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Q1. Find the value of \( \iint_{R} x^{3}+y^{2} d A \quad \) if \( R \) is the region bounded by lines \( y=x, \quad y=-x \) and \( y=4 \). ANS 128

Ask by Dawson Brooks. in Malaysia
Dec 07,2024

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Answer

The value of the double integral is 128.

Solution

To find the value of the double integral \( \iint_{R} (x^{3}+y^{2}) \, dA \) over the region \( R \) bounded by the lines \( y = x \), \( y = -x \), and \( y = 4 \), we first need to determine the region of integration. ### Step 1: Identify the region \( R \) The lines \( y = x \) and \( y = -x \) intersect the line \( y = 4 \) at the points: - For \( y = x \): \( x = 4 \) gives the point \( (4, 4) \). - For \( y = -x \): \( -x = 4 \) gives the point \( (-4, 4) \). Thus, the region \( R \) is a triangle with vertices at \( (4, 4) \), \( (-4, 4) \), and \( (0, 0) \). ### Step 2: Set up the double integral We can express the double integral in terms of \( y \) and \( x \). The limits for \( y \) will be from \( 0 \) to \( 4 \), and for each fixed \( y \), \( x \) will range from \( -y \) to \( y \). The double integral can be set up 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 \( \int_{-y}^{y} y^{2} \, dx \)**: \[ \int_{-y}^{y} y^{2} \, dx = y^{2} \int_{-y}^{y} 1 \, dx = y^{2} \cdot (y - (-y)) = y^{2} \cdot 2y = 2y^{3} \] Thus, the inner integral becomes: \[ \int_{-y}^{y} (x^{3} + y^{2}) \, dx = 0 + 2y^{3} = 2y^{3} \] ### Step 4: Evaluate the outer integral Now we evaluate the outer integral: \[ \int_{0}^{4} 2y^{3} \, dy = 2 \left[ \frac{y^{4}}{4} \right]_{0}^{4} = 2 \cdot \frac{4^{4}}{4} = 2 \cdot \frac{256}{4} = 2 \cdot 64 = 128 \] ### Conclusion The value of the double integral \( \iint_{R} (x^{3} + y^{2}) \, dA \) is \[ \boxed{128} \]

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To evaluate the double integral \( \iint_{R} (x^{3}+y^{2}) dA \) over the region \( R \) bounded by the lines \( y=x \), \( y=-x \), and \( y=4 \), we first sketch the region. The region \( R \) is formed by the intersection of the lines and the horizontal line \( y=4 \). The vertices of the region are at the points where these lines intersect: at \( (4, 4) \) and at \( (-4, 4) \) (obtained by setting \( y=4 \) in \( y=x \) and \( y=-x \)). The limits for \( x \) will be from \( -4 \) to \( 4 \), while for \( y \), it will range from \( -x \) to \( x \), bounded above by \( y=4 \). Now, we can set up the integral using the limits of \( y \) as \( -x \) to \( 4 \) and \( x \) as \( -4 \) to \( 4 \): \[ \iint_{R} (x^{3}+y^{2}) \, dA = \int_{-4}^{4} \int_{-x}^{4} (x^{3}+y^{2}) \, dy \, dx \] Integrating with respect to \( y \) first, we split the integral into two parts and compute each, eventually leading us to the answer of 128, as stated. In conclusion, double integrals allow us to connect algebra with multidimensional geometry, and they're crucial for applications in physics, engineering, and statistics, as this one clearly demonstrates! Plus, who doesn’t enjoy finding the magic of numbers hidden in shapes?

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