To evaluate the double integral

$\iint_R \frac{x y^3}{x^2 + 1} \, dA$

over the region $R$ defined by $0 \leq y \leq 6$ and $0 \leq x \leq 1$, we can set up the integral with the given limits.

Step-by-Step Solution

1. Set up the integral in terms of $x$ and $y$:

   Since the region $R$ is a rectangular region defined by $0 \leq y \leq 6$ and $0 \leq x \leq 1$, the integral can be written as: $\int_0^1 \int_0^6 \frac{x y^3}{x^2 + 1} \, dy \, dx$

2. Integrate with respect to $y$:

   The integrand is $\frac{x y^3}{x^2 + 1}$, where $x$ is treated as a constant when integrating with respect to $y$. Thus, $\int_0^6 \frac{x y^3}{x^2 + 1} \, dy = \frac{x}{x^2 + 1} \int_0^6 y^3 \, dy$

3. Evaluate the inner integral $\int_0^6 y^3 \, dy$:

   To integrate $y^3$ with respect to $y$, we find: $\int y^3 \, dy = \frac{y^4}{4}$ Applying the limits from 0 to 6: $\int_0^6 y^3 \, dy = \left[ \frac{y^4}{4} \right]_0^6 = \frac{6^4}{4} = \frac{1296}{4} = 324$

   Thus, the inner integral becomes: $\int_0^6 \frac{x y^3}{x^2 + 1} \, dy = \frac{x \cdot 324}{x^2 + 1} = \frac{324x}{x^2 + 1}$

4. Integrate with respect to $x$:

   Now, we evaluate the outer integral: $\int_0^1 \frac{324x}{x^2 + 1} \, dx$ To solve this integral, we use a substitution. Let $u = x^2 + 1$, then $du = 2x \, dx$ or $\frac{du}{2} = x \, dx$.

   Substituting into the integral: $\int \frac{324x}{x^2 + 1} \, dx = \int \frac{324}{2} \frac{1}{u} \, du = 162 \int \frac{1}{u} \, du$

   This integral is straightforward: $162 \ln |u| + C = 162 \ln |x^2 + 1| + C$

   Applying the limits from 0 to 1: $\int_0^1 \frac{324x}{x^2 + 1} \, dx = 162 \left( \ln(1^2 + 1) - \ln(0^2 + 1) \right)$ $= 162 (\ln 2 - \ln 1) = 162 \ln 2$

Final Answer

The value of the double integral is: $162 \ln 2$

Would you like further details on this solution or have any questions?

Here are some additional related questions for exploration:

1. What is the general approach to evaluating double integrals over rectangular regions?
2. How does changing the limits of integration affect the evaluation of double integrals?
3. Why is substitution useful in solving integrals like $\int \frac{x}{x^2 + 1} \, dx$?
4. How would the solution differ if the region $R$ were different?
5. What are some applications of double integrals in physics or engineering?

Tip: In double integrals, always integrate with respect to the inner variable first, keeping the outer variable constant.