Question

Evaluate ∫c yz dx + xz dy + xy dz over the line segment from (1, 1, 1) to (5, 2, 0).

Answer 1

To evaluate the line integral, we use the line integral formula and integrate each term separately. The result is 6.

Step-by-step explanation:

To evaluate the line integral ∫c yz dx + xz dy + xy dz over the line segment from (1, 1, 1) to (5, 2, 0), we can use the line integral formula:

∫c P dx + Q dy + R dz = ∫ P dx + ∫ Q dy + ∫ R dz

In this case, P = yz, Q = xz, and R = xy. We can integrate each term separately:

∫ yz dx = yzx |(1,1,1) to (5,2,0) = 6

∫ xz dy = xzy |(1,1,1) to (5,2,0) = -10

∫ xy dz = xyz |(1,1,1) to (5,2,0) = 10

Adding up the results, we get: 6 + (-10) + 10 = 6.

Answer 2

To evaluate the line integral over the line segment from (1, 1, 1) to (5, 2, 0), parameterize the path, calculate the differential elements, substitute into the integral, and evaluate from t = 0 to t = 1.

Step-by-step explanation:

To evaluate the given line integral, we need to parameterize the path from the point (1, 1, 1) to (5, 2, 0). This line segment can be described by the vector equation r(t) = (1 + 4t)i + (1 + t)j + (1 - t)k, where t varies from 0 to 1. The differential elements along this path are given by dr = (4dt)i + (dt)j + (-dt)k.

Now, we can rewrite the integral in terms of the parameter t as follows:

• dx = 4dt
• dy = dt
• dz = -dt
• yz = (1 + t)(1 - t)
• xz = (1 + 4t)(1 - t)
• xy = (1 + 4t)(1 + t)

Substituting these expressions into our integral expression, we can evaluate the integral from t = 0 to t = 1 using basic calculus operations:

∫ (from 0 to 1) [ (1 + t)(1 - t) * 4dt + (1 + 4t)(1 - t) * dt + (1 + 4t)(1 + t) * (-dt) ].

After performing the integral, you will find the value of the line integral over the given path.