To evaluate the given iterated integral by converting to polar coordinates, let's go through each step carefully.

The integral is:

$\int_{0}^{2} \int_{0}^{\sqrt{2x - x^2}} \frac{1}{6 \sqrt{x^2 + y^2}} \, dy \, dx$

Step 1: Set up Polar Coordinates

In polar coordinates:

• $x = r \cos \theta$
• $y = r \sin \theta$
• $x^2 + y^2 = r^2$
• The area element $dx \, dy$ becomes $r \, dr \, d\theta$.

Step 2: Convert the Integrand

The integrand $\frac{1}{6 \sqrt{x^2 + y^2}}$ can be rewritten in polar coordinates as: $\frac{1}{6 \sqrt{r^2}} = \frac{1}{6r}$

Step 3: Determine the Bounds in Polar Coordinates

• The region in Cartesian coordinates is defined by $0 \leq x \leq 2$ and $0 \leq y \leq \sqrt{2x - x^2}$.
• The curve $y = \sqrt{2x - x^2}$ represents a semicircle with radius $\sqrt{2}$, centered at $x = 1$ on the $x$-axis.

In polar coordinates, this region corresponds to:

• $r$ ranging from $0$ to $\sqrt{2}$,
• $\theta$ ranging from $0$ to $\pi$.

Step 4: Set Up the Integral in Polar Coordinates

Substitute these bounds and the integrand into the integral:

$\int_{0}^{\pi} \int_{0}^{\sqrt{2}} \frac{1}{6r} \cdot r \, dr \, d\theta$

Simplify the integrand: $= \int_{0}^{\pi} \int_{0}^{\sqrt{2}} \frac{1}{6} \, dr \, d\theta$

Step 5: Evaluate the Integral

First, integrate with respect to $r$:

$\int_{0}^{\pi} \int_{0}^{\sqrt{2}} \frac{1}{6} \, dr \, d\theta = \int_{0}^{\pi} \frac{1}{6} \cdot \left[ r \right]_{0}^{\sqrt{2}} \, d\theta$

$= \int_{0}^{\pi} \frac{1}{6} \cdot \sqrt{2} \, d\theta = \frac{\sqrt{2}}{6} \int_{0}^{\pi} d\theta$

Now, integrate with respect to $\theta$:

$= \frac{\sqrt{2}}{6} \cdot \left[ \theta \right]_{0}^{\pi} = \frac{\sqrt{2}}{6} \cdot \pi = \frac{\pi \sqrt{2}}{6}$

Final Answer

$\int_{0}^{2} \int_{0}^{\sqrt{2x - x^2}} \frac{1}{6 \sqrt{x^2 + y^2}} \, dy \, dx = \frac{\pi \sqrt{2}}{6}$

Would you like further details on any step, or do you have additional questions?

Here are some related questions you might explore:

1. What is the process of converting Cartesian integrals to polar coordinates?
2. How does the Jacobian affect integration in polar coordinates?
3. Can this method be extended to cylindrical or spherical coordinates?
4. What would change if the bounds were different?
5. How is integration over a semicircular region handled in other coordinate systems?

Tip: Always confirm the region of integration in polar coordinates to ensure correct bounds.