int 1 ^ 3 int 1/x ^ 1 int 0 ^ sqrt xy xyz dzdydx

Your integral expression involves multiple nested integrals over different variables. Let me break it down step by step:

\[
\int_1^3 \int_1^{\frac{1}{x}} \int_0^{\sqrt{xy}} xyz \, dz \, dy \, dx
\]

This represents a triple integral where:
- \( x \) ranges from 1 to 3,
- \( y \) ranges from 1 to \( \frac{1}{x} \),
- \( z \) ranges from 0 to \( \sqrt{xy} \),
- The integrand is \( xyz \).

### Step-by-step solution:

1. **Inner Integral (with respect to \( z \)):**
   \[
   \int_0^{\sqrt{xy}} xyz \, dz = xy \int_0^{\sqrt{xy}} z \, dz = xy \left[ \frac{z^2}{2} \right]_0^{\sqrt{xy}} = xy \cdot \frac{xy}{2} = \frac{(xy)^2}{2}
   \]

2. **Second Integral (with respect to \( y \)):**
   \[
   \int_1^{\frac{1}{x}} \frac{(xy)^2}{2} \, dy = \frac{x^2}{2} \int_1^{\frac{1}{x}} y^2 \, dy = \frac{x^2}{2} \left[ \frac{y^3}{3} \right]_1^{\frac{1}{x}}
   \]
   Now, compute the limits:
   \[
   \frac{x^2}{2} \left( \frac{\left( \frac{1}{x} \right)^3}{3} - \frac{1^3}{3} \right) = \frac{x^2}{2} \left( \frac{1}{3x^3} - \frac{1}{3} \right) = \frac{x^2}{6} \left( \frac{1 - x^3}{x^3} \right) = \frac{x^2(1 - x^3)}{6x^3} = \frac{x^2 - x^5}{6x^3} = \frac{1}{6} \left( \frac{1}{x} - x^2 \right)
   \]

3. **Outer Integral (with respect to \( x \)):**
   \[
   \int_1^3 \frac{1}{6} \left( \frac{1}{x} - x^2 \right) dx = \frac{1}{6} \left( \int_1^3 \frac{1}{x} \, dx - \int_1^3 x^2 \, dx \right)
   \]
   Compute the integrals:
   \[
   \int_1^3 \frac{1}{x} \, dx = \ln(3)
   \]
   \[
   \int_1^3 x^2 \, dx = \left[ \frac{x^3}{3} \right]_1^3 = \frac{27}{3} - \frac{1}{3} = 9 - \frac{1}{3} = \frac{26}{3}
   \]
   Therefore:
   \[
   \frac{1}{6} \left( \ln(3) - \frac{26}{3} \right) = \frac{\ln(3)}{6} - \frac{13}{9}
   \]

Thus, the final result is:

\[
\boxed{\frac{\ln(3)}{6} - \frac{13}{9}}
\]

Let me know if you'd like more details or clarifications.

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Here are 5 questions that could help extend your understanding:
1. How would the integral change if the limits of integration were different for \( y \)?
2. Can you generalize this result for different powers of \( x \), \( y \), or \( z \)?
3. How would you interpret this integral geometrically in 3D space?
4. What if the integrand was a different function, like \( e^{xyz} \)?
5. How does the logarithmic term arise from the integral of \( \frac{1}{x} \)?

**Tip:** Always verify the bounds of your integrals when solving multi-variable calculus problems, as they heavily affect the outcome.

This page provides a detailed step-by-step solution for evaluating a triple integral with variable limits: \( \int_1^3 \int_1^{\frac{1}{x}} \int_0^{\sqrt{xy}} xyz \, dz \, dy \, dx \). It explores multivariable calculus concepts, definite integrals, and their applications. Suitable for undergraduate students studying multivariable calculus.

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