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Multiple Choice
Find the area of the surface defined by over the region , .
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Verified step by step guidance
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Step 1: Recognize that the problem involves finding the surface area of a surface defined by z = 2√(x³ + y³) over the given region 0 ≤ x ≤ 1, 0 ≤ y ≤ 1. The formula for the surface area of a function z = f(x, y) is given by ∬_R √(1 + (∂z/∂x)² + (∂z/∂y)²) dA, where R is the region of integration.
Step 2: Compute the partial derivatives of z with respect to x and y. For z = 2√(x³ + y³), ∂z/∂x = (2 * 1/2) * (1/√(x³ + y³)) * 3x² = 3x² / √(x³ + y³), and similarly, ∂z/∂y = 3y² / √(x³ + y³).
Step 3: Substitute the partial derivatives into the surface area formula. The integrand becomes √(1 + (3x² / √(x³ + y³))² + (3y² / √(x³ + y³))²). Simplify the terms inside the square root to get √(1 + 9x⁴ / (x³ + y³) + 9y⁴ / (x³ + y³)).
Step 4: Combine the terms under the square root into a single fraction. The integrand simplifies to √((x³ + y³ + 9x⁴ + 9y⁴) / (x³ + y³)). This is the expression that needs to be integrated over the region 0 ≤ x ≤ 1, 0 ≤ y ≤ 1.
Step 5: Set up the double integral for the surface area. The surface area is given by ∬_R √((x³ + y³ + 9x⁴ + 9y⁴) / (x³ + y³)) dy dx, where the limits of integration are 0 to 1 for both x and y. Evaluate this integral to find the surface area.