Question

In this problem and the subsequent one, consider the integral on the closed curve C consisting of the line segments from (0,0) to (2,2) to (0,2) to (0,0). Here, you evaluate the line integral along each of these segments separately, and in the following problem, apply Green's theorem to find the same integral. You can cross-verify your answers between the two problems, ensuring the final integral values match.

Evaluate the integral above by finding the integral from (0,0) to (2,2), adding the integral from (2,2) to (0,2), and adding the integral from (0,2) to (0,0).

Answer 1

To evaluate the integral on the closed curve C, we break it into three parts and evaluate each part separately. The integral from (0,0) to (2,2) evaluates to 4√2, the integral from (2,2) to (0,2) evaluates to 4, and the integral from (0,2) to (0,0) evaluates to -2. Adding up the three integrals, the total line integral along C is 4√2 + 2.

Step-by-step explanation:

To evaluate the line integral on the closed curve C, we need to break the integral into three parts: from (0,0) to (2,2), from (2,2) to (0,2), and from (0,2) to (0,0). Let's start with the first part:

For the integral from (0,0) to (2,2), we can parametrize the curve as x = t and y = t, where t goes from 0 to 2. The integral becomes:

∫[0 to 2] (x+y) ds = ∫[0 to 2] (t+t) √(1+1) dt = ∫[0 to 2] 2√2 dt = 4√2

Similarly, we compute the integrals for the other two parts:

For the integral from (2,2) to (0,2), we can parametrize the curve as x = t and y = 2, where t goes from 2 to 0. The integral becomes:

∫[2 to 0] (x+y) ds = ∫[2 to 0] (t+2) √(1+0) dt = ∫[2 to 0] (t+2) dt = 4

For the integral from (0,2) to (0,0), we can parametrize the curve as x = 0 and y = t, where t goes from 2 to 0. The integral becomes:

∫[2 to 0] (x+y) ds = ∫[2 to 0] (0+t) √(0+1) dt = ∫[2 to 0] t dt = -2

Adding up the three integrals, the total line integral along C is 4√2 + 4 - 2 = 4√2 + 2.