To evaluate the integral

$\int_{-6}^{1} \int_{0}^{3} (2x^2 + y^2) \, dx \, dy,$

we will first evaluate the inner integral with respect to $x$, then the outer integral with respect to $y$.

Step 1: Inner Integral

The inner integral is

$\int_{0}^{3} (2x^2 + y^2) \, dx.$

We can split this into two separate integrals:

$\int_{0}^{3} 2x^2 \, dx + \int_{0}^{3} y^2 \, dx.$

1. Calculate $\int_{0}^{3} 2x^2 \, dx$:

$\int 2x^2 \, dx = \frac{2}{3}x^3 + C.$

Evaluating from 0 to 3:

[ \left[ \frac{2}{3}(3^3) \right] - \left[ \frac{2}{3}(0^3) \right] = \frac{2}{3}(27) = 18. ]

2. Calculate $\int_{0}^{3} y^2 \, dx$:

Since $y^2$ is treated as a constant with respect to $x$:

$\int y^2 \, dx = y^2 x \bigg|_{0}^{3} = y^2 (3 - 0) = 3y^2.$

Combining these results, the inner integral becomes:

$\int_{0}^{3} (2x^2 + y^2) \, dx = 18 + 3y^2.$

Step 2: Outer Integral

Now we evaluate the outer integral:

$\int_{-6}^{1} (18 + 3y^2) \, dy.$

This can also be split into two integrals:

$\int_{-6}^{1} 18 \, dy + \int_{-6}^{1} 3y^2 \, dy.$

1. Calculate $\int_{-6}^{1} 18 \, dy$:

$\int 18 \, dy = 18y \bigg|_{-6}^{1} = 18(1) - 18(-6) = 18 + 108 = 126.$

2. Calculate $\int_{-6}^{1} 3y^2 \, dy$:

$\int 3y^2 \, dy = y^3 \bigg|_{-6}^{1} = (1^3) - ((-6)^3) = 1 - (-216) = 1 + 216 = 217.$

Combining both results, we have:

$\int_{-6}^{1} (18 + 3y^2) \, dy = 126 + 217 = 343.$

Final Result

The value of the integral

$\int_{-6}^{1} \int_{0}^{3} (2x^2 + y^2) \, dx \, dy = 343.$

If you have any questions or need further details, let me know! Here are some related questions you might consider:

1. How would the result change if the limits of integration were different?
2. What would happen if we integrated a different function instead of $2x^2 + y^2$?
3. How can we visualize the area represented by this double integral?
4. What are other methods for evaluating double integrals?
5. Can this integral be solved using polar coordinates, and if so, how?

Tip: When working with double integrals, always consider splitting the integral into simpler parts for easier calculation.