To evaluate the cylindrical coordinate integral

$\int_0^{2\pi} \int_0^{\frac{\theta}{2\pi}} \int_0^{5 + 24r^2} r \, dz \, dr \, d\theta,$

we can proceed by integrating step-by-step with respect to each variable $z$, $r$, and $\theta$.

Step 1: Integrate with respect to $z$

The innermost integral is with respect to $z$, over the interval $z = 0$ to $z = 5 + 24r^2$. Since $r$ and $\theta$ are constants in this integral, we have:

$\int_0^{5 + 24r^2} r \, dz = r \cdot \left[ z \right]_0^{5 + 24r^2} = r \cdot (5 + 24r^2) = 5r + 24r^3.$

So the integral reduces to:

$\int_0^{2\pi} \int_0^{\frac{\theta}{2\pi}} (5r + 24r^3) \, dr \, d\theta.$

Step 2: Integrate with respect to $r$

Next, we integrate with respect to $r$, from $r = 0$ to $r = \frac{\theta}{2\pi}$:

$\int_0^{\frac{\theta}{2\pi}} (5r + 24r^3) \, dr.$

Breaking this into two integrals, we get:

$\int_0^{\frac{\theta}{2\pi}} 5r \, dr + \int_0^{\frac{\theta}{2\pi}} 24r^3 \, dr.$

First part:

$\int_0^{\frac{\theta}{2\pi}} 5r \, dr = 5 \cdot \frac{r^2}{2} \Big|_0^{\frac{\theta}{2\pi}} = 5 \cdot \frac{\left(\frac{\theta}{2\pi}\right)^2}{2} = \frac{5}{2} \cdot \frac{\theta^2}{4\pi^2} = \frac{5\theta^2}{8\pi^2}.$

Second part:

$\int_0^{\frac{\theta}{2\pi}} 24r^3 \, dr = 24 \cdot \frac{r^4}{4} \Big|_0^{\frac{\theta}{2\pi}} = 6 \cdot \left(\frac{\theta}{2\pi}\right)^4 = 6 \cdot \frac{\theta^4}{16\pi^4} = \frac{3\theta^4}{8\pi^4}.$

So after integrating with respect to $r$, we get:

$\int_0^{2\pi} \left(\frac{5\theta^2}{8\pi^2} + \frac{3\theta^4}{8\pi^4}\right) \, d\theta.$

Step 3: Integrate with respect to $\theta$

Now, we integrate each term with respect to $\theta$ from $0$ to $2\pi$:

$\int_0^{2\pi} \frac{5\theta^2}{8\pi^2} \, d\theta + \int_0^{2\pi} \frac{3\theta^4}{8\pi^4} \, d\theta.$

First part:

$\int_0^{2\pi} \frac{5\theta^2}{8\pi^2} \, d\theta = \frac{5}{8\pi^2} \int_0^{2\pi} \theta^2 \, d\theta.$

Using the formula for $\int \theta^2 \, d\theta = \frac{\theta^3}{3}$, we get:

$= \frac{5}{8\pi^2} \cdot \frac{(2\pi)^3}{3} = \frac{5}{8\pi^2} \cdot \frac{8\pi^3}{3} = \frac{5 \cdot 8\pi}{3} = \frac{40\pi}{3}.$

Second part:

$\int_0^{2\pi} \frac{3\theta^4}{8\pi^4} \, d\theta = \frac{3}{8\pi^4} \int_0^{2\pi} \theta^4 \, d\theta.$

Using the formula for $\int \theta^4 \, d\theta = \frac{\theta^5}{5}$, we get:

$= \frac{3}{8\pi^4} \cdot \frac{(2\pi)^5}{5} = \frac{3}{8\pi^4} \cdot \frac{32\pi^5}{5} = \frac{3 \cdot 32\pi}{5} = \frac{96\pi}{5}.$

Final Answer:

Combining both parts, we get:

$\frac{40\pi}{3} + \frac{96\pi}{5} = \frac{200\pi + 288\pi}{15} = \frac{488\pi}{15}.$

So, the value of the integral is:

$\frac{488\pi}{15}.$

Would you like more details on a specific step, or have any questions?

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1. How do cylindrical coordinates simplify volume integrals?
2. What are common applications of cylindrical integrals in physics?
3. How would this integral change if the upper limit of $z$ was a function of $\theta$ instead?
4. Can we use symmetry arguments in this integral for simplification?
5. How do cylindrical coordinates compare with spherical coordinates for integration?

Tip: When setting up cylindrical integrals, always ensure the limits match the geometry of the problem for an accurate result.