The question asks to rewrite the given double integral in polar coordinates:

$\int_{-2}^{2} \int_{0}^{\sqrt{4-y^2}} 3\sqrt{x^2 + y^2} \, dx \, dy$

Step 1: Convert to Polar Coordinates

• The integrand $\sqrt{x^2 + y^2}$ becomes $r$ in polar coordinates, because $r = \sqrt{x^2 + y^2}$.
• The differential element $dx \, dy$ converts to $r \, dr \, d\theta$.
• The limits of integration for $x$ and $y$ define a region that is a semicircle with radius 2, centered at the origin.

Step 2: Identify the Region of Integration

• The bounds for $y$ are from -2 to 2, and for each value of $y$, $x$ ranges from 0 to $\sqrt{4 - y^2}$, describing the right half of the circle $x^2 + y^2 = 4$ in Cartesian coordinates.
• In polar coordinates, this is the region where $r$ goes from 0 to 2, and $\theta$ goes from $0$ to $2\pi$ for the entire circle, but since we only want the right half, the bounds for $\theta$ are $0 \leq \theta \leq \pi$.

Step 3: Rewriting the Integral

Thus, the integral becomes:

$\int_{0}^{\pi} \int_{0}^{2} 3r^2 \, dr \, d\theta$

Step 4: Match with the Options

Looking at the options provided:

• Option (d) matches this form.

Therefore, the correct answer is (d).

Would you like further details or have any questions?

Here are some related questions to explore:

1. How do we convert rectangular coordinates to polar coordinates in general?
2. What does the Jacobian determinant represent in coordinate transformations?
3. Why does the integrand $\sqrt{x^2 + y^2}$ simplify to $r$ in polar coordinates?
4. Can you describe the limits of integration for a different region, such as a quarter-circle?
5. How would this problem change if the integrand were different, such as $1$ or $r^3$?

Tip: When converting to polar coordinates, always remember to multiply the integrand by the factor $r$, which comes from the Jacobian of the transformation.