# Question 0fc26

Dec 9, 2017

${W}_{1} = \setminus {\int}_{0}^{4} {x}^{2} \mathrm{dx} = \frac{64}{3}$.
${W}_{2} = \setminus {\int}_{x = 0}^{x = 4} \left({x}^{3} - 8 {x}^{2} + 12 x\right) \mathrm{dx} = - \frac{32}{3}$.
${W}_{3} = 0$

#### Explanation:

vecF(x,y)=x(x+y)\hati + xy^2\hatj;

Section 1: $\setminus \quad \left(0 , 0\right) \setminus \rightarrow \left(4 , 0\right)$

y=0; \qquad dy=0; \qquad vecF(x,y)=x^2\hati;\qquad dvecr = dx\hati;

${W}_{1} = \setminus {\int}_{0}^{4} \vec{F} \left(x , y\right) . \mathrm{dv} e c r = \setminus {\int}_{0}^{4} {x}^{2} \mathrm{dx} = \frac{64}{3}$.

Section 2: $\setminus \quad \left(4 , 0\right) \setminus \rightarrow \left(0 , 4\right)$
Equation of the line segment is: \quad y=4-x;
dvecr = dx\hati + dy\hatj; y=4-x; \qquad dy=-dx

$\vec{F} \left(x , y\right) . \mathrm{dv} e c r = x \left(x + y\right) \mathrm{dx} + x {y}^{2} \mathrm{dy} ,$
$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad = 4 x \mathrm{dx} - x {\left(4 - x\right)}^{2} \mathrm{dx} = - \left({x}^{3} - 8 {x}^{2} + 12 x\right) \mathrm{dx}$

${W}_{2} = \setminus {\int}_{\left(4 , 0\right)}^{\left(0 , 4\right)} \vec{F} \left(x , y\right) . \mathrm{dv} e c r ,$
$= - \setminus {\int}_{x = 4}^{x = 0} \left({x}^{3} - 8 {x}^{2} + 12 x\right) \mathrm{dx} = \setminus {\int}_{x = 0}^{x = 4} \left({x}^{3} - 8 {x}^{2} + 12 x\right) \mathrm{dx}$
$= {\left[{x}^{4} / 4 - \frac{8}{3} {x}^{3} + 6 {x}^{2}\right]}_{0}^{4} = - \frac{32}{3}$.

Section 3: $\setminus \quad \left(0 , 4\right) \setminus \rightarrow \left(0 , 0\right)$
x=0; dx=0; \qquad vecF(x,y)=vec0;#

${W}_{3} = 0$

Net Work Done:
${W}_{\text{net}} = {W}_{1} + {W}_{2} + {W}_{3} = \frac{64}{3} - \frac{32}{3} + 0 = \frac{32}{3}$.