Question

25 - 26 show that the line integral is independent of path and evaluate the integral.

1. $\int_{c} 2 x e^{-y} d x+(2 y-x^{2} e^{-y}) d y$,

$c$ is any path from $(1,0)$ to $(2,1)$

1. $\int_{c} \sin y d x+(x \cos y-\sin y) d y$,

$c$ is any path from $(2,0)$ to $(1, \pi)$

Explanation:

Step1: Check independence of path

For a line integral \(\int_{C}Pdx + Qdy\), if \(\frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x}\), the integral is path - independent.
For problem 25: \(P = 2xe^{-y}\), \(\frac{\partial P}{\partial y}=- 2xe^{-y}\); \(Q = 2y-x^{2}e^{-y}\), \(\frac{\partial Q}{\partial x}=-2xe^{-y}\).
For problem 26: \(P=\sin y\), \(\frac{\partial P}{\partial y}=\cos y\); \(Q=x\cos y-\sin y\), \(\frac{\partial Q}{\partial x}=\cos y\).

Step2: Find potential function \(f\)

Problem 25:

Since \(\frac{\partial f}{\partial x}=P = 2xe^{-y}\), integrate with respect to \(x\): \(f(x,y)=\int2xe^{-y}dx=x^{2}e^{-y}+g(y)\).
Differentiate \(f\) with respect to \(y\): \(\frac{\partial f}{\partial y}=-x^{2}e^{-y}+g^{\prime}(y)\).
Since \(\frac{\partial f}{\partial y}=Q = 2y - x^{2}e^{-y}\), then \(g^{\prime}(y)=2y\), integrate \(g(y)=\int2y dy=y^{2}+C\). Let \(C = 0\), so \(f(x,y)=x^{2}e^{-y}+y^{2}\).
Evaluate \(f(2,1)-f(1,0)\):

$$
LATEXBLOCK0
$$

Problem 26:

Since \(\frac{\partial f}{\partial x}=P=\sin y\), integrate with respect to \(x\): \(f(x,y)=x\sin y+g(y)\).
Differentiate \(f\) with respect to \(y\): \(\frac{\partial f}{\partial y}=x\cos y+g^{\prime}(y)\).
Since \(\frac{\partial f}{\partial y}=Q=x\cos y-\sin y\), then \(g^{\prime}(y)=-\sin y\), integrate \(g(y)=\cos y + C\). Let \(C = 0\), so \(f(x,y)=x\sin y+\cos y\).
Evaluate \(f(1,\pi)-f(2,0)\):

$$
LATEXBLOCK1
$$

Answer:

For problem 25, the value of the line integral is \(\frac{4}{e}\).
For problem 26, the value of the line integral is \(-2\).