Question

Compute the line integral y*2dx-x*2dy round the triangle whose vertices are (1,0)(0,1) and (-1,0) in the xy-plane

Answer 1

Answer:

i want this solution clearly

Answer 2

The line integral of y²dx - x²dy is -2/3.

Step-by-step Explanation:

Given: vertices of triangle = (1,0), (0,1) and (-1,0)

Line integral function = y²dx - x²dy

To Find: Line integral around the triangle in the xy-plane

Solution:

• Finding line integral of y²dx - x²dy around the triangle for vertices (1,0), (0,1) and (-1,0)  

The following integral can be solved by using Green's theorem of integration according to which,

$\int_C Mdx+Ndy = \int \int_R ( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} ) dx dy$

For y²dx - x²dy, we can write,

$\int_C y^{2} dx - x^{2} dy = \int \int_R ( \frac{\partial (-x^{2})}{\partial x} - \frac{\partial y^{2} }{\partial y} ) dx dy = -2 \int \int_R (x+y) dxdy$

To determine the limits of integration, we have to find equations of the line for the points (1,0) → (0,1) and (0,1) → (-1,0). Therefore, using the formula,

$(y - y_1) = \frac{(y_2 - y_1)}{(x_2 - x_1)} (x - x_1)$

For the points (1,0) → (0,1), we have,

$(y - 0) = \frac{(1 - 0)}{(0 - 1)} (x - 1) \Rightarrow x=1-y$

And, for the points (0,1) → (-1,0), we have,

$(y - 1) = \frac{(0 - 1)}{(-1 - 0)} (x - 0) \Rightarrow x=y-1$

Now, using Green's theorem, we have,

$\Rightarrow -2 \int_0^1 \int_{y-1}^{1-y} (x+y) dxdy$

$\Rightarrow -2 \int_0^1 (\frac{x^{2} }{2} +xy)_{y-1}^{1-y} dy$

$\Rightarrow -4 \int_0^1 (y-y^2) dy$

$\Rightarrow -4 \Big[ \frac{y^{2}}{2} - \frac{y^{3}}{3} \Big]_0^1 = -\frac{2}{3}$

Hence, the line integral of y²dx - x²dy around the triangle for vertices (1,0), (0,1) and (-1,0) is -2/3