Is My Quantum Mechanics Probability Calculation Correct?

[facenian]

Homework Statement

Let $$\psi(x,y,z)=\psi(\vec{r})$$ be the normalized wave function of a particle.Express in terms of $\psi(\vec{r})$ the probability for a simultaneous measurements o X y P_z to yield :
$$x_1 \leq x \leq x_2$$
$$p_z \geq 0$$

Homework Equations

$$<\vec{p}|\vec{r}>=\frac{1}{(2\pi\hbar)^{3/2}}e^{-i\vec{p}.\vec{r}/\hbar}$$
$$<\vec{p}|\psi>=\frac{1}{(2\pi\hbar)^{3/2}}\int \psi(\vec{r}) e^{-i\vec{p}.\vec{r}/\hbar} dr^3$$

The Attempt at a Solution

I have reached the following result:
$$\int_{-\infty}^{\infty}dz\int_{-\infty}^{\infty}dy\int_{x_1}^{x_2}dx \int_{-\infty}^{\infty}dp_x\int_{-\infty}^{\infty}dp_y\int_0^{\infty}dp_z <\vec{p}|\vec{r}>\psi(\vec{r})<\psi|\vec{p}>$$
I need to know two things: 1) is my result correct? 2) in case it is correct, is there any other more simple or concrete answer?

 

[elduderino]

Sorry, I am not answering your question. But could you explain how you arrived at the expression $$<\vec{p}|\vec{r}>\psi(\vec{r})<\psi|\vec{p}>$$

 

[facenian]

Sorry, I am not answering your question. But could you explain how you arrived at the expression $$<\vec{p}|\vec{r}>\psi(\vec{r})<\psi|\vec{p}>$$

I evalueted de expression $<\psi|P_2P_1|\psi>$ where P_1 and P_2 are the proyectors:
$$P_1=\int_{-\infty}^{\infty}dz\int_{-\infty}^{\infty}dy\int_{x_1}^{x_2}dx|x,y,z><x,y,z|$$
$$P_2=\int_{-\infty}^{\infty}dp_x\int_{-\infty}^{\infty}dp_y\int_0^{\infty}dp_z|p_x,p_y,p_z><p_x,p_y,p_z|$$
But I'm sure whether what I'm doing is correct