Integral of F on Curve C: Evaluate

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The integral of the vector field F = (2xy^4)i + (2x^2y^3)j over the specified curve C was evaluated using a double integral approach. The user initially calculated the integral to be 1/10 but later revised it to -1/10 after multiple attempts. There is confusion regarding the double integral setup and the evaluation process. The user seeks confirmation on the final result before submission, indicating a need for clarity in the calculations. Ultimately, the discussion highlights the challenges of evaluating line integrals and the importance of accuracy in mathematical submissions.
Tom McCurdy
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The problem:

Evaluate the integral
\oint_{c} Fdr for F = (2xy^4)i+(2x^2y^3)j on the curve C consisting of the x-axis from x=0 to x=1, the parabola y=1-x^2 up to the y-axis, and the y-axis down to the origin

Here is what I triedF(x,y)=<2xy^4,2x^2y^3>\int\int_{D} =[(2x^2y^3)*\frac{\partial}{\partial x}-(2xy^4)\frac{\partial}{\partial y}] dA= \int_{0}^{1} \int_{1-x^2}^{0} [4xy^3-8xy^3] dy dx=1/10
 
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Ok I have redone this problem... multiple times... and now I am getting

-1/10

can someone confirm this for me, I am out of tries on my submission so I need to be sure when I submit this problem
 
alright I solved it nevermind
 
I am so confused by the double integral
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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