You are looking to evaluate the following double integral:

$\int_{0}^{1} \int_{\sqrt{x}}^{x} (x^2 + y^2) \, dy \, dx$

Step-by-step solution:

1. Inner integral: First, evaluate the inner integral with respect to $y$, keeping $x$ constant.

$\int_{\sqrt{x}}^{x} (x^2 + y^2) \, dy = \int_{\sqrt{x}}^{x} x^2 \, dy + \int_{\sqrt{x}}^{x} y^2 \, dy$

2. First term: The first term is $\int_{\sqrt{x}}^{x} x^2 \, dy$. Since $x^2$ is independent of $y$, this becomes:

$x^2 \int_{\sqrt{x}}^{x} dy = x^2 \left( x - \sqrt{x} \right) = x^2 (x - \sqrt{x})$

3. Second term: Now evaluate $\int_{\sqrt{x}}^{x} y^2 \, dy$. The antiderivative of $y^2$ is $\frac{y^3}{3}$, so:

$\int_{\sqrt{x}}^{x} y^2 \, dy = \left[ \frac{y^3}{3} \right]_{\sqrt{x}}^{x} = \frac{x^3}{3} - \frac{(\sqrt{x})^3}{3} = \frac{x^3}{3} - \frac{x^{3/2}}{3}$

4. Combine results: Now combine the two terms:

$x^2 (x - \sqrt{x}) + \left( \frac{x^3}{3} - \frac{x^{3/2}}{3} \right)$ $= x^3 - x^{5/2} + \frac{x^3}{3} - \frac{x^{3/2}}{3}$

5. Simplify: Combine the like terms:

$\left( x^3 + \frac{x^3}{3} \right) - \left( x^{5/2} + \frac{x^{3/2}}{3} \right)$ $= \frac{4x^3}{3} - \left( x^{5/2} + \frac{x^{3/2}}{3} \right)$

6. Outer integral: Now evaluate the outer integral with respect to $x$:

$\int_{0}^{1} \left( \frac{4x^3}{3} - x^{5/2} - \frac{x^{3/2}}{3} \right) dx$

This can be broken down into three integrals:

$\frac{4}{3} \int_0^1 x^3 \, dx - \int_0^1 x^{5/2} \, dx - \frac{1}{3} \int_0^1 x^{3/2} \, dx$

7. Evaluate each integral:

$\frac{4}{3} \int_0^1 x^3 \, dx = \frac{4}{3} \cdot \frac{x^4}{4} \Bigg|_0^1 = \frac{4}{3} \cdot \frac{1}{4} = \frac{1}{3}$

$\int_0^1 x^{5/2} \, dx = \frac{x^{7/2}}{\frac{7}{2}} \Bigg|_0^1 = \frac{2}{7}$

$\frac{1}{3} \int_0^1 x^{3/2} \, dx = \frac{1}{3} \cdot \frac{x^{5/2}}{\frac{5}{2}} \Bigg|_0^1 = \frac{1}{3} \cdot \frac{2}{5} = \frac{2}{15}$

8. Final answer: Combine the results:

$\frac{1}{3} - \frac{2}{7} - \frac{2}{15}$

The common denominator for 3, 7, and 15 is 105, so rewrite each fraction:

$\frac{1}{3} = \frac{35}{105}, \quad \frac{2}{7} = \frac{30}{105}, \quad \frac{2}{15} = \frac{14}{105}$

Now subtract:

$\frac{35}{105} - \frac{30}{105} - \frac{14}{105} = \frac{-9}{105} = -\frac{3}{35}$

Thus, the value of the double integral is:

$-\frac{3}{35}$

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Would you like more details or have any questions?

Here are 5 related questions for further exploration:

1. What is the interpretation of this double integral in terms of area or volume?
2. How do different limits of integration affect the result of a double integral?
3. What methods can be used to simplify the evaluation of more complex double integrals?
4. How do Fubini's Theorem and the order of integration apply to this problem?
5. What are the potential applications of double integrals in physics or engineering?

Tip: When handling complex integrals, breaking the problem into smaller, manageable parts often simplifies the overall calculation process.