Question

Evaluate the line integral by the two following methods. xy dx + x2y3 dy C is counterclockwise around the triangle with vertices (0, 0), (1, 0), and (1, 2) (a) directly (b) using Green's Theorem

Answer 1

Answer:

a)

$\frac{2}{3}$

b)

$\frac{2}{3}$

Step-by-step explanation:

a) The first part requires that we use line integral to evaluate directly.

The line integral is

$\int_C xydx + {x}^{2} {y}^{3} dy$

where C is counterclockwise around the triangle with vertices (0, 0), (1, 0), and (1, 2)

The boundary of integration is shown in the attachment.

Our first line integral is

$L_1 = \int_ {(0,0)}^{(1,0)} xydx + {x}^{2} {y}^{3} dy$

The equation of this line is y=0, x varies from 0 to 1.

When we substitute y=0 every becomes zero.

$\therefore \: L_1 =0$

Our second line integral is

$L_2 = \int_ {(1,0)}^{(1,2)} xydx + {x}^{2} {y}^{3} dy$

The equation of this line is:

$x = 0 \implies \: dx = 0$

y varies from 1 to 2.

We substitute the boundary and the values to get:

$L_2 = \int_ {1}^{2}1 \cdot y(0) + {1}^{2} \cdot \: {y}^{3} dy$

$L_2 = \int_ {1}^2 {y}^{3} dy = \frac{8}{3}$

The 3rd line integral is:

$L_3 = \int_ {(1,2)}^{(0,0)} xydx + {x}^{2} {y}^{3} dy$

The equation of this line is

$y = 2x \implies \: dy = 2dx$

x varies from 0 to 1.

We substitute to get:

$L_3 = \int_ {1}^{0} x \cdot \: 2xdx + {x}^{2} {(2x)}^{3}(2 dx)$

$L_3 = \int_ {1}^{0} 8 {x}^{5} + 2 {x}^{2} dx = - 2$

The value of the line integral is

$L = L_1 + L_2 + L_3$

$L = 0 + \frac{8}{3} + - 2 = \frac{2}{3}$

b) The second part requires the use of Green's Theorem to evaluate:

$\int_C xydx + {x}^{2} {y}^{3} dy$

Since C is a closed curve with counterclockwise orientation, we can apply the Green's Theorem.

This is given by:

$\int_C \: Pdx +Q \: dy = \int \int_ R \: Q_y - P_x \: dA$

$\int_C \: xydx + {x}^{2} {y}^{3} \: dy = \int \int_ R \: 3 {x}^{2} {y}^{2} - y \: dA$

We choose our region of integration parallel to the y-axis.

$\int_C \: xydx + {x}^{2} {y}^{3} \: dy = \int_ 0^{1} \int_ 0^{2x} \: 3 {x}^{2} {y}^{2} - y \: dydx$

$\int_C \: xydx + {x}^{2} {y}^{3} \: dy = \int_ 0^{1} \: {x}^{2} {y}^{3} - \frac{1}{2} {y}^{2} |_ 0^{2x} dx$

$\int_C \: xydx + {x}^{2} {y}^{3} \: dy = \int_ 0^{1} \: 8{x}^{5} - 2 {x}^{2} dx = \frac{2}{3}$

Answer 2

The line integral in both cases is  $\dfrac{2}{3}$

a) The line integral is  $L=\int_{C}^{}xydx+x^2y^3dy$   where C is counterclockwise around the triangle with vertices $(0, 0), (1, 0)$  and  $(1, 2)$.

The boundaries of integration is shown in the attached image.

Our first line integral is

$L_{1}=\int_{(0,0)}^{(1,0)}xydx+x^2y^3dy$

The equation of this line is  $y=0$,  $x$ varies from $0$ to $1$.

$L_{1}=0$

Our second line integral is

$L_{2}=\int_{(1,0)}^{(1,2)}xydx+x^2y^3dy$

The equation of this line is $x=0$  and $y$ varies from $1$ to $2$.

$L_{2}=\dfrac{8}{3}$

The third line integral is

$L_{3}=\int_{1}^{0}xydx+x^2y^3dy$

The equation of this line is $y=2x$  and $x$ varies from $0$ to $1$.

$L_{3}=-2$

The value of the line integral is

$L=L_{1}+L_{2}+L_{3}=0+\frac{8}{3}+(-2)=\frac{2}{3}$

b) The second part requires the use of Green's Theorem to evaluate:

This is given by:

$\int_{C}^{}P\;dx+Q\;dy=\int \int_{R}^{}Q_{y}-P_{x}\;dA\\ \\ \int_{C}^{}xy\;dx+x^2y^3\;dy=\int \int_{R}^{}3x^2y^2-y\;dA$

We choose our region of integration parallel to the $y$-axis.

$\int_{C}^{}xy\;dx+x^2y^3\;dy\\\\=\int_{0}^{1}\int_{0}^{2x}3x^2y^2-y\;dydx\\ \\ =\int_{0}^{1}x^2y^3-\frac{1}{2}\left | y^2 \right |_{0}^{2x}\;dx\\ \\ =\int_{0}^{1}8x^5-2x^2\;dx=\frac{2}{3}$

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