$$ \int_0^1\int_0^{\sqrt{1-y^2}}\sqrt{x^2+y^2}\differentialD\text{xdy} $$

The given double integral is:

\[
\int_0^1 \int_0^{\sqrt{1-y^2}} \sqrt{x^2 + y^2} \, dx \, dy
\]

This represents the integral of $\sqrt{x^2 + y^2}$ over a region bounded by the curve $x = 0$ to $x = \sqrt{1 - y^2}$, and $y$ ranging from 0 to 1. Let's evaluate this step by step.

### Step 1: Understand the Region of Integration

The region described by the limits is part of a quarter circle of radius 1 in the first quadrant.

### Step 2: Change to Polar Coordinates

To simplify the calculation, let's convert to polar coordinates:

- $x = r \cos \theta$
- $y = r \sin \theta$
- The Jacobian of the transformation is $r$.

The limits in polar coordinates are:

- $r$ ranges from 0 to 1.
- $\theta$ ranges from 0 to $\frac{\pi}{2}$.

The integrand $\sqrt{x^2 + y^2}$ in polar coordinates simplifies to $r$.

### Step 3: Rewrite the Integral in Polar Coordinates

\[
\int_0^{\frac{\pi}{2}} \int_0^1 r \cdot r \, dr \, d\theta = \int_0^{\frac{\pi}{2}} \int_0^1 r^2 \, dr \, d\theta
\]

### Step 4: Evaluate the Inner Integral with respect to $r$

\[
\int_0^1 r^2 \, dr = \left[\frac{r^3}{3}\right]_0^1 = \frac{1}{3}
\]

### Step 5: Evaluate the Outer Integral with respect to $\theta$

\[
\int_0^{\frac{\pi}{2}} \frac{1}{3} \, d\theta = \frac{1}{3} \left[\theta\right]_0^{\frac{\pi}{2}} = \frac{1}{3} \cdot \frac{\pi}{2} = \frac{\pi}{6}
\]

Thus, the value of the integral is:

\[
\boxed{\frac{\pi}{6}}
\]

Would you like more details on any of these steps or have any other questions?

### Related Questions:
1. How do you determine the region of integration for double integrals?
2. What are the steps to convert Cartesian coordinates to polar coordinates in integration?
3. How can you evaluate integrals involving $\sqrt{x^2 + y^2}$ without polar coordinates?
4. What are the benefits of using polar coordinates in double integration?
5. How does the Jacobian affect the transformation of integrals?

**Tip:** For regions with circular symmetry, converting to polar coordinates can often simplify the integration process significantly!

Learn how to evaluate the double integral of sqrt(x^2 + y^2) over a specified region using Cartesian and polar coordinates. Step-by-step solution provided with Cartesian to polar coordinate transformation, integral evaluation, and detailed calculation. Suitable for advanced undergraduate students and those studying calculus.

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