\\ = & \int_1^e (32zy^2)_0^{\sqrt{\ln z}} dz\\ = & \int_1^e 32z\ln z dz\\ = & 16\int_1^e \ln z dz^2\\ = & (16z^2\ln z)_1^e - 16\int_1^e z^2 d\ln z\\ = & 16e^2 - 16\int_1^e zdz\\ = & 16e^2 - (8z^2)_1^e\\ = & 16e^2 - 8e^2 + 8\\ = & 8(e^2+1) \end{align}$$
Addition: $$\begin{align} & \int_1^e 32z\ln z dz\\ = & \int_1^e 16\ln z\cdot 2zdz \\ = & \int_1^e 16\ln z\cdot dz^2 \\ = & 16\int_1^e \ln z\cdot dz^2 \end{align}$$
