A particle moves along line segments from the origin to the points (2, 0, 0), (2, 4, 1), (0, 4, 1), and back to the origin under the influence of the force field. F(x, y, z) = z^2i + 3xyj + 2y^2k. Find the work done.

The work is equal to the line integral of [tex]\vec F[/tex] over each line segment.Parameterize the pathsfrom (0, 0, 0) to (2, 0, 0) by [tex]\vec r_1(t)=t\,\vec\imath[/tex] with [tex]0\le t\le2[/tex],from (2, 0, 0) to (2, 4, 1) by [tex]\vec r_2(t)=2\,\vec\imath+4t\,\vec\jmath+t\,\vec k[/tex] with [tex]0\le t\le1[/tex],from (2, 4, 1) to (0, 4, 1) by [tex]\vec r_3(t)=(2-t)\,\vec\imath+4\,\vec\jmath+\vec k[/tex] with [tex]0\le t\le2[/tex], andfrom (0, 4, 1) to (0, 0, 0) by [tex]\vec r_4(t)=(4-4t)\,\vec\jmath+(1-t)\,\vec k[/tex] with [tex]0\le t\le1[/tex]The work done by [tex]\vec F[/tex] over each segment (call them [tex]C_1,\ldots,C_4[/tex]) is[tex]\displaystyle\int_{C_1}\vec F\cdot\mathrm d\vec r_1=\int_0^2\vec0\cdot\vec\imath\,\mathrm dt=0[/tex][tex]\displaystyle\int_{C_2}\vec F\cdot\mathrm d\vec r_2=\int_0^1(t^2\,\vec\imath+24t\,\vec\jmath+32t^2\,\vec k)\cdot(4\,\vec\jmath+\vec k)\,\mathrm dt=\int_0^1(96t+32t^2)\,\mathrm dt=\frac{176}3[/tex][tex]\displaystyle\int_{C_3}\vec F\cdot\mathrm d\vec r_3=\int_0^2(\vec\imath+(24-12t)\,\vec\jmath+32\,\vec k)\cdot(-\vec\imath)\,\mathrm dr=-\int_0^2\mathrm dt=-2[/tex][tex]\displaystyle\int_{C_4}\vec F\cdot\mathrm d\vec r_4=\int_0^1((1-t)^2\,\vec\imath+2(4-4t)^2\,\vec k)\cdot(-4\,\vec\jmath-\vec k)\,\mathrm dt=-2\int_0^1(4-4t)^2\,\mathrm dt=-\frac{32}3[/tex]Then the total work done by [tex]\vec F[/tex] over the particle's path is 46.