Compute the work done by the force F = sin(x + y), xy, (x^2)z> in moving an object along the trajectory that is the line segment from (1, 1, 1) to (2, 2, 2) followed by the line segment from (2, 2, 2) to (3, 6, 8) when force is measured in Newtons and distance in meters.

Parameterize the line segments by[tex]\vec r(t)=(1-t)\langle1,1,1\rangle+t\langle2,2,2\rangle=\langle1+t,1+t,1+t\rangle[/tex]and[tex]\vec s(t)=(1-t)\langle2,2,2\rangle+t\langle3,6,8\rangle=\langle2+t,2+4t,2+6t\rangle[/tex]both with [tex]0\le t\le1[/tex]. Then[tex]\vec r'(t)=\langle1,1,1\rangle[/tex][tex]\vec s'(t)=\langle1,4,6\rangle[/tex]so that the work done by [tex]\vec F(x,y,z)=\langle\sin(x+y),xy,x^2z\rangle[/tex] over the respective line segments is given by[tex]\displaystyle\int_{C_1}\vec F\cdot\mathrm d\vec r=\int_0^1\langle\sin(2+2t),(1+t)^2,(1+t)^3\rangle\cdot\langle1,1,1\rangle\,\mathrm dt[/tex][tex]=\displaystyle\int_0^1\sin(2+2t)+(1+t)^2+(1+t)^3\,\mathrm dt=\boxed{\frac{73+6\cos2-6\cos4}{12}}[/tex]and[tex]\displaystyle\int_{C_2}\vec F\cdot\mathrm d\vec s=\int_0^1\langle\sin(4+5t),(2+t)(2+4t),(2+t)^2(2+6t)\rangle\cdot\langle1,4,6\rangle\,\mathrm dt[/tex][tex]=\displaystyle\int_0^1\sin(4+5t)+4(2+t)(2+4t)+6(2+t)^2(2+6t)\,\mathrm dt=\boxed{\frac{3695+3\cos4-3\cos9}{15}}[/tex](both measured in Newton-meters)