$F=y^{2}\mathbf{i}+x^{3}\mathbf{j}$ curve C is counterclockwise path around triangle with vertices $(0,0)$, $(1,0)$, and $(0,4)$ Find the work done by $F$ over the curve in the direction of increasing t.

$$
F=y^{2}\mathbf{i}+x^{3}\mathbf{j};
$$
curve $C$ is counterclockwise path around triangle with vertices $(0,0)$, $(1,0)$, and $(0,4)$ Thanks!
Hints

Along the segment $(0,0)$ to $(1,0)$, $y=0$.

Along the segment $(0,0)$ to $(0,4)$, $x=0$.

The hypotenuse can be parameterize as
$$
 t(0,1)+(t-1)(4,0), \quad 0 \le t \le 1.
$$

Solution

A diagram of the contour's three components against the vector field:



Quick check: is this vector field $F=P(x,y)\mathbf{i} + Q(x,y)\mathbf{j}$ conservative? Does $P_{y}=Q_{x}$ If yes, the closed contour is $0$:

$$
P_{y} = 2y, \quad Q_{x} = 3x^{2} \qquad \Rightarrow \qquad P_{y} \color{red}{\ne} Q_{x}
$$

Evaluate the pieces.

Leg 1: $y = 0$

$$\int_{0}^{1} f(x,y) dx = \int_{0}^{1} \left( 0, x^{3} \right) dx = \left( 0, \frac{1}{4} \right)$$

Leg 2: $y=4-x$
$$
\int_{0}^{1} f(x,y) dx = \int_{0}^{1} \left( \left( 4-4x \right)^{2}, x^{3} \right) dx = 
\left(\frac{16}{3},\frac{1}{4}\right)
$$

Leg 3: $x = 0$
$$
\int_{0}^{1} f(x,y) dy = \int_{0}^{1} \left( y^{2}, 0 \right) dx = 
\left( \frac{1}{3}, 0 \right)
$$

$$
I_{C} = I_{1} + I_{2} + I_{3} = 
\left( 0, \frac{1}{4} \right) +
\left(\frac{16}{3},\frac{1}{4}\right) +
\left( \frac{1}{3}, 0 \right) =
\mathbf{\left( \frac{17}{3}, \frac{1}{2} \right)}
$$
Hints

Along the segment $(0,0)$ to $(1,0)$, $y=0$.

Along the segment $(0,0)$ to $(0,4)$, $x=0$.

The hypotenuse can be parameterize as
$$
 t(0,1)+(t-1)(4,0), \quad 0 \le t \le 1.
$$

Solution

A diagram of the contour's three components against the vector field:



Quick check: is this vector field $F=P(x,y)\mathbf{i} + Q(x,y)\mathbf{j}$ conservative? Does $P_{y}=Q_{x}$ If yes, the closed contour is $0$:

$$
P_{y} = 2y, \quad Q_{x} = 3x^{2} \qquad \Rightarrow \qquad P_{y} \color{red}{\ne} Q_{x}
$$

Evaluate the pieces.

Leg 1: $y = 0$

$$\int_{0}^{1} f(x,y) dx = \int_{0}^{1} \left( 0, x^{3} \right) dx = \left( 0, \frac{1}{4} \right)$$

Leg 2: $y=4-x$
$$
\int_{0}^{1} f(x,y) dx = \int_{0}^{1} \left( \left( 4-4x \right)^{2}, x^{3} \right) dx = 
\left(\frac{16}{3},\frac{1}{4}\right)
$$

Leg 3: $x = 0$
$$
\int_{0}^{1} f(x,y) dy = \int_{0}^{1} \left( y^{2}, 0 \right) dx = 
\left( \frac{1}{3}, 0 \right)
$$

$$
I_{C} = I_{1} + I_{2} + I_{3} = 
\left( 0, \frac{1}{4} \right) +
\left(\frac{16}{3},\frac{1}{4}\right) +
\left( \frac{1}{3}, 0 \right) =
\mathbf{\left( \frac{17}{3}, \frac{1}{2} \right)}
$$