This integral involves triple integration in spherical coordinates. Let’s carefully evaluate it.

Given integral:

$\int_0^6 \int_{-2\pi}^0 \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \rho^3 \sin(2\varphi) \, d\varphi \, d\theta \, d\rho.$

Step 1: Analyze the integrand and order of integration

• $\rho$: This represents the radial coordinate, and it ranges from $0$ to $6$.
• $\theta$: This is the azimuthal angle, and it ranges from $-2\pi$ to $0$.
• $\varphi$: This is the polar angle (or colatitude), and it ranges from $\frac{\pi}{4}$ to $\frac{\pi}{2}$.
• The integrand is $\rho^3 \sin(2\varphi)$.

We’ll integrate in the following order: first $d\varphi$, then $d\theta$, and finally $d\rho$.

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Step 2: Integrate with respect to $\varphi$

We need to evaluate:

$\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \sin(2\varphi) \, d\varphi.$

The antiderivative of $\sin(2\varphi)$ is:

$-\frac{1}{2} \cos(2\varphi).$

Evaluating this at the limits:

[ -\frac{1}{2} \left[ \cos\left(2 \cdot \frac{\pi}{2}\right) - \cos\left(2 \cdot \frac{\pi}{4}\right) \right] = -\frac{1}{2} \left[ \cos(\pi) - \cos\left(\frac{\pi}{2}\right) \right]. ]

Using $\cos(\pi) = -1$ and $\cos(\pi/2) = 0$, this becomes:

$-\frac{1}{2} \left[ -1 - 0 \right] = \frac{1}{2}.$

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Step 3: Integrate with respect to $\theta$

Now we need to evaluate:

$\int_{-2\pi}^0 \frac{1}{2} \, d\theta.$

This simplifies to:

$\frac{1}{2} \left[ \theta \right]_{-2\pi}^0 = \frac{1}{2} \left( 0 - (-2\pi) \right) = \frac{1}{2} (2\pi) = \pi.$

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Step 4: Integrate with respect to $\rho$

Now we integrate:

$\int_0^6 \rho^3 \pi \, d\rho.$

Factor out the $\pi$:

$\pi \int_0^6 \rho^3 \, d\rho.$

The antiderivative of $\rho^3$ is:

$\frac{\rho^4}{4}.$

Evaluating at the limits:

$\frac{1}{4} \left[ 6^4 - 0^4 \right] = \frac{1}{4} (1296) = 324.$

Thus:

$\pi \cdot 324 = 324\pi.$

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Step 5: Final Answer

The value of the integral is:

$324\pi.$

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Would you like more details on any of these steps?

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Here are 5 related questions you might find interesting:

1. How are spherical coordinates defined and converted to Cartesian coordinates?
2. How do you determine the limits of integration in triple integrals?
3. What are some common uses of spherical coordinates in physics?
4. How does the integration of trigonometric functions differ from polynomial functions?
5. What are alternative methods for evaluating triple integrals?

Tip: When working with integrals involving trigonometric functions, keep symmetry in mind—it can often simplify calculations!