We are tasked with converting the given double integral from Cartesian to polar coordinates. Let's go through the process systematically.

The original integral:

$\int_0^{\sqrt{4}} \int_{-x}^{x} dy \, dx$

Step 1: Conversion to polar coordinates

Recall that in polar coordinates:

• $x = r \cos(\theta)$
• $y = r \sin(\theta)$
• $dx\,dy = r \, dr\,d\theta$

We need to convert the limits of integration to polar coordinates. The given region is bounded by $0 \leq x \leq \sqrt{4}$ and $-x \leq y \leq x$.

This describes the region in the first quadrant, where:

• $r$ ranges from $0$ to $2$ (since $\sqrt{4} = 2$),
• $\theta$ ranges from $-\frac{\pi}{4}$ to $\frac{\pi}{4}$.

Step 2: Rewrite the limits

With polar coordinates, the bounds and integrand are adjusted as follows: $\int_0^{\sqrt{4}} \int_{-x}^{x} dy \, dx = \int_0^{2} \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} r \, d\theta \, dr$

Step 3: Perform the integration

1. Integrate with respect to $\theta$: $\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} r \, d\theta = r \left[ \theta \right]_{-\frac{\pi}{4}}^{\frac{\pi}{4}} = r \left( \frac{\pi}{4} - \left(-\frac{\pi}{4}\right) \right) = r \cdot \frac{\pi}{2}$

2. Integrate with respect to $r$: $\int_0^2 r \cdot \frac{\pi}{2} \, dr = \frac{\pi}{2} \int_0^2 r \, dr = \frac{\pi}{2} \left[ \frac{r^2}{2} \right]_0^2 = \frac{\pi}{2} \cdot \frac{4}{2} = \frac{\pi \cdot 2}{2} = \pi$

Thus, the value of the integral is $\pi$.

Filling in the blanks:

With this information, the corresponding values are:

• $a = 0$
• $b = 2$
• $c = -\frac{\pi}{4}$
• $d = \frac{\pi}{4}$

$\int_0^{\sqrt{4}} \int_{-x}^{x} dy \, dx = \int_0^2 \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} r \, d\theta \, dr = \pi$

Would you like more details on any part of this process?

Related Questions:

1. How do you generally convert Cartesian integrals to polar coordinates?
2. Why do we multiply by $r$ when changing to polar coordinates?
3. What happens if the region covers more than the first quadrant?
4. How does the Jacobian factor relate to the transformation from Cartesian to polar?
5. Can this integral be solved directly without converting to polar?

Tip:

Always sketch the region of integration when converting to polar coordinates to ensure proper limits.