Evaluate the following line integral: ∮C[(4x^2+3x+5y)dx+(6x^2+5x+3y)dy] where C is the path around the square with vertices (0,0),(2,0),(2,2) and (0,2).
from point 1 to point 2 dy = 0, y = 0 and x goes from 0 to 2
so integral one = ∮[(4x^2+3x)dx =(4/3)x^3 + (3/2)x^2 = 32/3+6 = 50/3
from point 2 to point 3, dx = 0 , x = 2 and y from 0 to 2
so integral 2 = ∮(34+3y)dy] from 0 to 2
similar for the last two lines, then add the results. Remember direction for example for integal 3 y does not change but x is from 2 to 0, not 0 to 2
To evaluate the line integral, we need to find a parametrization of the path C and then calculate the integral using that parametrization.
Step 1: Parametrize the path C
From the given information, we can see that the path C forms a square. We can parametrize this square by dividing it into four sides.
Side 1: From (0,0) to (2,0)
We can parametrize this side as r(t) = (t, 0), where t varies from 0 to 2.
Side 2: From (2,0) to (2,2)
We can parametrize this side as r(t) = (2, t), where t varies from 0 to 2.
Side 3: From (2,2) to (0,2)
We can parametrize this side as r(t) = (t, 2), where t varies from 2 to 0 (in reverse order).
Side 4: From (0,2) to (0,0)
We can parametrize this side as r(t) = (0, t), where t varies from 2 to 0 (in reverse order).
Step 2: Calculate the line integral
Using the parametrizations from step 1, we can evaluate the line integral ∮C[(4x^2+3x+5y)dx+(6x^2+5x+3y)dy].
Let's evaluate each part separately:
For the first integral, ∫[(4x^2+3x+5y)dx] along the four sides:
Side 1: ∫[(4t^2+3t+0) dt], where t varies from 0 to 2.
Side 2: ∫[(4(2)^2+3(2)+5t) dt], where t varies from 0 to 2.
Side 3: ∫[(4t^2+3t+5(2))] dt, where t varies from 2 to 0 (in reverse order).
Side 4: ∫[(4(0)^2+3(0)+5t)] dt, where t varies from 2 to 0 (in reverse order).
Similarly, for the second integral, ∫[(6x^2+5x+3y)dy] along the four sides:
Side 1: ∫[(6t^2+5t+0) dt], where t varies from 0 to 2.
Side 2: ∫[(6(2)^2+5(2)+3t) dt], where t varies from 0 to 2.
Side 3: ∫[(6t^2+5t+3(2))] dt, where t varies from 2 to 0 (in reverse order).
Side 4: ∫[(6(0)^2+5(0)+3t)] dt, where t varies from 2 to 0 (in reverse order).
Finally, add up the answers to all the integrals to get the total value of the line integral: ∮C[(4x^2+3x+5y)dx+(6x^2+5x+3y)dy].