Not much can be done without knowing what
![\mathbf F(x,y,z)](https://tex.z-dn.net/?f=%5Cmathbf%20F%28x%2Cy%2Cz%29)
is, but at the least we can set up the integral.
First parameterize the pieces of the contour:
![C_1:\mathbf r_1(t_1)=(2\sin t_1,2\cos t_1,0)](https://tex.z-dn.net/?f=C_1%3A%5Cmathbf%20r_1%28t_1%29%3D%282%5Csin%20t_1%2C2%5Ccos%20t_1%2C0%29)
![C_2:\mathbf r_2(t_2)=(1-t_2)(2,0,0)+t_2(3,3,3)=(2+t_2, 3t_2, 3t_2)](https://tex.z-dn.net/?f=C_2%3A%5Cmathbf%20r_2%28t_2%29%3D%281-t_2%29%282%2C0%2C0%29%2Bt_2%283%2C3%2C3%29%3D%282%2Bt_2%2C%203t_2%2C%203t_2%29)
where
![0\le t_1\le\dfrac\pi2](https://tex.z-dn.net/?f=0%5Cle%20t_1%5Cle%5Cdfrac%5Cpi2)
and
![0\le t_2\le1](https://tex.z-dn.net/?f=0%5Cle%20t_2%5Cle1)
. You have
![\mathrm d\mathbf r_1=(2\cos t_1,-2\sin t_1,0)\,\mathrm dt_1](https://tex.z-dn.net/?f=%5Cmathrm%20d%5Cmathbf%20r_1%3D%282%5Ccos%20t_1%2C-2%5Csin%20t_1%2C0%29%5C%2C%5Cmathrm%20dt_1)
![\mathrm d\mathbf r_2=(1,3,3)\,\mathrm dt_2](https://tex.z-dn.net/?f=%5Cmathrm%20d%5Cmathbf%20r_2%3D%281%2C3%2C3%29%5C%2C%5Cmathrm%20dt_2)
and so the work is given by the integral
![\displaystyle\int_C\mathbf F(x,y,z)\cdot\mathrm d\mathbf r](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cint_C%5Cmathbf%20F%28x%2Cy%2Cz%29%5Ccdot%5Cmathrm%20d%5Cmathbf%20r)
![=\displaystyle\int_0^{\pi/2}\mathbf F(2\sin t_1,2\cos t_1,0)\cdot(2\cos t_1,-2\sin t_1,0)\,\mathrm dt_1](https://tex.z-dn.net/?f=%3D%5Cdisplaystyle%5Cint_0%5E%7B%5Cpi%2F2%7D%5Cmathbf%20F%282%5Csin%20t_1%2C2%5Ccos%20t_1%2C0%29%5Ccdot%282%5Ccos%20t_1%2C-2%5Csin%20t_1%2C0%29%5C%2C%5Cmathrm%20dt_1)
Length of AB: 8 - - 6 = 14
Mid point at half length: 14/2 =7
—>Point G
1008 square inches. Just multiply length times width to find a 2 dimensional area.
Answer:
Its 6/3
Step-by-step explanation: