(40) It looks like the line integral is
[tex]\displaystyle \int_C (x^2-2xy)\,\mathrm dx + (x^2y+3)\,\mathrm dy[/tex]
where C is the boundary of the region D,
[tex]D = \left\{(x,y) \mid \dfrac{y^2}8\le x\le2\text{ and }-4\le y\le4\right\}[/tex]
(a) To evaluate the line integral directly, split up C into two paths C₁ and C₂, parameterized by
• C₁ : x = t ²/8 and y = -t, where -4 ≤ t ≤ 4
(the y component is negative to make this path have a positive/counterclockwise orientation)
• C₂ : x = 2 and y = t, where -4 ≤ t ≤ 4
Then the line integral over C is the sum of the line integrals over C₁ and C₂ :
[tex]\displaystyle \int_C = \int_C \left(\left(x(t)^2-2x(t)y(t)\right)\dfrac{\mathrm dx}{\mathrm dt} + \left(x(t)^2y(t)-3\right)\dfrac{\mathrm dy}{\mathrm dt}\right)\,\mathrm dt \\\\ = \int_{-4}^4 \left(\left(\frac{t^3}4+\frac{t^4}{64}\right)\frac t4 - \left(3-\frac{t^5}{64}\right)\right)\,\mathrm dt + \int_{-4}^4 (4t+3)\,\mathrm dt \\\\ = \int_{-4}^4 \left(4t+\dfrac{t^4}{16}+\dfrac{5t^5}{256}\right)\,\mathrm dt = \boxed{\frac{128}5}[/tex]
(b) Using Green's theorem, we have
[tex]\displaystyle \int_C (x^2-2xy)\,\mathrm dx + (x^2+3)\,\mathrm dy = \iint_D \frac{\partial(x^2y+3)}{\partial x} - \frac{\partial(x^2-2xy)}{\partial y}\,\mathrm dx\,\mathrm dy \\\\ = \int_{-4}^4 \int_{y^2/8}^2 (2xy+2x)\,\mathrm dx\,\mathrm dy = \boxed{\frac{128}5}[/tex]
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(41) I assume you meant to use θ in the parameterization, and not 6, so that C is parameterized by x = θ - sin(θ) and y = 1 - cos(θ). There's no range given for θ, so I'll just assume 0 ≤ θ ≤ π.
Then ... (for some reason, the math text won't render properly. I've attached the computation as an image)
(where s = sin(θ) and c = cos(θ))
