Evaluate the cylindrical coordinate integral ModifyingAbove ModifyingBelow Integral from nothing to nothing With 2 pi width 0 ModifyingAbove ModifyingBelow Integral from nothing to nothing With theta divided by 2 pi width 0 ModifyingAbove ModifyingBelow Integral from nothing to nothing With 5 plus 24 r squared width 0 r dz dr d theta
.

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?

---

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.

Evaluate the cylindrical coordinate integral ∫∫∫ r dz dr dθ with limits 0 to 2π for θ, 0 to θ/2π for r, and 0 to 5 + 24r² for z.

Learn how to evaluate a triple integral in cylindrical coordinates with variable limits, covering step-by-step integration over z, r, and θ. This detailed solution explains using cylindrical coordinates to simplify volume integrals, suitable for college-level calculus students.

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