Convert \int_{0}^{1} \int_{0}^{\pi/4} \int_{0}^{2\pi} {\rho^{3}\sin{(2\phi)} \,d\theta \,d\phi \,d\rho} to cartesian coordinates

To convert the given triple integral from cylindrical coordinates to Cartesian coordinates, we need to follow these steps:

### Step 1: Recognize the coordinate system.
The original triple integral is expressed in **cylindrical coordinates**. Cylindrical coordinates \((\rho, \phi, \theta)\) are related to Cartesian coordinates \((x, y, z)\) through the following relations:

- \(x = \rho \cos(\theta)\)
- \(y = \rho \sin(\theta)\)
- \(z = \rho \sin(\phi)\)
- \(\rho = \sqrt{x^2 + y^2}\)
- \(\tan(\theta) = \frac{y}{x}\)

Here, \(\rho\) represents the radial distance in the xy-plane, \(\phi\) is the angle between the radius and the z-axis (the "polar angle"), and \(\theta\) is the azimuthal angle (rotation in the xy-plane).

### Step 2: Rewriting the integrand.
The given integral is:
\[
\int_{0}^{1} \int_{0}^{\pi/4} \int_{0}^{2\pi} \rho^{3} \sin(2\phi) \,d\theta \,d\phi \,d\rho
\]
We want to express the integrand \(\rho^3 \sin(2\phi)\) and the limits of integration in Cartesian coordinates.

### Step 3: Convert the integrand.
In Cartesian coordinates:
- \(\rho = \sqrt{x^2 + y^2}\), so \(\rho^3\) becomes \((x^2 + y^2)^{3/2}\).
- The function \(\sin(2\phi)\) depends on \(\phi\), which is the angle between \(\rho\) and the z-axis. In Cartesian coordinates, we can express \(\phi\) through the relation:
  \[
  \sin(\phi) = \frac{z}{\sqrt{x^2 + y^2 + z^2}}
  \]
  To express \(\sin(2\phi)\), we use the double-angle identity:
  \[
  \sin(2\phi) = 2\sin(\phi)\cos(\phi)
  \]
  where:
  \[
  \cos(\phi) = \frac{\sqrt{x^2 + y^2}}{\sqrt{x^2 + y^2 + z^2}} \quad \text{and} \quad \sin(\phi) = \frac{z}{\sqrt{x^2 + y^2 + z^2}}
  \]
  Hence, we get:
  \[
  \sin(2\phi) = 2 \left(\frac{z}{\sqrt{x^2 + y^2 + z^2}}\right) \left(\frac{\sqrt{x^2 + y^2}}{\sqrt{x^2 + y^2 + z^2}}\right)
  \]
  Simplifying this expression, we have:
  \[
  \sin(2\phi) = \frac{2z\sqrt{x^2 + y^2}}{x^2 + y^2 + z^2}
  \]

### Step 4: Convert the limits of integration.
The limits for \(\rho\), \(\phi\), and \(\theta\) must also be converted:
- \(\rho\) ranges from 0 to 1, which implies \(x^2 + y^2 + z^2 \leq 1\) in Cartesian coordinates. This describes the interior of a unit sphere.
- \(\phi\) ranges from 0 to \(\pi/4\), meaning the angle between the z-axis and the radial direction ranges from 0 to \(\pi/4\). This restricts the z-coordinate such that \(z \geq \frac{1}{\sqrt{2}}\) when \(\rho = 1\).
- \(\theta\) ranges from 0 to \(2\pi\), representing a full rotation in the xy-plane.

### Step 5: Combine everything.
Thus, the integral in Cartesian coordinates becomes:
\[
\int_{V} \frac{2z(x^2 + y^2)}{(x^2 + y^2 + z^2)^{5/2}} \, dx\, dy\, dz
\]
where \(V\) is the region of integration corresponding to \(x^2 + y^2 + z^2 \leq 1\) and \(z \geq \frac{1}{\sqrt{2}}\).

Would you like a detailed explanation of how to solve this integral or have any questions? 

Here are 5 related questions for further understanding:
1. How do cylindrical coordinates differ from spherical coordinates?
2. What are the steps to converting triple integrals between different coordinate systems?
3. How would you evaluate the integral in spherical coordinates instead?
4. What is the physical interpretation of the coordinate transformation?
5. How do you apply Jacobian determinants when converting between coordinate systems?

**Tip**: Always consider the geometric interpretation of limits in integrals when converting between coordinate systems, as it simplifies understanding the region of integration.

Learn how to convert the triple integral \int_{0}^{1} \int_{0}^{\pi/4} \int_{0}^{2\pi} {\rho^{3}\sin{(2\phi)} \,d\theta \,d\phi \,d\rho} from cylindrical to Cartesian coordinates. This problem covers coordinate transformations, including the conversion of limits and integrand expressions. Suitable for undergraduate math students.

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