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Split up the boundary of C (which I denote ∂C throughout) into the parabolic segment from (1, 1) to (0, 0) (the part corresponding to y = x ²), and the line segment from (1, 1) to (0, 0) (the part of ∂C on the line y = x).
Parameterize these pieces respectively by
r(t) = x(t) i + y(t) j = t i + t ² j
and
s(t) = x(t) i + y(t) j = (1 - t ) i + (1 - t ) j
both with 0 ≤ t ≤ 1.
The circulation of F around ∂C is given by the line integral with respect to arc length,
[tex]\displaystyle \int_{\partial C}\mathbf F\cdot\mathbf T \,\mathrm ds[/tex]
where T denotes the tangent vector to ∂C. Split up the integral over each piece of ∂C :
• on the parabolic segment, we have
T = dr/dt = i + 2t j
• on the line segment,
T = ds/dt = -i - j
Then the circulation is
[tex]\displaystyle \int_{\partial C}\mathbf F\cdot\mathbf T\,\mathrm ds = \int_0^1 (7t^3\,\mathbf i+5t^4\,\mathbf j)\cdot(\mathbf i+2t\,\mathbf j)\,\mathrm dt + \int_0^1 (7(1-t)^2\,\mathbf i+5(1-t)^2\,\mathbf j)\cdot(-\mathbf i-\mathbf j)\,\mathrm dt \\\\ = \int_0^1 (7t^3+10t^5)\,\mathrm dt - 12 \int_0^1 (1-t)^2\,\mathrm dt =\boxed{-\frac7{12}}[/tex]
Alternatively, we can use Green's theorem to compute the circulation, as
[tex]\displaystyle\int_{\partial C}\mathbf F\cdot\mathbf T\,\mathrm ds = \iint_C\frac{\partial(5y^2)}{\partial x} - \frac{\partial(7xy)}{\partial y}\,\mathrm dx\,\mathrm dy \\\\ = -7\int_0^1\int_{x^2}^x x\,\mathrm dx \\\\ = -7\int_0^1 xy\bigg|_{y=x^2}^{y=x}\,\mathrm dx \\\\ =-7\int_0^1(x^2-x^3)\,\mathrm dx = -\frac7{12}[/tex]
The flux of F across ∂C is
[tex]\displaystyle \int_{\partial C}\mathbf F\cdot\mathbf N \,\mathrm ds[/tex]
where N is the normal vector to ∂C. While T = x'(t) i + y'(t) j, the normal vector is N = y'(t) i - x'(t) j.
• on the parabolic segment,
N = 2t i - j
• on the line segment,
N = - i + j
So the flux is
[tex]\displaystyle \int_{\partial C}\mathbf F\cdot\mathbf N\,\mathrm ds = \int_0^1 (7t^3\,\mathbf i+5t^4\,\mathbf j)\cdot(2t\,\mathbf i-\mathbf j)\,\mathrm dt + \int_0^1 (7(1-t)^2\,\mathbf i+5(1-t)^2\,\mathbf j)\cdot(-\mathbf i+\mathbf j)\,\mathrm dt \\\\ = \int_0^1 (14t^4-5t^4)\,\mathrm dt - 2 \int_0^1 (1-t)^2\,\mathrm dt =\boxed{\frac{17}{15}}[/tex]