The integral provided is:

$\int_{0}^{\pi} \int_{0}^{\frac{\theta}{2\pi}} \int_{0}^{6+29r^2} r \, dz \, dr \, d\theta$

To evaluate this integral in cylindrical coordinates, we proceed by evaluating each part step-by-step.

Step 1: Integrate with respect to $z$

The inner integral with respect to $z$ is:

$\int_{0}^{6+29r^2} r \, dz$

Since $r$ is independent of $z$, this simplifies to:

$= r \cdot z \Big|_{0}^{6+29r^2} = r(6 + 29r^2) = 6r + 29r^3$

Step 2: Substitute into the remaining integrals

Now we substitute $6r + 29r^3$ for the $z$-integral:

$\int_{0}^{\pi} \int_{0}^{\frac{\theta}{2\pi}} (6r + 29r^3) \, dr \, d\theta$

Step 3: Integrate with respect to $r$

Next, we evaluate the integral with respect to $r$:

$\int_{0}^{\frac{\theta}{2\pi}} (6r + 29r^3) \, dr$

Breaking this into two separate integrals:

$= \int_{0}^{\frac{\theta}{2\pi}} 6r \, dr + \int_{0}^{\frac{\theta}{2\pi}} 29r^3 \, dr$

Evaluating each part:

1. For $\int 6r \, dr$: $= 6 \cdot \frac{r^2}{2} \Big|_{0}^{\frac{\theta}{2\pi}} = 3r^2 \Big|_{0}^{\frac{\theta}{2\pi}} = 3 \left(\frac{\theta}{2\pi}\right)^2 = \frac{3\theta^2}{4\pi^2}$

2. For $\int 29r^3 \, dr$: $= 29 \cdot \frac{r^4}{4} \Big|_{0}^{\frac{\theta}{2\pi}} = \frac{29}{4} \left(\frac{\theta}{2\pi}\right)^4 = \frac{29\theta^4}{64\pi^4}$

Adding these results together:

$= \frac{3\theta^2}{4\pi^2} + \frac{29\theta^4}{64\pi^4}$

Step 4: Integrate with respect to $\theta$

Now, we evaluate the remaining integral with respect to $\theta$:

$\int_{0}^{\pi} \left( \frac{3\theta^2}{4\pi^2} + \frac{29\theta^4}{64\pi^4} \right) d\theta$

Breaking this into two parts:

1. For $\int \frac{3\theta^2}{4\pi^2} \, d\theta$: $= \frac{3}{4\pi^2} \cdot \frac{\theta^3}{3} \Big|_{0}^{\pi} = \frac{3}{4\pi^2} \cdot \frac{\pi^3}{3} = \frac{3\pi}{4}$

2. For $\int \frac{29\theta^4}{64\pi^4} \, d\theta$: $= \frac{29}{64\pi^4} \cdot \frac{\theta^5}{5} \Big|_{0}^{\pi} = \frac{29}{64\pi^4} \cdot \frac{\pi^5}{5} = \frac{29\pi}{64} \cdot \frac{\pi}{5} = \frac{29\pi^5}{320\pi^4} = \frac{29\pi}{320}$

Adding these two results together:

$= \frac{3\pi}{4} + \frac{29\pi}{320} = \frac{240\pi + 29\pi}{320} = \frac{269\pi}{320}$

Final Answer

$\int_{0}^{\pi} \int_{0}^{\frac{\theta}{2\pi}} \int_{0}^{6+29r^2} r \, dz \, dr \, d\theta = \frac{269\pi}{320}$

Would you like further steps explained or additional assistance?

Here are 5 related questions that expand on this:

1. How does converting to cylindrical coordinates help in evaluating triple integrals?
2. What are common applications of cylindrical coordinate systems?
3. How do boundary conditions change when using spherical coordinates instead of cylindrical?
4. How do we apply cylindrical coordinates for volume calculations?
5. What are other ways to simplify integrals involving powers of $r$ and $\theta$?

Tip: Always verify the order of integration limits and boundaries for accurate computation in triple integrals.