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Multiple Choice
Evaluate the line integral , where is the curve given by , , for .
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Verified step by step guidance
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Step 1: Understand the problem. You are tasked with evaluating the line integral ∫_C y^3 ds, where C is a curve parameterized by x = t^3 and y = t, for 0 ≤ t ≤ 3. The integral involves the arc length differential ds, which depends on the parameterization of the curve.
Step 2: Recall the formula for ds in terms of the parameter t. The arc length differential is given by ds = √((dx/dt)^2 + (dy/dt)^2) dt. Compute dx/dt and dy/dt using the parameterization x = t^3 and y = t.
Step 3: Compute dx/dt and dy/dt. For x = t^3, dx/dt = 3t^2. For y = t, dy/dt = 1. Substitute these into the formula for ds: ds = √((3t^2)^2 + (1)^2) dt = √(9t^4 + 1) dt.
Step 4: Substitute the parameterization of y and ds into the integral. Since y = t, y^3 = t^3. The integral becomes ∫_C y^3 ds = ∫_0^3 t^3 √(9t^4 + 1) dt.
Step 5: Set up the integral for evaluation. The integral is now ∫_0^3 t^3 √(9t^4 + 1) dt. This integral can be solved using numerical methods or advanced techniques, but the setup is complete.