I'll only look at (37) here, since
• (38) was addressed in 24438105
• (39) was addressed in 24434477
• (40) and (41) were both addressed in 24434541
In both parts, we're considering the line integral
![\displaystyle \int_C (3x^2-8y^2)\,\mathrm dx + (4y-6xy)\,\mathrm dy](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint_C%20%283x%5E2-8y%5E2%29%5C%2C%5Cmathrm%20dx%20%2B%20%284y-6xy%29%5C%2C%5Cmathrm%20dy)
and I assume <em>C</em> has a positive orientation in both cases
(a) It looks like the region has the curves <em>y</em> = <em>x</em> and <em>y</em> = <em>x</em> ² as its boundary***, so that the interior of <em>C</em> is the set <em>D</em> given by
![D = \left\{(x,y) \mid 0\le x\le1 \text{ and }x^2\le y\le x\right\}](https://tex.z-dn.net/?f=D%20%3D%20%5Cleft%5C%7B%28x%2Cy%29%20%5Cmid%200%5Cle%20x%5Cle1%20%5Ctext%7B%20and%20%7Dx%5E2%5Cle%20y%5Cle%20x%5Cright%5C%7D)
• Compute the line integral directly by splitting up <em>C</em> into two component curves,
<em>C₁ </em>: <em>x</em> = <em>t</em> and <em>y</em> = <em>t</em> ² with 0 ≤ <em>t</em> ≤ 1
<em>C₂</em> : <em>x</em> = 1 - <em>t</em> and <em>y</em> = 1 - <em>t</em> with 0 ≤ <em>t</em> ≤ 1
Then
![\displaystyle \int_C = \int_{C_1} + \int_{C_2} \\\\ = \int_0^1 \left((3t^2-8t^4)+(4t^2-6t^3)(2t))\right)\,\mathrm dt \\+ \int_0^1 \left((-5(1-t)^2)(-1)+(4(1-t)-6(1-t)^2)(-1)\right)\,\mathrm dt \\\\ = \int_0^1 (7-18t+14t^2+8t^3-20t^4)\,\mathrm dt = \boxed{\frac23}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint_C%20%3D%20%5Cint_%7BC_1%7D%20%2B%20%5Cint_%7BC_2%7D%20%5C%5C%5C%5C%20%3D%20%5Cint_0%5E1%20%5Cleft%28%283t%5E2-8t%5E4%29%2B%284t%5E2-6t%5E3%29%282t%29%29%5Cright%29%5C%2C%5Cmathrm%20dt%20%5C%5C%2B%20%5Cint_0%5E1%20%5Cleft%28%28-5%281-t%29%5E2%29%28-1%29%2B%284%281-t%29-6%281-t%29%5E2%29%28-1%29%5Cright%29%5C%2C%5Cmathrm%20dt%20%5C%5C%5C%5C%20%3D%20%5Cint_0%5E1%20%287-18t%2B14t%5E2%2B8t%5E3-20t%5E4%29%5C%2C%5Cmathrm%20dt%20%3D%20%5Cboxed%7B%5Cfrac23%7D)
*** Obviously this interpretation is incorrect if the solution is supposed to be 3/2, so make the appropriate adjustment when you work this out for yourself.
• Compute the same integral using Green's theorem:
![\displaystyle \int_C (3x^2-8y^2)\,\mathrm dx + (4y-6xy)\,\mathrm dy = \iint_D \frac{\partial(4y-6xy)}{\partial x} - \frac{\partial(3x^2-8y^2)}{\partial y}\,\mathrm dx\,\mathrm dy \\\\ = \int_0^1\int_{x^2}^x 10y\,\mathrm dy\,\mathrm dx = \boxed{\frac23}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint_C%20%283x%5E2-8y%5E2%29%5C%2C%5Cmathrm%20dx%20%2B%20%284y-6xy%29%5C%2C%5Cmathrm%20dy%20%3D%20%5Ciint_D%20%5Cfrac%7B%5Cpartial%284y-6xy%29%7D%7B%5Cpartial%20x%7D%20-%20%5Cfrac%7B%5Cpartial%283x%5E2-8y%5E2%29%7D%7B%5Cpartial%20y%7D%5C%2C%5Cmathrm%20dx%5C%2C%5Cmathrm%20dy%20%5C%5C%5C%5C%20%3D%20%5Cint_0%5E1%5Cint_%7Bx%5E2%7D%5Ex%2010y%5C%2C%5Cmathrm%20dy%5C%2C%5Cmathrm%20dx%20%3D%20%5Cboxed%7B%5Cfrac23%7D)
(b) <em>C</em> is the boundary of the region
![D = \left\{(x,y) \mid 0\le x\le 1\text{ and }0\le y\le1-x\right\}](https://tex.z-dn.net/?f=D%20%3D%20%5Cleft%5C%7B%28x%2Cy%29%20%5Cmid%200%5Cle%20x%5Cle%201%5Ctext%7B%20and%20%7D0%5Cle%20y%5Cle1-x%5Cright%5C%7D)
• Compute the line integral directly, splitting up <em>C</em> into 3 components,
<em>C₁</em> : <em>x</em> = <em>t</em> and <em>y</em> = 0 with 0 ≤ <em>t</em> ≤ 1
<em>C₂</em> : <em>x</em> = 1 - <em>t</em> and <em>y</em> = <em>t</em> with 0 ≤ <em>t</em> ≤ 1
<em>C₃</em> : <em>x</em> = 0 and <em>y</em> = 1 - <em>t</em> with 0 ≤ <em>t</em> ≤ 1
Then
![\displaystyle \int_C = \int_{C_1} + \int_{C_2} + \int_{C_3} \\\\ = \int_0^1 3t^2\,\mathrm dt + \int_0^1 (11t^2+4t-3)\,\mathrm dt + \int_0^1(4t-4)\,\mathrm dt \\\\ = \int_0^1 (14t^2+8t-7)\,\mathrm dt = \boxed{\frac53}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint_C%20%3D%20%5Cint_%7BC_1%7D%20%2B%20%5Cint_%7BC_2%7D%20%2B%20%5Cint_%7BC_3%7D%20%5C%5C%5C%5C%20%3D%20%5Cint_0%5E1%203t%5E2%5C%2C%5Cmathrm%20dt%20%2B%20%5Cint_0%5E1%20%2811t%5E2%2B4t-3%29%5C%2C%5Cmathrm%20dt%20%2B%20%5Cint_0%5E1%284t-4%29%5C%2C%5Cmathrm%20dt%20%5C%5C%5C%5C%20%3D%20%5Cint_0%5E1%20%2814t%5E2%2B8t-7%29%5C%2C%5Cmathrm%20dt%20%3D%20%5Cboxed%7B%5Cfrac53%7D)
• Using Green's theorem:
![\displaystyle \int_C (3x^2-8y^2)\,\mathrm dx + (4y-6xy)\,\mathrm dx = \int_0^1\int_0^{1-x}10y\,\mathrm dy\,\mathrm dx = \boxed{\frac53}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint_C%20%283x%5E2-8y%5E2%29%5C%2C%5Cmathrm%20dx%20%2B%20%284y-6xy%29%5C%2C%5Cmathrm%20dx%20%3D%20%5Cint_0%5E1%5Cint_0%5E%7B1-x%7D10y%5C%2C%5Cmathrm%20dy%5C%2C%5Cmathrm%20dx%20%3D%20%5Cboxed%7B%5Cfrac53%7D)