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Let C be the positively oriented square with vertices ( 0, 0 ), ( 3, 0 ), ( 3, 3 ), ( 0, 3 ). Use Green's Theorem to evaluate the line integral.

∫(y²x dx + 6x²y dy)

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User Idania
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1 Answer

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Final Answer:

The line integral ∫(y²x dx + 6x²y dy) over the positively oriented square C with vertices (0, 0), (3, 0), (3, 3), and (0, 3) using Green's Theorem is 54.

Step-by-step explanation:

Green's Theorem states that for a positively oriented region R with piecewise smooth boundary C, if P(x, y) and Q(x, y) have continuous partial derivatives on an open region containing R, then the line integral of P and Q along C is equal to the double integral of (∂Q/∂x - ∂P/∂y) over the region R. In this case, P(x, y) = y²x and Q(x, y) = 6x²y.

Applying Green's Theorem, we calculate the double integral of (6x² - 2xy) over the region R, where 0 ≤ x ≤ 3 and 0 ≤ y ≤ 3. The result is ∫[0 to 3] ∫[0 to 3] (6x² - 2xy) dy dx, which evaluates to 54.

Using Green's Theorem in this context allows us to transform a line integral into a double integral over the region R. It provides a powerful tool for simplifying certain types of integrals and relates line integrals to the flux across a region. In this example, the line integral is efficiently evaluated by calculating the double integral over the specified square region, resulting in the final answer of 54.

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User Boby
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