# Find the value of the line integral.F · dr (Hint: If F is conservative, the integration may be easier on an alternative path.) F(x,y) = ye^xyi + xe^xyj (a) r1(t) = ti − (t − 8)j, 0 ≤ t ≤ 8 ?

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(b) the closed path consisting of line segments from (0, 8) to (0, 0), from (0, 0) to (8, 0), and then from (8, 0) to (0, 8)

(b) the closed path consisting of line segments from (0, 8) to (0, 0), from (0, 0) to (8, 0), and then from (8, 0) to (0, 8)

##### 1 Answer

#### Answer:

Zero in both cases

#### Explanation:

A field **potential function**

#bbF = - nabla phi#

Or, in 2-D:

#bbF = - (: phi_x, phi_y:) #

This can be solved directly here, because:

So for:

**Path a)**

#bbr_1(t) = t bbi − (t − 8) bbj, qquad 0 le t le 8#

This is along line

**Path b)**

Regardless of path, this is closed path from

**Going instead with the hint** , the curl of a conserved vector field is zero. This follows from the existence of a **potential function** and the **gradient theorem**.

**Curl of the Field:**

The field is conservative / path-independent

Path **a)**

This is along line

Following the hint, break this into

#{(bb1, x = 0, dx = 0, y in [8,0]),(bb2, y = 0, dy = 0, x in [0,8]):}#

Path

Path

Path **b)**

This is clearly zero as it is a closed path in a conserved field.

FINALLY:

Noting that

#bbr(t) : {(x = t),(y = 8 - t):} qquad bbr'(t) : {( x' = 1),( y' = -1):}#