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.