QUIZ 2 Sample ANSWER ALL QUESTIONS Q1. FSCE 1993 (ENG MATH 2) and $$ y=4 $$. ANS 128 of $$ \iint_{R} x^{3}+y^{2} d A \quad $$ if $$ R $$ is the region bounded by lines $$ y=x, y=-x $$

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To evaluate the double integral

$$
\iint_{R} (x^{3} + y^{2}) \, dA
$$

where $$ R $$ is the region bounded by the lines $$ y = x $$ and $$ y = -x $$ and the line $$ y = 4 $$, we first need to determine the region $$ R $$.

### Step 1: Identify the region $$ R $$

The lines $$ y = x $$ and $$ y = -x $$ intersect the line $$ y = 4 $$ at the points:

1. For $$ y = x $$: $$ x = 4 $$ gives the point $$ (4, 4) $$.
2. For $$ y = -x $$: $$ -x = 4 $$ gives the point $$ (-4, 4) $$.

Thus, the region $$ R $$ is bounded by the lines $$ y = x $$, $$ y = -x $$, and the horizontal line $$ y = 4 $$. The vertices of the region are $$ (4, 4) $$, $$ (-4, 4) $$, $$ (0, 0) $$.

### Step 2: Set up the double integral

The region $$ R $$ can be described in terms of $$ y $$ from $$ 0 $$ to $$ 4 $$ and for each fixed $$ y $$, $$ x $$ ranges from $$ -y $$ to $$ y $$. Therefore, we can set up the double integral as follows:

$$
\iint_{R} (x^{3} + y^{2}) \, dA = \int_{0}^{4} \int_{-y}^{y} (x^{3} + y^{2}) \, dx \, dy
$$

### Step 3: Evaluate the inner integral

We first evaluate the inner integral:

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

1. **Evaluate $$ \int_{-y}^{y} x^{3} \, dx $$**:

$$
\int_{-y}^{y} x^{3} \, dx = \left[ \frac{x^{4}}{4} \right]_{-y}^{y} = \frac{y^{4}}{4} - \frac{(-y)^{4}}{4} = \frac{y^{4}}{4} - \frac{y^{4}}{4} = 0
$$

2. **Evaluate $$ \int_{-y}^{y} y^{2} \, dx $$**:

$$
\int_{-y}^{y} y^{2} \, dx = y^{2} \int_{-y}^{y} 1 \, dx = y^{2} [x]_{-y}^{y} = y^{2} (y - (-y)) = y^{2} (2y) = 2y^{3}
$$

Combining these results, we have:

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

### Step 4: Evaluate the outer integral

Now we evaluate the outer integral:

$$
\int_{0}^{4} 2y^{3} \, dy = 2 \left[ \frac{y^{4}}{4} \right]_{0}^{4} = 2 \left( \frac{4^{4}}{4} - 0 \right) = 2 \left( \frac{256}{4} \right) = 2 \cdot 64 = 128
$$

### Final Answer

Thus, the value of the double integral is

$$
\boxed{128}
$$
The value of the double integral is 128.

QUIZ 2 Sample ANSWER ALL QUESTIONS Q1. FSCE 1993 (ENG MATH 2) and \( y=4 \). ANS 128 of \( \iint_{R} x^{3}+y^{2} d A \quad \) if \( R \) is the region bounded by lines \( y=x, y=-x \)

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